Free Math Worksheets

- Can you break apart the number you are subtracting? - First subtract enough to reach the previous multiple of ten. - How much do you still need to subtract?

1. For \(232 - 5\), subtract \(2\) to reach \(230\), then subtract the remaining \(3\): \(230 - 3 = 227\). 2. For \(463 - 7\), subtract \(3\) to reach \(460\), then subtract the remaining \(4\): \(460 - 4 = 456\). 3. For \(851 - 6\), subtract \(1\) to reach \(850\), then subtract the remaining \(5\): \(850 - 5 = 845\). 4. For \(544 - 8\), subtract \(4\) to reach \(540\), then subtract the remaining \(4\): \(540 - 4 = 536\).

a) \(227\) b) \(456\) c) \(845\) d) \(536\)

- First subtract enough to reach the previous multiple of \(10\). - How much is left to subtract in the second step? - Number pairs that make \(10\) may help.

1. For part a, subtract \(3\) to reach \(450\), then subtract the remaining amount: \(450 - 2 = 448\), \(450 - 4 = 446\), \(450 - 6 = 444\), and \(450 - 1 = 449\). 2. For part b, subtract \(4\) to reach \(810\), then subtract the remaining amount: \(810 - 2 = 808\), \(810 - 4 = 806\), \(810 - 1 = 809\), and \(810 - 5 = 805\).

a) \(448\), \(446\), \(444\), \(449\) b) \(808\), \(806\), \(809\), \(805\)

- First find the value on the left. - Then compare it with the number on the right. - Choose the symbol that shows whether the left side is less than, greater than, or equal to the right side.

1. For a), \(504 - 7 = 497\), so \(497 = 497\). 2. For b), \(712 - 6 = 706\), so \(706 < 708\). 3. For c), \(230 - 50 = 180\), so \(180 = 180\). 4. For d), \(941 - 9 = 932\), so \(932 < 933\).

a) \(=\) b) \(<\) c) \(=\) d) \(<\)

- What two-digit number is formed by the tens and ones digits? - Solve the related two-digit subtraction first. - Then include the hundreds digit in the difference.

1. Use \(24 - 8 = 16\). Keeping the \(5\) hundreds gives \(524 - 8 = 516\). 2. Use \(11 - 5 = 6\). Keeping the \(3\) hundreds gives \(311 - 5 = 306\). 3. Use \(52 - 6 = 46\). Keeping the \(8\) hundreds gives \(852 - 6 = 846\). 4. Use \(33 - 7 = 26\). Keeping the \(1\) hundred gives \(133 - 7 = 126\).

a) \(516\) b) \(306\) c) \(846\) d) \(126\)

- Look for a connection between the first table and the second table. - Think about how multiplying a whole number by \(10\) changes its digits.

1. For the pairs that sum to \(100\): \(100 - 60 = 40\), \(100 - 20 = 80\), \(100 - 45 = 55\), and \(100 - 15 = 85\). 2. For the pairs that sum to \(1000\): \(1000 - 600 = 400\), \(1000 - 200 = 800\), \(1000 - 450 = 550\), and \(1000 - 150 = 850\). 3. Each missing number in the second table is ten times the corresponding missing number in the first table.

Sum of \(100\): \(40\), \(80\), \(55\), \(85\). Sum of \(1000\): \(400\), \(800\), \(550\), \(850\).

- Which parts of \(457\) have already been subtracted? - Which place value has not yet been used? - Continue from the last intermediate result.

1. Paul has already subtracted the \(4\) hundreds and \(5\) tens in \(457\). 2. He still needs to subtract the \(7\) ones. 3. The final step is \(374 - 7 = 367\).

The final step is \(374 - 7 = 367\). The difference is \(367\).

- Decide whether it is easier to subtract in parts or count up from the smaller number. - When subtracting in parts, subtract the tens and then the ones. - When counting up, first reach the next hundred.

1. For \(400 - 72\), subtract in parts: \(400 - 70 = 330\), then \(330 - 2 = 328\). 2. For \(900 - 36\), subtract in parts: \(900 - 30 = 870\), then \(870 - 6 = 864\). 3. For \(600 - 58\), subtract in parts: \(600 - 50 = 550\), then \(550 - 8 = 542\). 4. For \(200 - 19\), count up: \(19 + 1 = 20\), then \(20 + 180 = 200\). The total increase is \(1 + 180 = 181\), so the difference is \(181\).

a) \(328\) b) \(864\) c) \(542\) d) \(181\)

- Is the number being subtracted close to a multiple of \(100\)? - When two numbers are close together, try counting up to find the difference. - Can you break the subtraction into easier steps?

1. For a), subtract in parts: \(800 - 70 = 730\), then \(730 - 2 = 728\). 2. For b), use compensation: \(534 - 200 = 334\), then add back \(1\). The difference is \(335\). 3. For c), count up from \(395\): it takes \(5\) to reach \(400\) and \(12\) more to reach \(412\). Therefore, \(5 + 12 = 17\). 4. For d), subtract by place value: \(900 - 400 = 500\), \(60 - 50 = 10\), and \(7 - 2 = 5\). Then \(500 + 10 + 5 = 515\).

a) \(728\) b) \(335\) c) \(17\) d) \(515\)

- Find each difference separately. - Compare the hundreds, tens, and ones digits of the results. - Check whether any two results are equal.

1. Lucas: \(900 - 450 = 450\). 2. Sarah: \(1000 - 555 = 445\). 3. Max: \(812 - 362 = 450\). 4. Compare the results: \(445 < 450 = 450\).

Sarah: \(445\); Lucas: \(450\); Max: \(450\). \(445 < 450 = 450\)

- Identify the total and the part that rides the bus. - Subtract the known part from the total. - You can decompose \(280\) into hundreds and tens. - Check that your answer is less than \(630\).

1. Subtract the number of bus riders from the total: \(630 - 280\). 2. Subtract in parts: \(630 - 200 = 430\), then \(430 - 80 = 350\).

\(350\) students walk to school.

- Use subtraction to find the difference between the capacity and the current number of books. - Count up to the next hundred first. - Then count the remaining hundreds to the target.

1. Subtract the number of books on the shelf from its capacity: \(600 - 380 = 220\).

\(220\) more books can fit on the shelf.

- Decide whether the number of daffodils must be less than or greater than the number of tulips. - The phrase “fewer than” indicates subtraction. - Decompose \(370\) into hundreds and tens.

1. Subtract the difference from the number of tulips: \(850 - 370\). 2. Subtract in parts: \(850 - 300 = 550\), then \(550 - 70 = 480\).

The garden center planted \(480\) daffodils.

- Identify the total number of books. - Identify the part that is nonfiction. - Subtract the known part from the total. - Try subtracting the hundreds and tens separately.

1. Subtract the nonfiction books from the total: \(780 - 350\). 2. Subtract by place value: \(700 - 300 = 400\) and \(80 - 50 = 30\). 3. Combine the parts: \(400 + 30 = 430\).

There are \(430\) fiction books.

- What is the total distance of the trail? - How far has the class already walked? - Use subtraction to find the remaining distance. - Count up to the next hundred or subtract in parts.

1. Subtract the distance already walked from the total distance: \(940\,\text{m} - 460\,\text{m}\). 2. Subtract in parts: \(940 - 400 = 540\), then \(540 - 60 = 480\). 3. The remaining distance is \(480\,\text{m}\).

\(480\,\text{m}\) remain.

- Subtract the known part from the whole. - Decompose \(245\) into hundreds, tens, and ones. - Check your answer with a related addition equation.

1. Subtract the number of bears from the total: \(730 - 245\). 2. Decompose \(245\): \(730 - 200 = 530\), \(530 - 40 = 490\), and \(490 - 5 = 485\).

There are \(485\) stuffed dogs.

- Notice whether you are subtracting ones or tens. - Could you first go back to a multiple of \(10\) or \(100\)? - Compare parts a and b. How does subtracting \(4\) differ from subtracting \(40\)?

1. \(670 - 4 = 666\). 2. \(670 - 40 = 630\). 3. \(910 - 7 = 903\). 4. \(910 - 70 = 840\). 5. \(340 - 6 = 334\).

a) \(666\) b) \(630\) c) \(903\) d) \(840\) e) \(334\)

- Subtract the tens first, then the ones. - For a problem with \(1000\), you can count up from the smaller number to \(1000\). - When subtracting the ones, check whether you cross a multiple of \(10\).

1. \(1000 - 430 = 570\). 2. Count up from \(760\) to \(1000\), or subtract in parts: \(1000 - 760 = 240\). 3. Subtract \(5\) tens: \(684 - 50 = 634\). 4. Subtract in parts: \(457 - 30 = 427\), then \(427 - 8 = 419\). 5. Subtract in parts: \(832 - 20 = 812\), then \(812 - 5 = 807\).

a) \(570\) b) \(240\) c) \(634\) d) \(419\) e) \(807\)

- Find the difference on the left first. - Compare that result with the number on the right. - The open side of the comparison symbol faces the greater value.

1. \(430 - 70 = 360\), and \(360 > 350\). 2. \(650 - 80 = 570\), and \(570 < 580\). 3. \(910 - 50 = 860\), and \(860 = 860\). 4. \(500 - 33 = 467\), and \(467 < 477\). 5. \(1000 - 12 = 988\), and \(988 > 978\).

a) \(>\) b) \(<\) c) \(=\) d) \(<\) e) \(>\)

- Use the inverse operation to find a missing number. - Think about the distance between the two known numbers. - You may add or subtract in place-value steps.

1. a) Find the missing addend by subtracting: \(360 - 356 = 4\). 2. b) Find the missing number being subtracted: \(680 - 673 = 7\). 3. c) Add: \(445 + 55 = 500\). 4. d) Subtract: \(710 - 40 = 670\). 5. e) Find the missing addend: \(600 - 350 = 250\).

a) \(4\) b) \(7\) c) \(500\) d) \(670\) e) \(250\)

- Identify which class collected more. - Decide whether the unknown amount must be less than or greater than \(385\). - Subtract the difference from the greater amount.

1. The other class collected fewer acorns, so subtract the difference: \(385 - 120 = 265\).

The other class collected \(265\) acorns.

- Find each difference separately. - Regroup carefully when subtracting tens. - Compare all four results.

1. For a), \(320 - 50 = 270\). 2. For b), \(350 - 80 = 270\). 3. For c), \(310 - 60 = 250\). 4. For d), \(440 - 70 = 370\). 5. Problems a) and b) both have a result of \(270\).

a) and b); both equal \(270\).

- Use addition to undo the subtraction in part 1. - In part 2, find the distance from \(550\) to \(820\). - Check each answer in the original statement.

1. Add to find the missing first number: \(450 + 180 = 630\). 2. Subtract to find the missing number being subtracted: \(820 - 550 = 270\).

1. \(630\) 2. \(270\)

- Write the starting number before the minus sign. - Subtract to find the result. - Recall the name of the result of subtraction.

1. Write the subtraction expression: \(900 - 560\). 2. Calculate: \(900 - 560 = 340\). 3. The result of a subtraction equation is called the difference.

\(900 - 560 = 340\). The result is called the difference.

- Read the starting and ending value of each segment. - Subtract the starting value from the ending value to find a length. - Determine the value of each small interval before reading an endpoint. - Add the two individual lengths for part d).

1. Segment \(A\) starts at \(220\) and ends at \(580\), so its length is \(580 - 220 = 360\) units. 2. Segment \(B\) starts at \(140\) and ends at \(460\), so its length is \(460 - 140 = 320\) units. 3. Segment \(C\) starts at \(400\) and ends at \(820\), so its length is \(820 - 400 = 420\) units. 4. The combined length of \(B\) and \(C\) is \(320 + 420 = 740\) units. 5. Since \(320 < 360 < 420\), the order is \(B, A, C\).

a) \(360\) units b) \(320\) units c) \(420\) units d) \(740\) units e) \(B, A, C\)

- How are the starting marker, distance traveled, and ending marker related? - Use the inverse operation to find the start of the movement.

1. The ending marker is \(810\,\text{m}\), and the distance traveled is \(345\,\text{m}\). 2. Subtract the distance from the ending marker: \(810\,\text{m} - 345\,\text{m}\). 3. Subtract in steps: \(810 - 300 = 510\), \(510 - 40 = 470\), and \(470 - 5 = 465\). 4. The starting marker was \(465\,\text{m}\).

The hiker started at the \(465\,\text{m}\) marker because \(810\,\text{m} - 345\,\text{m} = 465\,\text{m}\).

- Subtract the right neighbor from the left neighbor. - Work upward one row at a time.

1. The second row is \(60 - 25 = 35\), \(25 - 10 = 15\), and \(10 - 2 = 8\). 2. The third row is \(35 - 15 = 20\) and \(15 - 8 = 7\). 3. The top is \(20 - 7 = 13\).

Second row: \(35\), \(15\), \(8\) Third row: \(20\), \(7\) Top: \(13\)

- A three-digit number cannot begin with \(0\). - Use place value to arrange the digits for the least and greatest numbers. - Subtract the smaller number from the greater number.

1. A three-digit number cannot begin with \(0\). To make the least number, place \(1\) in the hundreds place, then arrange the remaining digits in increasing order: \(104\). 2. To make the greatest number, arrange the digits in decreasing order: \(410\). 3. Subtract: \(410-104=306\).

\(306\)

- Break the second number into tens and ones. - It may help to subtract to a multiple of \(10\) or \(100\) first. - Rewrite each problem as two smaller subtractions.

1. For \(263 - 47\), subtract the tens and then the ones: \(263 - 40 = 223\), then \(223 - 7 = 216\). 2. For \(514 - 38\), subtract the tens and then the ones: \(514 - 30 = 484\), then \(484 - 8 = 476\). 3. For \(821 - 56\), subtract the tens and then the ones: \(821 - 50 = 771\), then \(771 - 6 = 765\).

a) \(216\) b) \(476\) c) \(765\)

- Break the subtrahend into hundreds, tens, and ones. - Subtract the hundreds first, then the tens, and finally the ones. - Record each intermediate result.

1. For \(452 - 168\): \(452 - 100 = 352\), \(352 - 60 = 292\), and \(292 - 8 = 284\). 2. For \(731 - 354\): \(731 - 300 = 431\), \(431 - 50 = 381\), and \(381 - 4 = 377\). 3. For \(615 - 279\): \(615 - 200 = 415\), \(415 - 70 = 345\), and \(345 - 9 = 336\).

a) \(284\) b) \(377\) c) \(336\)

- Break apart the number you are subtracting. - First subtract enough to reach the previous multiple of \(100\). - Then subtract the remaining amount.

1. For \(302 - 4\), subtract \(2\) to reach \(300\), then subtract the remaining \(2\): \(300 - 2 = 298\). 2. For \(605 - 8\), subtract \(5\) to reach \(600\), then subtract the remaining \(3\): \(600 - 3 = 597\). 3. For \(103 - 6\), subtract \(3\) to reach \(100\), then subtract the remaining \(3\): \(100 - 3 = 97\). 4. For \(901 - 5\), subtract \(1\) to reach \(900\), then subtract the remaining \(4\): \(900 - 4 = 896\).

a) \(298\) b) \(597\) c) \(97\) d) \(896\)

- Break the second number into tens and ones. - Subtract the tens first. - Then subtract the remaining ones.

1. \(452 - 70 = 382\), then \(382 - 8 = 374\). 2. \(631 - 50 = 581\), then \(581 - 4 = 577\). 3. \(215 - 30 = 185\), then \(185 - 9 = 176\).

a) \(374\) b) \(577\) c) \(176\)

- Break the subtrahend into hundreds, tens, and ones. - Subtract the hundreds, then the tens, and finally the ones. - Watch for crossings of a multiple of \(10\) or \(100\).

1. For part a, \(824 - 300 = 524\), \(524 - 50 = 474\), and \(474 - 1 = 473\). 2. For part b, \(415 - 100 = 315\), \(315 - 60 = 255\), and \(255 - 2 = 253\).

a) \(473\) b) \(253\)

- Break each subtrahend into hundreds, tens, and ones. - Subtract one place-value part at a time. - Compare the final differences carefully.

1. For part a, \(903 - 500 = 403\), \(403 - 40 = 363\), and \(363 - 6 = 357\). 2. For part b, \(721 - 300 = 421\), \(421 - 80 = 341\), and \(341 - 4 = 337\). 3. For part c, \(612 - 200 = 412\), \(412 - 70 = 342\), and \(342 - 5 = 337\). 4. The differences in parts b and c are equal.

a) \(357\) b) \(337\) c) \(337\) The differences in parts b and c are equal.

- How many tens must you subtract to reach the previous hundred? - Can you break apart the number you subtract into two parts? - What happens to the hundreds digit when you cross the hundred?

1. For part a, first subtract \(20\) to reach \(300\). Then subtract the remaining amount: \(300 - 20 = 280\), \(300 - 40 = 260\), and \(300 - 70 = 230\). 2. For part b, first subtract \(10\) to reach \(600\). Then subtract the remaining amount: \(600 - 20 = 580\), \(600 - 40 = 560\), and \(600 - 70 = 530\).

a) \(280\), \(260\), \(230\) b) \(580\), \(560\), \(530\)

- Break the second number into hundreds, tens, and ones. - Subtract the largest place-value part first. - Rewrite each problem as two or three smaller subtractions.

1. For \(462 - 38\): \(462 - 30 = 432\), then \(432 - 8 = 424\). 2. For \(462 - 125\): \(462 - 100 = 362\), \(362 - 20 = 342\), and \(342 - 5 = 337\). 3. For \(462 - 247\): \(462 - 200 = 262\), \(262 - 40 = 222\), and \(222 - 7 = 215\). 4. For \(462 - 356\): \(462 - 300 = 162\), \(162 - 50 = 112\), and \(112 - 6 = 106\).

a) \(424\) b) \(337\) c) \(215\) d) \(106\)

- Watch for regrouping when subtracting. - Check a difference with the inverse operation, addition. - Identify the hundreds, tens, and ones in each subtrahend.

1. \(821 - 300 = 521\), \(521 - 40 = 481\), and \(481 - 5 = 476\). 2. \(534 - 200 = 334\), \(334 - 60 = 274\), and \(274 - 7 = 267\). 3. \(712 - 400 = 312\), \(312 - 50 = 262\), and \(262 - 8 = 254\). 4. \(943 - 500 = 443\), \(443 - 70 = 373\), and \(373 - 6 = 367\).

a) \(476\) b) \(267\) c) \(254\) d) \(367\)

- Work from left to right, one arrow at a time. - Each landing number is the starting number for the next arrow. - Check every intermediate result because it affects all later steps.

1. First arrow: \(815 - 200 = 615\), then \(615 - 40 = 575\). 2. Second arrow: \(575 - 80 = 495\), then \(495 - 7 = 488\). 3. Third arrow: \(488 - 300 = 188\), \(188 - 50 = 138\), and \(138 - 6 = 132\).

\(815 \xrightarrow{-240} 575 \xrightarrow{-87} 488 \xrightarrow{-356} 132\)

- Which place-value part should you subtract first? - Break \(245\) into hundreds, tens, and ones. - Record the result after each step.

1. Subtract the hundreds: \(673 - 200 = 473\). 2. Subtract the tens: \(473 - 40 = 433\). 3. Subtract the ones: \(433 - 5 = 428\).

\(428\)

- What happens when you subtract more than \(18\) from \(218\)? - Break the tens apart so you can first reach \(200\). - How many tens must you subtract altogether?

1. Subtract the hundreds: \(518 - 300 = 218\). 2. Subtract the tens: \(218 - 60 = 158\). One way to show the crossing is \(218 - 18 = 200\), then \(200 - 42 = 158\). 3. Subtract the ones: \(158 - 2 = 156\).

\(156\)

- Which nearby multiple of \(100\) would be easier to subtract? - If you subtract too much, how can you adjust the difference? - Decide whether your first subtraction removed too much or too little.

1. Subtract \(500\): \(832 - 500 = 332\). Since this subtracts \(1\) too much, add \(1\): \(332 + 1 = 333\). 2. Subtract \(300\): \(546 - 300 = 246\). Since this subtracts \(2\) too much, add \(2\): \(246 + 2 = 248\). 3. Subtract \(200\): \(715 - 200 = 515\). Since this subtracts \(3\) too much, add \(3\): \(515 + 3 = 518\).

a) \(333\) b) \(248\) c) \(518\)

- Which nearby number is easier to subtract than \(198\)? - If you subtract \(200\) instead of \(198\), do you remove too many or too few? - How much must you add back?

1. Write the subtraction: \(462 - 198\). 2. Subtract the nearby multiple of \(100\): \(462 - 200 = 262\). 3. Subtracting \(200\) removes \(2\) too many books, so add \(2\): \(262 + 2 = 264\).

\(264\) books are left on the display.

- Compare the subtrahend with the multiple of \(100\) shown in the equation. - What is the difference between those two numbers? - Add back the amount that was subtracted in excess.

1. The difference between \(399\) and \(400\) is \(1\). Since \(924 - 400 = 524\), add \(1\): \(524 + 1 = 525\). 2. The difference between \(298\) and \(300\) is \(2\). Since \(637 - 300 = 337\), add \(2\): \(337 + 2 = 339\). 3. The difference between \(196\) and \(200\) is \(4\). Since \(512 - 200 = 312\), add \(4\): \(312 + 4 = 316\).

a) \(924 - 399 = 924 - 400 + 1 = 525\) b) \(637 - 298 = 637 - 300 + 2 = 339\) c) \(512 - 196 = 512 - 200 + 4 = 316\)

- Which nearby multiple of \(100\) is easier to subtract? - If you subtract more than needed, add the extra amount back. - How far is each subtrahend from the next hundred?

1. Use \(453 - 200 = 253\). Since \(200\) is \(1\) more than \(199\), add \(1\): \(253 + 1 = 254\). 2. Use \(816 - 400 = 416\). Since \(400\) is \(2\) more than \(398\), add \(2\): \(416 + 2 = 418\). 3. Use \(627 - 300 = 327\). Since \(300\) is \(3\) more than \(297\), add \(3\): \(327 + 3 = 330\).

a) \(254\) b) \(418\) c) \(330\)

- Break the second number into hundreds, tens, and ones. - Subtract the hundreds first, then the tens, and finally the ones. - Record the result after every step.

1. For part a, \(623 - 300 = 323\), \(323 - 50 = 273\), and \(273 - 8 = 265\). 2. For part b, \(841 - 400 = 441\), \(441 - 60 = 381\), and \(381 - 7 = 374\). 3. For part c, \(512 - 200 = 312\), \(312 - 70 = 242\), and \(242 - 5 = 237\).

a) \(265\) b) \(374\) c) \(237\)

- Is there a nearby number that is easier to subtract? - If you subtract too much, what adjustment must you make? - Could you count up from the smaller number to the larger number?

1. Use compensation: \(754 - 400 = 354\). Since \(400\) is \(1\) more than \(399\), add \(1\): \(354 + 1 = 355\). 2. Subtract in parts: \(600 - 100 = 500\), \(500 - 20 = 480\), and \(480 - 6 = 474\). 3. Use compensation: \(432 - 300 = 132\). Since \(300\) is \(2\) more than \(298\), add \(2\): \(132 + 2 = 134\).

a) \(355\) b) \(474\) c) \(134\)

- Use addition when the first number is missing. - To find a missing number being subtracted, find the difference between the first number and the known result. - Check each completed equation.

1. For a), subtract in parts: \(950 - 200 = 750\), then \(750 - 70 = 680\). 2. For b), use the inverse operation: \(445 + 155 = 600\), so \(600 - 155 = 445\). 3. For c), find the missing number being subtracted: \(824 - 549 = 275\), so \(824 - 275 = 549\).

a) \(680\) b) \(600\) c) \(275\)

- You are looking for pairs with one fixed difference. - For each number, look for another number that is \(7\) greater or \(7\) less. - Numbers near the same multiple of \(10\) or \(100\) may be useful to compare. - Compare two numbers by thinking about their distance on a number line.

1. Compare numbers that are close in value. 2. The matching pairs give \(135-128=7\), \(676-669=7\), \(403-396=7\), \(812-805=7\), and \(250-243=7\). 3. No other pair from the list has a difference of \(7\).

\(135-128=7\) \(676-669=7\) \(403-396=7\) \(812-805=7\) \(250-243=7\)

- Break \(275\) into hundreds, tens, and ones. - What remains after subtracting only the hundreds? - Continue by subtracting the tens and then the ones.

1. Subtract the hundreds: \(632 - 200 = 432\). 2. Subtract the tens: \(432 - 70 = 362\). 3. Subtract the ones: \(362 - 5 = 357\).

\(632 - 200 = 432\) \(432 - 70 = 362\) \(362 - 5 = 357\) The difference is \(357\).

- \(199\) is close to which multiple of \(100\)? - If you subtract too much, how do you correct the result? - When two numbers are close, count up from the smaller number to the larger number.

1. For part a, subtract \(200\): \(753 - 200 = 553\). Since \(200\) is \(1\) more than \(199\), add \(1\): \(553 + 1 = 554\). 2. For part b, count up \(2\) from \(398\) to \(400\), then \(2\) more from \(400\) to \(402\). The total difference is \(2 + 2 = 4\).

a) \(753 - 200 + 1 = 554\) b) \(398 + 2 = 400\), \(400 + 2 = 402\); the difference is \(4\).

- Choose a step-by-step method that is easy to track. - You can break the subtrahend into hundreds, tens, and ones. - In part c, pay close attention when crossing tens and hundreds.

1. For part a, \(876 - 300 = 576\), \(576 - 40 = 536\), and \(536 - 2 = 534\). 2. For part b, \(524 - 60 = 464\), then \(464 - 7 = 457\). 3. For part c, \(911 - 400 = 511\), \(511 - 50 = 461\), and \(461 - 8 = 453\).

a) \(534\) b) \(457\) c) \(453\)

- Find a missing number being subtracted by subtracting the known result from the first number. - Work from the first line to the last. - Check that each completed equation is true.

1. The first missing number being subtracted is \(1000 - 720 = 280\). 2. The second missing number being subtracted is \(720 - 485 = 235\). 3. The final difference is \(485 - 139 = 346\).

The missing numbers, in order, are \(280\), \(235\), and \(346\).

- Look for pairs of digits whose difference is \(3\). - Try to make the difference \(3\) in the hundreds, tens, and ones places. - Check that no digit repeats within an equation.

1. To make a difference of \(333\) without regrouping, choose a pair of digits with a difference of \(3\) in each place. 2. Using the pairs \((4,1)\), \((5,2)\), and \((6,3)\) gives \(456-123=333\). 3. Using the pairs \((7,4)\), \((8,5)\), and \((9,6)\) gives \(987-654=333\).

\(456-123=333\) \(987-654=333\)

- Is one number close to a multiple of \(100\)? - Which problem requires regrouping in more than one place? - Could you count up or use compensation instead of subtracting directly?

1. Problem A is efficient to solve mentally because \(395\) is close to \(400\). Use compensation: \(700 - 400 + 5 = 305\). 2. Problem B requires regrouping in more than one place, so a written subtraction method is easier to track. The difference is \(642 - 278 = 364\).

Problem A is easier to solve mentally because \(395\) is close to \(400\). A: \(305\) B: \(364\)

- Find all four differences first. - Use a subtraction strategy such as counting back to a nearby hundred. - Compare the hundreds digits of the results before comparing tens and ones.

1. Find the differences: a) \(920 - 350 = 570\), b) \(640 - 180 = 460\), c) \(810 - 270 = 540\), and d) \(758 - 490 = 268\). 2. Order the results: \(268 < 460 < 540 < 570\).

a) \(570\) b) \(460\) c) \(540\) d) \(268\) Order: \(268 < 460 < 540 < 570\)

- Break the number you subtract into two parts. - Could you first subtract enough to reach a multiple of \(100\)? - Think about the tens and ones separately.

1. \(520 - 20 = 500\), then \(500 - 40 = 460\). 2. \(340 - 40 = 300\), then \(300 - 40 = 260\). 3. \(600 - 40 = 560\), then \(560 - 2 = 558\). 4. \(900 - 70 = 830\), then \(830 - 5 = 825\). 5. \(1000 - 10 = 990\), then \(990 - 3 = 987\).

a) \(460\) b) \(260\) c) \(558\) d) \(825\) e) \(987\)

- For b), find the total distance from \(780\) to \(820\), then take half. - For c), count up from \(780\) to \(950\). - A number line may help you see the distances.

1. For a), \(780 - 50 = 730\). 2. For b), the distance from \(780\) to \(820\) is \(40\). Half of \(40\) is \(20\), and \(780 + 20 = 800\). 3. For c), \(950 - 780 = 170\).

a) \(730\) b) \(800\) c) \(170\)

- Order the three numbers from least to greatest. - When a question asks how many more or how many are needed, find a difference. - Pay attention when the subtraction crosses a ten or a hundred. - Check each answer by adding the difference to the smaller amount.

1. Order the amounts: \(820 > 640 > 560\). Lisa has the most, and Tim has the fewest. 2. Lisa and Tim: \(820 - 560 = 260\). 3. Lisa and Paul: \(820 - 640 = 180\). 4. Paul and Tim: \(640 - 560 = 80\).

a) Most: Lisa; fewest: Tim b) \(260\) marbles c) \(180\) marbles d) \(80\) marbles

- Break the amount being subtracted into hundreds, tens, and ones. - Subtract the hundreds first, then the tens, and finally the ones. - Watch for regrouping across tens or hundreds.

1. Decrease by \(15\): \(432 - 10 = 422\), then \(422 - 5 = 417\). 2. Decrease by \(150\): \(432 - 100 = 332\), then \(332 - 50 = 282\). 3. Decrease by \(215\): \(432 - 200 = 232\), \(232 - 10 = 222\), then \(222 - 5 = 217\).

\(417\), \(282\), \(217\)

- Decide which inverse operation will isolate each missing number. - In part 4, combine the two known addends first. - Check each value in the original equation.

1. Subtract to find the missing addend: \(610 - 480 = 130\). 2. Find the missing number being subtracted: \(820 - 740 = 80\). 3. Add to find the missing first number: \(410 + 190 = 600\). 4. First combine the known addends: \(350 + 100 = 450\). Then subtract: \(700 - 450 = 250\).

1. \(130\) 2. \(80\) 3. \(600\) 4. \(250\)

- Use the relationship: starting number minus number subtracted equals difference. - When the starting number is missing, add the number subtracted and the difference. - When the number subtracted is missing, subtract the difference from the starting number.

1. For a), subtract to find the difference: \(840 - 365 = 475\). 2. For b), add the number subtracted and the difference to find the starting number: \(210 + 450 = 660\). 3. For c), subtract the difference from the starting number to find the number subtracted: \(910 - 280 = 630\).

a) \(475\) b) \(660\) c) \(630\)

- Write a subtraction equation with a box for the missing value. - The starting number comes before the minus sign. - Use addition or subtraction as an inverse operation when needed.

1. Part a: \(625 - 148 = 477\). 2. Part b: Add the difference and the number subtracted: \(230 + 370 = 600\). 3. Part c: Subtract the difference from the starting number: \(800 - 555 = 245\).

a) \(477\) b) \(600\) c) \(245\)

- If you subtract a larger round number, you remove too much. - How much extra did you subtract? - Add that extra amount back at the end.

1. Subtracting \(400\) instead of \(396\) removes \(4\) too much. Since \(502 - 400 = 102\), add \(4\): \(102 + 4 = 106\). 2. Subtracting \(300\) instead of \(298\) removes \(2\) too much. Since \(701 - 300 = 401\), add \(2\): \(401 + 2 = 403\). 3. Subtracting \(500\) instead of \(495\) removes \(5\) too much. Since \(804 - 500 = 304\), add \(5\): \(304 + 5 = 309\).

a) \(502 - 396 = 502 - 400 + 4 = 106\) b) \(701 - 298 = 701 - 300 + 2 = 403\) c) \(804 - 495 = 804 - 500 + 5 = 309\)

- Decompose the numbers into hundreds and tens if helpful. - Notice whether the minuend and subtrahend change by the same amount. - Check a difference with a related addition equation.

1. For a), \(450 - 180 = 270\) and \(540 - 270 = 270\), so the values are equal. 2. For b), \(720 - 350 = 370\) and \(810 - 440 = 370\), so the values are equal. 3. For c), \(620 - 280 = 340\) and \(750 - 390 = 360\), so \(340 < 360\).

a) \(=\) b) \(=\) c) \(<\)

- Calculate and record each difference. - Regroup carefully across tens and hundreds when needed. - Compare all five results.

1. Calculate the differences: A) \(724 - 385 = 339\), B) \(613 - 274 = 339\), C) \(801 - 462 = 339\), D) \(542 - 193 = 349\), and E) \(930 - 591 = 339\). 2. A, B, C, and E all have a result of \(339\). D has a result of \(349\).

D) \(542 - 193 = 349\)

- Which subtrahend is closest to a multiple of \(100\)? - Why are multiples of \(100\) easier to subtract mentally? - If you subtract \(200\) instead of \(199\), how must you adjust?

1. The subtrahends \(312\) and \(245\) are not especially close to a multiple of \(100\). The subtrahend \(199\) is only \(1\) less than \(200\). 2. Choose C. Subtract \(200\): \(631 - 200 = 431\). 3. Since \(200\) is \(1\) more than \(199\), add \(1\): \(431 + 1 = 432\).

C is best because \(199\) is close to \(200\). \(631 - 200 + 1 = 432\).

- Which number is closer to a multiple of \(100\)? - Change one number to make the first calculation easier. - Then compensate for exactly how much you changed it.

1. Adjust the minuend from \(302\) to \(300\): \(300 - 65 = 235\). Add back the \(2\) removed from the minuend: \(235 + 2 = 237\). 2. Adjust the subtrahend from \(97\) to \(100\): \(514 - 100 = 414\). Since \(3\) too much was subtracted, add \(3\): \(414 + 3 = 417\). 3. Adjust the subtrahend from \(198\) to \(200\): \(705 - 200 = 505\). Since \(2\) too much was subtracted, add \(2\): \(505 + 2 = 507\). 4. Adjust the minuend from \(603\) to \(600\): \(600 - 47 = 553\). Add back the \(3\) removed from the minuend: \(553 + 3 = 556\).

a) \(237\) b) \(417\) c) \(507\) d) \(556\)

- Find the original difference first. - Determine the new number being subtracted. - Consider what happens when less is subtracted from the same starting number.

1. Part a: \(950 - 380 = 570\). 2. Part b: The new number subtracted is \(380 - 20 = 360\). 3. The new difference is \(950 - 360 = 590\). 4. Decreasing the number subtracted by \(20\) increases the difference by \(20\).

a) The difference is \(570\). b) The new difference is \(590\), which is \(20\) greater than the original difference.

- Work downward with the inverse relationship. - Start by asking, “What number minus \(10\) equals \(25\)?”

1. For the left value in the second row, solve \(x - 10 = 25\), so \(x = 35\). 2. For the bottom-left value, solve \(y - 15 = 35\), so \(y = 50\). 3. For the bottom-right value, solve \(15 - z = 10\), so \(z = 5\).

Bottom row: \(50\), \(15\), \(5\) Second row: \(35\), \(10\) Top: \(25\)

- Find a connected group with two known values. - Ask what must be subtracted from the lower-left value to get the upper value. - Reverse a subtraction equation with addition when the missing value is below.

1. For the middle bottom value, solve \(x - 12 = 23\), so \(x = 23 + 12 = 35\). 2. The left brick in the second row is \(78 - 35 = 43\). 3. The top is \(43 - 23 = 20\).

Bottom row: \(78\), \(35\), \(12\) Second row: \(43\), \(23\) Top: \(20\)

- Use the inverse relationship when a bottom value is missing. - Start where two connected values are already given. - Verify each subtraction from bottom to top.

1. The second bottom value is \(95 - 75 = 20\). 2. The missing bricks in the second row are \(20 - 10 = 10\) and \(10 - 5 = 5\). 3. The third row is \(75 - 10 = 65\) and \(10 - 5 = 5\). 4. The top is \(65 - 5 = 60\).

Bottom row: \(95\), \(20\), \(10\), \(5\) Second row: \(75\), \(10\), \(5\) Third row: \(65\), \(5\) Top: \(60\)

- Always subtract the right value from the left value. - Work upward one row at a time. - Check that the values become smaller as you move upward in this wall.

1. The second row is \(200 - 80 = 120\), \(80 - 40 = 40\), \(40 - 20 = 20\), and \(20 - 10 = 10\). 2. The third row is \(120 - 40 = 80\), \(40 - 20 = 20\), and \(20 - 10 = 10\). 3. The fourth row is \(80 - 20 = 60\) and \(20 - 10 = 10\). 4. The top is \(60 - 10 = 50\).

Bottom row: \(200\), \(80\), \(40\), \(20\), \(10\) Second row: \(120\), \(40\), \(20\), \(10\) Third row: \(80\), \(20\), \(10\) Fourth row: \(60\), \(10\) Top: \(50\)

- First determine what must be subtracted from \(20\) to get \(15\). - Then work from the bottom row upward. - Remember: lower-left minus lower-right gives the brick above.

1. For the bottom-right value, solve \(20 - x = 15\), so \(x = 5\). 2. The left brick in the second row is \(55 - 20 = 35\). 3. The top is \(35 - 15 = 20\).

Bottom row: \(55\), \(20\), \(5\) Second row: \(35\), \(15\) Top: \(20\)

- Begin with a connected group that has two known values. - You can work downward from the top by reversing subtraction.

1. For the bottom-left value, solve \(x - 50 = 70\), so \(x = 120\). 2. For the right brick in the second row, solve \(70 - y = 40\), so \(y = 30\). 3. For the bottom-right value, solve \(50 - z = 30\), so \(z = 20\).

Bottom row: \(120\), \(50\), \(20\) Second row: \(70\), \(30\) Top: \(40\)

- The result must satisfy two inequalities: it is greater than \(5\) and less than \(10\). - List the whole-number results that are greater than \(5\) and less than \(10\). - Look for pairs of numbers that are close together. - Sorting the numbers by size may make the pairs easier to find.

1. The possible results are \(6, 7, 8,\) or \(9\). 2. Comparing nearby numbers gives \(605-597=8\), \(411-403=8\), \(904-895=9\), and \(715-706=9\). 3. The pair \(302\) and \(298\) has a difference of \(4\), so it does not meet the condition. 4. No other pair from the list has a difference from \(6\) through \(9\).

\(605-597=8\) \(411-403=8\) \(904-895=9\) \(715-706=9\)

- Sort the numbers to find the least positive difference efficiently. - Use the greatest and least numbers for the greatest difference. - For part c), estimate promising sums and compare their exact distances from \(500\).

1. Sort the numbers: \(34, 98, 156, 207, 441, 523, 615, 782\). The adjacent differences are \(64, 58, 51, 234, 82, 92,\) and \(167\). The least is \(51\), from \(207-156\). 2. The greatest difference comes from the greatest and least numbers: \(782-34=748\). 3. Compare pair sums by their distance from \(500\). The closest candidates include \(441+34=475\), which is \(25\) away, and \(441+98=539\), which is \(39\) away. Checking all pairs confirms that \(475\) is closest.

a) \(207-156=51\) b) \(782-34=748\) c) \(441+34=475\), which is \(25\) away from \(500\)

- Find a connected group in which two of the three values are known. - To recover a missing lower-left value, you may need to add even though the wall rule uses subtraction.

1. For the third bottom value, solve \(x - 30 = 70\), so \(x = 100\). 2. The middle value in the second row is \(250 - 100 = 150\). 3. The two values in the third row are \(250 - 150 = 100\) and \(150 - 70 = 80\). 4. The top is \(100 - 80 = 20\). 5. For the first bottom value, solve \(y - 250 = 250\), so \(y = 500\).

Bottom row: \(500\), \(250\), \(100\), \(30\) Second row: \(250\), \(150\), \(70\) Third row: \(100\), \(80\) Top: \(20\)

- For a missing lower value, ask which subtraction equation the three connected bricks represent. - Start where two values in a connected group are already known.

1. The middle value in the second row is \(200 - 80 = 120\). 2. For the bottom-left value, solve \(x - 200 = 300\), so \(x = 500\). 3. The left value in the third row is \(300 - 120 = 180\). 4. For the right value in the second row, solve \(120 - y = 75\), so \(y = 45\). 5. For the bottom-right value, solve \(80 - z = 45\), so \(z = 35\). 6. The top is \(180 - 75 = 105\).

Bottom row: \(500\), \(200\), \(80\), \(35\) Second row: \(300\), \(120\), \(45\) Third row: \(180\), \(75\) Top: \(105\)

- Move from known values to connected blanks one at a time. - Reverse subtraction with addition when a lower-left value is missing. - Check the rule in every row after completing the wall.

1. For the second bottom value, solve \(x - 200 = 250\), so \(x = 450\). 2. The left brick in the second row is \(800 - 450 = 350\). 3. The left brick in the third row is \(350 - 250 = 100\). 4. For the right brick in the second row, solve \(250 - y = 100\), so \(y = 150\). 5. For the bottom-right value, solve \(200 - z = 150\), so \(z = 50\). 6. The top is \(100 - 100 = 0\).

Bottom row: \(800\), \(450\), \(200\), \(50\) Second row: \(350\), \(250\), \(150\) Third row: \(100\), \(100\) Top: \(0\)

- Start with a connected group that has two known values. - Add to recover a missing lower-left value. - Check every difference from bottom to top.

1. For the second bottom value, solve \(x - 60 = 90\), so \(x = 150\). 2. The right brick in the second row is \(60 - 20 = 40\). 3. The left brick in the second row is \(400 - 150 = 250\). 4. The third row is \(250 - 90 = 160\) and \(90 - 40 = 50\). 5. The top is \(160 - 50 = 110\).

Bottom row: \(400\), \(150\), \(60\), \(20\) Second row: \(250\), \(90\), \(40\) Third row: \(160\), \(50\) Top: \(110\)

- For the greatest difference, maximize the first number and minimize the second. - For the least positive difference, begin with hundreds digits that are as close as possible. - Then make the greater number as small as possible and the lesser number as great as possible.

1. For the greatest difference, make the greatest possible minuend and the least possible subtrahend: \(654-123=531\). 2. For the least positive difference, the hundreds digits must be as close as possible. Using \(4\) and \(3\), make the number beginning with \(4\) as small as possible and the number beginning with \(3\) as great as possible. 3. This gives \(412-365=47\), which is the least possible positive difference.

a) \(654-123=531\) b) \(412-365=47\)