- In a prime factorization, every factor must be prime. - Check whether every number in each choice is prime. - You can also evaluate each choice and compare it with \(720\).

1. Decompose \(720\) as \(72\times10\). 2. Factor each part into primes: \(72=8\times9=2\times2\times2\times3\times3\), and \(10=2\times5\). 3. Combine repeated prime factors: \(2\times2\times2\times2\times3\times3\times5=2^4\times3^2\times5\).

d) \(2^4\times3^2\times5\)

- Start with any factor pair for \(18\). - Continue factoring any factor that is composite. - Stop when every factor is prime.

1. Split \(18\) into \(2\times9\). 2. Factor \(9\) as \(3\times3\). 3. All factors are prime, so \(18=2\times3\times3=2\times3^2\).

\(18=2\times3^2\)

- Check whether each factor shown is prime. - If a factor can be split into smaller whole-number factors, keep going. - Multiply the final prime factors to check the original number.

1. The factor \(9\) is not prime because \(9=3\times3\). 2. Substitute to obtain \(18=2\times3\times3\).

\(2\times3\times3\)

- Group identical factors. - Count how often each different factor appears. - Keep the two different prime bases separate.

1. The factor \(3\) appears three times, so write \(3^3\). 2. The factor \(5\) appears twice, so write \(5^2\). 3. The product is \(3^3\times5^2\).

\(3^3\times5^2\); the exponents count three copies of \(3\) and two copies of \(5\).

- Both products multiply to \(30\), so the final value alone does not settle the question. - Look at each factor, one at a time, rather than the product as a whole. - Try finding a factor pair for any factor you are unsure about.

1. Both products equal \(30\). 2. Mia's factors \(2\), \(3\), and \(5\) are all prime. 3. Ben's product contains the composite factor \(10\), so only Mia's product is a prime factorization.

Mia's \(2\times3\times5\) is the prime factorization. Ben's product contains the composite factor \(10\).

- Factor each number until every factor is prime. - How can repeated equal prime factors be written using an exponent? - Begin with the smallest prime numbers.

1. Factor each prime power: \(81=3\times3\times3\times3=3^4\), \(125=5\times5\times5=5^3\), and \(64=2\times2\times2\times2\times2\times2=2^6\). 2. Factor the remaining numbers: \(200=2\times2\times2\times5\times5=2^3\times5^2\), and \(450=2\times3\times3\times5\times5=2\times3^2\times5^2\).

1) \(81=3^4\); \(125=5^3\); \(64=2^6\) 2) \(200=2^3\times5^2\); \(450=2\times3^2\times5^2\)

- Rewrite each power as repeated multiplication. - Count the total number of factors equal to \(2\). - Use that count as the exponent in the single power.

1. Expand the powers: \(2^3\times2^2=(2\times2\times2)(2\times2)\). 2. There are five factors of \(2\), so the product is \(2^5\). 3. Evaluate \(2^5=32\).

\(2^3\times2^2=2^5=32\)

- Break the number into powers of the two listed primes. - Use each exponent to count repeated factors. - Multiply the expanded factors to check the result.

1. Factor \(200=8\times25\). 2. Write \(8=2^3\) and \(25=5^2\). 3. Therefore, \(a=3\), \(b=2\), and the expanded product is \(2\times2\times2\times5\times5\).

a) \(a=3\), \(b=2\) b) \(2\times2\times2\times5\times5\)

- Count how many times each prime factor repeats. - Use that count as an exponent. - Evaluate the exponent before multiplying by the remaining prime factor.

1. The prime factor \(2\) appears three times, so \(2\times2\times2=2^3\). 2. The factorization is \(2^3\times5\). 3. Evaluate: \(2^3\times5=8\times5=40\).

a) \(2^3\times5\) b) \(40\)

- Work out each power as repeated multiplication. - Multiply the two prime-power values. - Check that the result contains exactly the requested prime copies.

1. Evaluate the powers: \(2^3=8\) and \(3^2=9\). 2. Multiply \(8\times9=72\).

\(2^3=8\) and \(3^2=9\), so \(2^3\times3^2=8\times9=72\).

- Find the value of the student's product first. - Write the actual number as a product of primes. - Compare the copies of each prime in the two products.

1. The student's product is \(2\times3\times25=150\), not \(90\). 2. The correct factorization is \(90=2\times3\times3\times5=2\times3^2\times5\). 3. The student's product has one too many factors of \(5\) and one too few factors of \(3\).

The claimed product is \(150\), not \(90\). The student used two factors of \(5\) instead of one and only one factor of \(3\) instead of two. Correct: \(90=2\times3^2\times5\).

- Recall the definition of a prime number. - Decompose each number until every factor is prime. - Count how many times each prime factor appears. - Compare the three lists of prime factors.

1. Factor \(60\): \(60=2\times30=2\times2\times15=2^2\times3\times5\). 2. Factor \(126\): \(126=2\times63=2\times3\times21=2\times3^2\times7\). 3. Factor \(270\): \(270=2\times135=2\times3\times45=2\times3^2\times15=2\times3^3\times5\). 4. The prime factors \(2\) and \(3\) appear in all three factorizations.

\(60=2^2\times3\times5\) \(126=2\times3^2\times7\) \(270=2\times3^3\times5\) The common prime factors are \(2\) and \(3\).

- What happens to a prime factorization when the number is multiplied by another prime number? - Which ones digits indicate divisibility by \(5\)? - How many positive factors does a prime number have?

1. Evaluate the factorization: \(n=2^2\times3\times7=4\times3\times7=84\). 2. Multiplying \(n\) by \(5\) adds one factor of \(5\). The new prime factorization is \(2^2\times3\times5\times7\). 3. Every number ending in \(5\) is divisible by \(5\). If the number is greater than \(5\), then it has at least the positive factors \(1\), \(5\), and itself, so it is not prime.

a) \(n=84\) b) \(2^2\times3\times5\times7\) c) A number greater than \(5\) that ends in \(5\) is divisible by \(5\), so it has more than two positive factors and is not prime.

- Start from the prime factorization so your factor-pair list is complete. - For part a, compare the sum of each generated factor pair with \(21\). - For part b, group the prime factors into three different positive factors. - Check that every reported factor comes from the prime factorization.

1. For a), \(108=2^2\times3^3\). Its positive factor pairs are \((1,108),(2,54),(3,36),(4,27),(6,18),(9,12)\). 2. The factor pair whose sum is \(21\) is \(9\) and \(12\). 3. For b), \(180=2^2\times3^2\times5\). One grouping of the prime factors gives \(2\times9\times10=180\), and the factors \(2,9,10\) are different.

a) \(9\) and \(12\) b) For example, \(2\), \(9\), and \(10\)

- Break each composite number into smaller factors until every factor is prime. - Collect repeated prime factors with exponents. - Prime-factorize \(42\) before comparing it with the other factorizations. - A number is divisible by \(42\) only if its prime factorization contains all prime factors required by \(42\).

1. \(3600=36\times100=(2^2\times3^2)(2^2\times5^2)=2^4\times3^2\times5^2\). 2. \(294=2\times147=2\times3\times49=2\times3\times7^2\). 3. Since \(42=2\times3\times7\), a number must contain at least one factor each of \(2\), \(3\), and \(7\) in its prime factorization to be divisible by \(42\). 4. The factorization of \(3600\) has no factor \(7\), so \(42\) is not a factor of \(3600\). 5. The factorization of \(294\) contains \(2\), \(3\), and \(7^2\), so \(42\) is a factor of \(294\).

a) \(3600=2^4\times3^2\times5^2\) b) \(294=2\times3\times7^2\) c) \(42\) is not a factor of \(3600\), but it is a factor of \(294\).

- Continue factoring until every factor is prime. - Each positive factor can be made by multiplying some of the prime factors. - Remember to include \(1\) and the number itself. - Organize the products systematically so that none are missed.

1. Factor the number: \(130=2\times65=2\times5\times13\). 2. Form every possible product using each prime factor either once or not at all: \(1,2,5,13,2\times5,2\times13,5\times13,\) and \(2\times5\times13\). 3. In numerical order, the positive factors are \(1,2,5,10,13,26,65,\) and \(130\). Therefore, \(130\) has \(8\) positive factors.

a) \(130=2\times5\times13\) b) \(1,2,5,10,13,26,65,130\) c) \(8\) positive factors

- List all positive factors of each number. - Compare the prime factors in the two factorizations. - How is multiplying two equal primes different from multiplying two distinct primes?

1. The positive factors of \(6\) are \(1,2,3,\) and \(6\), so \(6\) has \(4\) positive factors. 2. The positive factors of \(9\) are \(1,3,\) and \(9\), so \(9\) has \(3\) positive factors. 3. The prime factors of \(6\) are distinct, so each creates a different factor. In \(9=3^2\), the two prime factors are the same, so they do not create two different one-prime factors.

a) \(4\) positive factors: \(1,2,3,6\) b) \(3\) positive factors: \(1,3,9\) c) The number \(6\) has two distinct prime factors, while \(9\) repeats the same prime factor. Repetition produces fewer distinct products.

- Continue factoring until every factor is prime. - Positive factors are made by multiplying allowable numbers of the prime factors. - Try different combinations of the prime factors. - Check systematically which products are between \(10\) and \(30\).

1. Factor repeatedly: \(72=2\times36=2\times2\times18=2\times2\times2\times9=2^3\times3^2\). 2. Form factor products in the requested interval: \(2^2\times3=12\), \(2\times3^2=18\), and \(2^3\times3=24\). 3. Other nearby products, such as \(2^3=8\), \(3^2=9\), and \(2^2\times3^2=36\), are outside the interval. Therefore, the requested factors are \(12,18,\) and \(24\).

a) \(72=2^3\times3^2\) b) \(12,18,24\)

- What must be true about the prime factors of a factor of a number? - Find the prime factorizations of \(15\), \(20\), and \(25\). - Check whether \(150\) contains every required prime factor enough times.

1. Since \(15=3\times5\), and \(150\) contains both required prime factors, \(15\) is a factor of \(150\). 2. Since \(20=2^2\times5\), it requires two factors of \(2\). The factorization of \(150\) contains only one factor of \(2\), so \(20\) is not a factor. 3. Since \(25=5^2\), and \(150\) contains two factors of \(5\), \(25\) is a factor.

a) True, because \(15=3\times5\), and both prime factors occur in \(150\). b) False, because \(20=2^2\times5\), but \(150\) contains only one factor of \(2\). c) True, because \(25=5^2\), and \(150\) contains two factors of \(5\).

- Multiply the factors in stages to find \(n\). - Which individual prime factors or products of factors are between \(10\) and \(20\)? - Find the prime factorization of \(9\). - Compare how many factors of \(3\) are required with how many occur in \(n\).

1. Evaluate the factorization: \(n=2^2\times3\times11=4\times3\times11=132\). 2. The prime factor \(11\) is between \(10\) and \(20\). Another possible factor is \(2^2\times3=12\). 3. Since \(9=3^2\), a number divisible by \(9\) must contain at least two factors of \(3\). The factorization of \(n\) contains only one factor of \(3\), so \(9\) is not a factor of \(n\).

a) \(n=132\) b) Possible answers include \(11\) and \(12\). c) Since \(9=3^2\) but \(n\) contains only one factor of \(3\), \(9\) is not a factor of \(n\).

- Break \(756\) into factors that are easier to factor again. - Continue until every factor is prime. - Use an exponent to count repeated copies of the same prime.

1. Factor \(756=4\times189\). 2. Write \(4=2^2\) and \(189=27\times7=3^3\times7\). 3. Therefore, \(756=2^2\times3^3\times7\), so the prime \(3\) occurs three times. 4. Check: \(2^2\times3^3\times7=4\times27\times7=756\).

\(756=2^2\times3^3\times7\); the prime \(3\) occurs three times.

- Evaluate each exponent before multiplying the factors. - Use the exponents to count repeated prime factors. - Count prime bases, not exponents, for the distinct-factor question.

1. Evaluate the prime powers: \(2^4=16\) and \(5^2=25\). 2. Multiply: \(16\times3\times25=1200\). 3. Counting repeats gives \(4+1+2=7\) prime factors, and the distinct primes are \(2,3,5\), so there are \(3\) distinct prime factors.

a) \(1200\) b) \(7\) c) \(3\)

- Factor both the divisor and the quotient. - Combine all prime factors before evaluating the number. - Check by dividing the final number by the given divisor.

1. Write \(9=3^2\) and \(77=7\times11\). 2. Combine the prime factors to get \(3^2\times7\times11\). 3. The number is \(9\times77=693\).

Prime factorization: \(3^2\times7\times11\); number: \(693\)

- Evaluate the powers before multiplying. - Compare the final whole-number values. - Add exponents to count repeated prime factors.

1. Evaluate \(A=8\times9=72\). 2. Evaluate \(B=4\times3\times5=60\). 3. Therefore, \(A>B\). 4. Expression \(A\) has \(3+2=5\) prime factors, while \(B\) has \(2+1+1=4\), so \(A\) has more.

a) \(A=72\), \(B=60\) b) \(>\) c) \(A\)

- Factor each possible divisor into primes. - Compare how many copies of each prime are required. - A missing repeated prime factor prevents divisibility.

1. The factorization is \(N=2^3\times3\times7\). 2. Since \(24=2^3\times3\), all its prime factors occur in \(N\), so \(N\) is divisible by \(24\). 3. Since \(28=2^2\times7\), all its prime factors occur in \(N\), so \(N\) is divisible by \(28\). 4. Since \(18=2\times3^2\) requires two factors of \(3\) but \(N\) has one, \(N\) is not divisible by \(18\).

a) Yes. b) Yes. c) No.

- Translate each repeated-factor count into an exponent. - Evaluate the prime powers before multiplying. - Factor the comparison number to identify the mismatch.

1. The prime factorization is \(2^3\times3^2\times5\). 2. Its value is \(8\times9\times5=360\). 3. Since \(180=2^2\times3^2\times5\), it has only two factors of \(2\), not three.

a) \(2^3\times3^2\times5\) b) \(360\) c) \(180\) has only \(2^2\), so it is missing one factor of \(2\).

- Use a divisibility test to choose a small prime factor. - Continue dividing until the quotient is no longer divisible by that prime. - Rewrite repeated primes with exponents and multiply to check.

1. The digit sum is \(9\), so divide repeatedly by \(3\): \(1323=3\times441=3\times3\times147=3^3\times49\). 2. Write \(49=7^2\). 3. Therefore, \(1323=3^3\times7^2\). 4. Verification: \(27\times49=1323\).

\(1323=3^3\times7^2\)

- Multiply the known prime factors first. - Use the original number to find the missing factor. - Check the factor against the definition of a prime number.

1. The known factors multiply to \(2^2\times3=12\). 2. The missing factor is \(132\div12=11\). 3. The number \(11\) has only factors \(1\) and \(11\), so it is prime.

a) \(p=11\) b) \(11\) has exactly two positive factors, \(1\) and \(11\).

- Expand the exponents mentally as repeated prime factors. - Cancel only the copies contained in the divisor. - Evaluate the remaining prime powers last.

1. Remove two factors of \(2\) from the five and one factor of \(3\) from the two. 2. The quotient is \(2^3\times3\times5\). 3. Its value is \(8\times3\times5=120\).

a) \(2^3\times3\times5\) b) \(120\)

- Factor the multiplier before combining it with the given factorization. - Group equal prime factors together. - Compare the original and final prime bases and exponents.

1. The prime factorization of \(15\) is \(3\times5\). 2. Combine factors: \(15K=2^3\times3^2\times3\times5=2^3\times3^3\times5\). 3. The exponent on \(3\) increases from \(2\) to \(3\), and the new prime base \(5\) appears.

a) \(15=3\times5\) b) \(15K=2^3\times3^3\times5\) c) The exponent on \(3\) increases by one, and a factor of \(5\) is added.

- Evaluate each allowed exponent separately. - Compare consecutive values rather than all three at once. - Interpret what increasing an exponent does to repeated factors.

1. For \(a=1\), the number is \(2\times3=6\). 2. For \(a=2\), the number is \(4\times3=12\); for \(a=3\), it is \(8\times3=24\). 3. Each increase of one in \(a\) adds another factor of \(2\), so the value doubles and the prime \(2\) repeats more often.

a) \(6, 12, 24\) b) Each number doubles. c) The prime factor \(2\)

- Check both the numerical value and whether every base is prime. - Evaluate each expression before deciding. - A product can equal \(864\) and still fail to be a prime factorization.

1. Choice A equals \(32\times27=864\), and both bases \(2\) and \(3\) are prime. 2. Choice B equals \(16\times27=432\), so it does not have the correct value. 3. Choice C equals \(32\times27=864\), but \(27\) is composite, so the expression is not fully factored into primes.

A: \(2^5\times3^3\)

- Identify the factors shared by both numbers. - Evaluate each prime-power product carefully. - Compare the one factor that differs.

1. Evaluate \(A=4\times3\times13=156\). 2. The factorization of \(B\) is \(2^2\times3\times5\), so \(B=4\times3\times5=60\). 3. Number \(A\) is greater because the common factor \(2^2\times3\) is multiplied by \(13\) instead of \(5\).

a) \(A=156\), \(B=60\) b) \(2^2\times3\times5\) c) \(A\) is greater because \(13>5\) while the remaining factors are identical.

- Write each denominator as a product of prime factors. - A power of \(10\) has equal numbers of factors of \(2\) and \(5\). - Look for a denominator containing a prime factor other than \(2\) or \(5\).

1. For a), \(8=2^3\). Since \(100=2^2\times5^2\), \(8\) is not a factor of \(100\). Since \(1000=2^3\times5^3\), \(8\) is a factor of \(1000\). 2. For b), \(40=2^3\times5\). The smallest power of \(10\) containing at least three factors of \(2\) and one factor of \(5\) is \(10^3=1000\). 3. For c), one answer is \(12\), because \(12=2^2\times3\) contains a factor of \(3\), and powers of \(10\) contain only factors of \(2\) and \(5\).

a) \(8\) is a factor of \(1000\), but not of \(100\). b) \(1000\) c) For example, \(12\), because it contains the prime factor \(3\).

- A divisor of \(2^3\times5^3\) can use only the primes \(2\) and \(5\), with exponents no larger than \(3\). - For part b, express \(8\) as a power of \(2\) and remove those prime factors from \(1000\). - A common factor with \(125\) must involve which prime?

1. Every divisor of \(1000=2^3\times5^3\) has the form \(2^a5^b\) with \(0\le a,b\le3\). 2. Keeping only values at most \(100\) gives \(1,2,4,5,8,10,20,25,40,50,100\). 3. Since \(8=2^3\), dividing the prime factorization of \(1000\) by \(8\) leaves \(5^3=125\). Thus \(m=125\). 4. A fraction with denominator \(125=5^3\) is in simplest form when its numerator is not divisible by \(5\). Examples are \(\frac{1}{125},\frac{2}{125},\frac{3}{125}\).

a) Every divisor has the form \(2^a5^b\) with \(0\le a,b\le3\). Those at most \(100\) are \(1,2,4,5,8,10,20,25,40,50,100\). b) \(m=125=5^3\); for example, \(\frac{1}{125},\frac{2}{125},\frac{3}{125}\). The numerator must not be divisible by \(5\).

- Combine the mixed number and fraction into one fraction with denominator \(12\). - Prime-factorize \(12\) using exponent notation. - A numerator divisible by \(12=2^2\times3\) must be divisible by both \(4\) and \(3\). - Use the allowed range of \(\triangle\) to narrow the possible numerator values.

1. Write the expression with denominator \(12\): \( 3\frac{\triangle}{12}+\frac56=\frac{36+\triangle+10}{12}=\frac{46+\triangle}{12}. \) 2. Prime-factorize the denominator: \(12=2^2\times3\). Therefore, \(46+\triangle\) must be divisible by both \(4\) and \(3\). 3. Since \(1\le\triangle\le11\), the numerator \(46+\triangle\) lies from \(47\) through \(57\). 4. In this interval, the only number divisible by both \(4\) and \(3\) is \(48\). 5. Thus \(46+\triangle=48\), so \(\triangle=2\). The value of the expression is \(48/12=4\).

\(\triangle=2\), and the expression equals \(4\).

- Write the total number of tiles as a product of prime factors. - Think about the factors each dimension must contain to satisfy the divisibility conditions. - How can you account for every possible use of the remaining factors?

1. The prime factorization is \(360=2^3\times3^2\times5\). 2. The rows must use at least \(2^3\) and the columns must use at least \(5\). These required factors use \(8\times5=40\) of the product. 3. The remaining factor is \(3^2=9\). It can be split between rows and columns as \(1\times9\), \(3\times3\), or \(9\times1\). 4. This gives \((8,45)\), \((24,15)\), and \((72,5)\). Each pair has product \(360\), and there are no other ways to distribute the remaining factors.

\((8,45)\), \((24,15)\), and \((72,5)\). Since \(360=2^3\times3^2\times5\), the required row factor \(8\) and column factor \(5\) leave \(3^2\), which can be divided between the dimensions in exactly three ways.

- How can you combine two products of prime factors into one product? - Identify which primes appear in either factorization, then consider the least prime absent from both. - If two numbers share a factor, what does that tell you about their sum?

1. Evaluate each factorization: \(A=4\times3\times5=60\) and \(B=2\times9\times5=90\). 2. Combine like prime factors in the product: \(A\times B=(2^2\times3\times5)(2\times3^2\times5)=2^{2+1}\times3^{1+2}\times5^{1+1}=2^3\times3^3\times5^2\). 3. The primes \(2\), \(3\), and \(5\) occur in at least one of the factorizations. The next prime, \(7\), occurs in neither. 4. Both \(A\) and \(B\) contain a factor of \(5\), so both are multiples of \(5\). The sum of two multiples of \(5\) is also divisible by \(5\).

a) \(A=60\), \(B=90\) b) \(2^3\times3^3\times5^2\) c) \(7\) d) Yes. Both numbers contain a factor of \(5\), so their sum is divisible by \(5\).

- First list the multiples of \(3\) in the interval. - Write the prime factorization of each candidate. - Think about how many exponent choices each prime factor allows when building a positive factor.

1. The multiples of \(3\) between \(30\) and \(50\) are \(33,36,39,42,45,48\). 2. Their prime factorizations are \(33=3\times11\), \(36=2^2\times3^2\), \(39=3\times13\), \(42=2\times3\times7\), \(45=3^2\times5\), and \(48=2^4\times3\). 3. For a prime factorization, a factor can use each prime exponent from \(0\) through its exponent in the number. Thus \(33\) and \(39\) each have \(2\times2=4\) factors; \(36\) has \(3\times3=9\); \(42\) has \(2\times2\times2=8\); \(45\) has \(3\times2=6\); and \(48\) has \(5\times2=10\). 4. The factors of \(45\) are \(1,3,5,9,15,45\), so \(x=45\).

\(x=45\); its factors are \(1,3,5,9,15,45\).

- Translate divisibility by \(4\) and \(9\) into minimum prime exponents. - Use the exponent rule for counting positive factors. - Ask whether any extra prime factor or larger exponent could keep the factor count at exactly nine.

1. Divisibility by \(4=2^2\) and \(9=3^2\) means the number’s prime factorization must contain at least \(2^2\times3^2\). 2. A number with prime factorization \(p^a q^b\) has \((a+1)(b+1)\) positive factors. 3. The required exponents \(2\) and \(2\) already give \((2+1)(2+1)=9\) factors. 4. Increasing either exponent or adding another prime factor would make the factor count greater than \(9\). 5. Therefore, the number is exactly \(2^2\times3^2=36\), which is less than \(100\).

\(36\)

- Think of each exponent as a count of repeated prime factors. - Divide each repeated-factor count into two equal groups. - Multiply one group to find the number being squared.

1. Evaluate \(N=16\times9=144\). 2. Pair equal prime factors: \(2^4\times3^2=(2^2\times3)^2=12^2\). 3. Each even exponent splits into two equal groups of prime factors.

a) \(144\) b) \(N=12^2\) c) Even exponents allow every repeated prime factor to be paired equally between the two copies of \(12\).

- Organize the factors by how many of the four primes are used. - How many ways can you choose none, one, two, three, or all four letters? - A tree diagram or table can help you find every combination.

1. Group the factors by the number of prime factors used. Using none gives \(1\). Using one gives \(a,b,c,d\). Using two gives \(ab,ac,ad,bc,bd,cd\). Using three gives \(abc,abd,acd,bcd\). Using all four gives \(abcd\). 2. The number of factors is \(1+4+6+4+1=16\). 3. For \(210=2\times3\times5\times7\), the positive factors are \(1,2,3,5,6,7,10,14,15,21,30,35,42,70,105,\) and \(210\). This list has \(16\) factors, confirming the result.

a) \(1\); \(a,b,c,d\); \(ab,ac,ad,bc,bd,cd\); \(abc,abd,acd,bcd\); \(abcd\) b) \(16\) positive factors c) \(1,2,3,5,6,7,10,14,15,21,30,35,42,70,105,210\)