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OK so I have this math problem that I am not sure on how to solve. Bellow is the question.

  • Find the greatest common divisor of $100!$ and $(3072\times 7^{10}\times 23^5 \times 59^2)$.

Give the answer as a product of powers of primes.

OK so I know that I have to collect the common prime factors with the smallest element. So how to I do that especially with $100!$ being a very large number.

Any assistants would be appreciated.

Community
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Phillip Mc
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  • This problem is actually pretty straightforward. Where did you get stuck? – Clayton Apr 11 '17 at 02:26
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    Also, are you sure you want assistants? If you show your effort, many would be more than willing to offer assistance, but I don't think they're on here to be assistants... :) – Clayton Apr 11 '17 at 02:27
  • Factor everything in the second number into prime factors (it's almost but not quite completely factored) and see which prime factors and how many of each fit into the $100!$ - e.g. do 10 sevens fit into $100!$? How about 5 twenty-threes? etc. The product of these prime factors, each with the correct multiplicity, is the GCD. – NickD Apr 11 '17 at 02:36
  • Ok so would i get gcd = 2^107^323^2*59? – Phillip Mc Apr 11 '17 at 03:15
  • Instead of adding "[Solved]" to the title choose the answer that helped you the most and click on the check-mark near that answer to mark it as solved. – kingW3 Apr 11 '17 at 18:42
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    Why did you change the title to one that is more cryptic? At least the prime factorization of the number allows it to be worked easily. Furthermore, writing an integer in scientific notation obscures the fact that the value is exact. – robjohn Apr 11 '17 at 19:16

2 Answers2

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Hint: Look at the primes in $3072 \cdot 7^{10} \cdot 23^5\cdot 59^2$ and their exponents in the factorization of $100!$. Note that $3072 = 3 \cdot2^{10}$.

lhf
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  • Yes 3027 is a product of primes 3027 = 3*2^10 – Phillip Mc Apr 11 '17 at 02:28
  • Don't I have to list the primes of 100! to see the primes in the factorization in 100! ? – Phillip Mc Apr 11 '17 at 02:37
  • @PhillipMc, no, you only need to consider the primes that appear in the other numbers, because their gcd is composed by the common primes with the correct exponent. – lhf Apr 11 '17 at 02:38
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We can use Legendre's Formula here.

The number of factors of $2$ in $100!$ is $\frac{100-3}{2-1}=97$.

The number of factors of $3$ in $100!$ is $\frac{100-4}{3-1}=48$.

The number of factors of $7$ in $100!$ is $\frac{100-4}{7-1}=16$.

The number of factors of $23$ in $100!$ is $\frac{100-12}{23-1}=4$.

The number of factors of $59$ in $100!$ is $\frac{100-42}{59-1}=1$.

Thus, $2^{97}\cdot3^{48}\cdot7^{16}\cdot23^4\cdot59^1$ divides $100!$ Now, using the minimum of the exponents gives $$ \begin{align} \gcd(100!,\,\overbrace{2^{10}\cdot3}^{3072}\cdot7^{10}\cdot23^5\cdot59^2) &=2^{10}\cdot3\cdot7^{10}\cdot23^4\cdot59\\ &=14,\!327,\!320,\!206,\!855,\!570,\!432 \end{align} $$

robjohn
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