Is all the numbers of the form \$2^n3^m7^k35^ell\$ with \$nin,...,7, min,...,4,kin,...,3\$ and \$ellin,...,5\$ and that gives \$8cdot 5cdot 4cdot 6\$ possibilities.

You are watching: How many different positive integer factors does (64)(81)(125) have?. Generally, if

\$\$n = p_1^alpha_1·p_2^alpha_2ldots p_k^alpha_k\$\$

Then the amount of positive divisors of \$n\$ is:

\$\$d(n) = (alpha_1+1)(alpha_2+1)ldots (alpha_k+1)\$\$

In your case, it has \$(7+1)(4+1)(3+1)(5+1) = 8·5·4·6 = 960\$ Note the following:

\$gcd(2,3)=1\$\$gcd(2,7)=1\$\$gcd(2,23)=1\$\$gcd(3,7)=1\$\$gcd(3,23)=1\$\$gcd(7,23)=1\$

Therefore, the number of different positive integer divisors of \$(2^7)(3^4)(7^3)(23^5)\$ is:

\$\$(7+1)cdot(4+1)cdot(3+1)cdot(5+1)=960\$\$ What you are looking for is the number of the positive divisors of \$2^7cdot 3^4cdot 7^3cdot 23^5\$. Then, the answer is\$\$(7+1)(4+1)(3+1)(5+1)=8cdot 5cdot 4cdot 6=960.\$\$

In general, if\$\$N=p_1^colorredq_1cdot p_2^colorredq_2cdots p_k^colorredq_k\$\$where \$q_i,kinlifwynnfoundation.orgbb N\$ and \$p_1lt p_2ltcdotslt p_k\$ are primes, then the number of the positive divisors of \$N\$ can be represented as\$\$(colorredq_1+1)(colorredq_2+1)cdots (colorredq_k+1).\$\$ Thanks for contributing an answer to lifwynnfoundation.orgematics Stack Exchange!

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