kevincan wrote:
How many positive three digit integers have exactly 7 factors?
(A) none
(B) one
(C) two
(D) three
(E) more than three
Theory:The total number of factors of integer n with prime factorization \(n=a^x*b^y*c^z\) is equal to \((x+1)(y+1)(z+1)\); this includes 1 and \(n\). In other words, to find the total number of factors of integer \(n\), make the prime factorization of \(n\) and then take the product of (each exponent +1).
Back to the question:Assume we have a generic integer \(n\) with prime factorization \(n=a^x*b^y*c^z\)
We are told that the total number of factors is 7; this implies that \((x+1)(y+1)(z+1) = 7\)
Now, when we inspect that equation, we see that 7 is a PRIME number, which means it's only factors are "one and itself". Thus, the only way to have the product equal 7 is for one of \(x, y, z\) to equal 6 while the other two are zero; that way we get the required \((6+1)(0+1)(0+1)=7*1*1=7\).
The key inference is thus that our generic integer \(n\) has only one prime factor whose exponent is 6.
Now, if \(n=a^6\), we just need to see how many prime numbers \(a\) fit the given conditions: \(100\)
≤ \(a^6\)
≤ \(999\)
\(
2^6= 64\) --> no good; only two digits
\(
3^6 = 27^2=729\) --> good; three digits
\(
5^6 = 625*5*5=3125*5\) --> no good; four digits and counting
Thus, there is only ONE. Answer B.
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