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Fun factoid:lheiannie07 wrote:$$When\ 5^{11}\ and\ 2^n7^2\ have\ the\ same\ number\ of\ factors,\ what\ is\ the\ value\ of\ n?$$
A. 2
B. 3
C. 4
D. 5
E. 6
Can someone show me how to solve this?
OA B
Thanks a lot!DavidG@VeritasPrep wrote:Fun factoid:lheiannie07 wrote:$$When\ 5^{11}\ and\ 2^n7^2\ have\ the\ same\ number\ of\ factors,\ what\ is\ the\ value\ of\ n?$$
A. 2
B. 3
C. 4
D. 5
E. 6
Can someone show me how to solve this?
OA B
If $$T = a^x * b^y * c^z$$, where a, b, and c are distinct prime bases, then we can find the number of factors in T, by calculating (x + 1)(y +1)(z+1). Put another way, we can find the number of factors of any number by first taking the prime factorization of that number, then adding one to each exponent and multiplying the results.
So we know that 5^11 has 11+1 = 12 factors.
Similarly, 2^n * 7^2 will have (n +1)( 2+1) factors. We want this to equal 12, so (n +1)(2+1) = 12, and (n+1)(3) = 12, or n+1 = 4, and n = 3. The answer is B
The total number of factors 5^11 has is 11 + 1 = 12 and the total number of factors 2^n x 7^2 has is (n + 1)(2 + 1) = 3(n + 1). Thus we can say:lheiannie07 wrote:$$When\ 5^{11}\ and\ 2^n7^2\ have\ the\ same\ number\ of\ factors,\ what\ is\ the\ value\ of\ n?$$
A. 2
B. 3
C. 4
D. 5
E. 6
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