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Que: If both 7 and \(3^3\) are factors of \(p\cdot4^4\cdot6^2\cdot5^4\) , then what is the smallest.....

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by Max@Math Revolution » Tue Jun 29, 2021 9:18 pm

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Que: If both 7 and \(3^3\) are factors of \(p\cdot4^4\cdot6^2\cdot5^4\), then what is the smallest possible value of p?

(A) 3
(B) 7
(C) 21
(D) 189
(E) 243
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Re: Que: If both 7 and \(3^3\) are factors of \(p\cdot4^4\cdot6^2\cdot5^4\) , then what is the smallest.....

by gmatbyexample » Thu Jul 01, 2021 12:06 pm
The first step would be to prime factorize p*(4^4)*(6^2)*(5^2) = p * 2^8 * (2^2 * 3^2) * 5^2 = p * 2^10 * 3^2 * 5^2

For both 7 and 3^3 to be factors of this number - there should be atleast one 7 (which p could fulfill) and three 3s - we already see that 3^2 is in the factors - which means one additional 3 is missing (a gap which p could fulfill)

No p could have many more factors - however per the question at a "minimum" p must have a 7 and a 3 as factors.

So value of p = 7*3 = 21

(C)
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Re: Que: If both 7 and \(3^3\) are factors of \(p\cdot4^4\cdot6^2\cdot5^4\) , then what is the smallest.....

by Max@Math Revolution » Fri Jul 02, 2021 7:45 pm
Solution: Express the given number \(p\cdot4^4\cdot6^2\cdot5^4\) in prime factors as \(p\cdot2^8\cdot2^2\cdot3^2\cdot5^4\).

=> For 7 to be the factor it should be contained in it. If we remove p then we don’t have 7. Hence p = 7.

\(3^2\) is contained already and hence we need 1 more 3 so that \(3^3\) is the factor and hence 3 should also be in p.

=> Thus, smallest value of p = 3 * 7 = 21

Therefore, C is the correct answer.

Answer C
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