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2007 iTest Problems/Problem 24

Revision as of 18:49, 30 June 2018 by Rockmanex3 (talk | contribs) (Solution to Problem 24)
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Problem

Let $N$ be the smallest positive integer such that $2008N$ is a perfect square and $2007N$ is a perfect cube. Find the remainder when $N$ is divided by $25$.

$\text{(A) }0 \quad \text{(B) }1 \quad \text{(C) }2 \quad \text{(D) }3 \quad \text{(E) }4 \quad \text{(F) }5 \quad \text{(G) }6 \quad \text{(H) }7 \quad \text{(I) } 8\quad$

$\text{(J) }9 \quad \text{(K) }10 \quad \text{(L) }11 \quad \text{(M) }12 \quad \text{(N) }13 \quad \text{(O) }14 \quad \text{(P) }15 \quad \text{(Q) }16 \quad$

$\text{(R) }17 \quad \text{(S) }18 \quad \text{(T) }19 \quad \text{(U) }20 \quad \text{(V) }21 \quad \text{(W) }22 \quad \text{(X) }23$

Solution

The prime factorization of $2008$ is $2^3 \cdot 251$, and the prime factorization of $2007$ is $3^2 \cdot 223$. Since $2008N$ is a perfect square and $2007N$ is a perfect cube, the exponents in the prime factorization of $2008N$ are even, and the exponents in the prime factorization of $2007N$ are a multiple of three.

To make the exponents of $2008N$ even, the exponent of $2$ in $N$ is at least $1$, but since there are no powers of $2$ in $2007$, the exponent of $2$ in $N$ is at least $3$. Similarly, in $N$, the exponent of $251$ is at least $3$, the exponent of $3$ is at least $4$, and the exponent of $223$ is at least $2$.

Thus, $N$, the minimum positive integer that satisfies the criteria, equals $2^3 \cdot 3^4 \cdot 251^3 \cdot 223^2$. Using modular arithmetic to find the remainder, \[N \equiv 8 \cdot 6 \cdot 1^3 \cdot (-2)^2 \pmod{25}\] \[N \equiv 192 \pmod{25}\] \[N \equiv 17 \pmod{25}\] The remainder when $N$ is divided by $25$ is $\boxed{\textbf{(R) } 17}$.

See Also

2007 iTest (Problems, Answer Key)
Preceded by:
Problem 23 		Followed by:
Problem 25
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