# 2007 iTest Problems/Problem 24

## 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) Preceded by:Problem 23 Followed by:Problem 25 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 • 51 • 52 • 53 • 54 • 55 • 56 • 57 • 58 • 59 • 60 • TB1 • TB2 • TB3 • TB4
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