Difference between revisions of "2007 iTest Problems/Problem 24"
m (Created page with "== Problem == Let <math>N</math> be the smallest positive integer such that <math>2008N</math> is a perfect square and <math>2007N</math> is a perfect cube. Find the remainder w...") |
Rockmanex3 (talk | contribs) (Solution to Problem 24) |
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− | == Problem == | + | ==Problem== |
Let <math>N</math> be the smallest positive integer such that <math>2008N</math> is a perfect square and <math>2007N</math> is a perfect cube. | Let <math>N</math> be the smallest positive integer such that <math>2008N</math> is a perfect square and <math>2007N</math> is a perfect cube. | ||
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\text{(G) }6 \quad | \text{(G) }6 \quad | ||
\text{(H) }7 \quad | \text{(H) }7 \quad | ||
− | \text{(I) } 8\quad | + | \text{(I) } 8\quad</math> |
<math>\text{(J) }9 \quad | <math>\text{(J) }9 \quad | ||
Line 21: | Line 21: | ||
\text{(O) }14 \quad | \text{(O) }14 \quad | ||
\text{(P) }15 \quad | \text{(P) }15 \quad | ||
− | \text{(Q) }16 \quad | + | \text{(Q) }16 \quad</math> |
<math>\text{(R) }17 \quad | <math>\text{(R) }17 \quad | ||
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\text{(X) }23 </math> | \text{(X) }23 </math> | ||
− | == Solution == | + | ==Solution== |
+ | |||
+ | The prime factorization of <math>2008</math> is <math>2^3 \cdot 251</math>, and the [[prime factorization]] of <math>2007</math> is <math>3^2 \cdot 223</math>. Since <math>2008N</math> is a perfect square and <math>2007N</math> is a perfect cube, the exponents in the prime factorization of <math>2008N</math> are even, and the exponents in the prime factorization of <math>2007N</math> are a multiple of three. | ||
+ | |||
+ | To make the exponents of <math>2008N</math> even, the exponent of <math>2</math> in <math>N</math> is at least <math>1</math>, but since there are no powers of <math>2</math> in <math>2007</math>, the exponent of <math>2</math> in <math>N</math> is at least <math>3</math>. Similarly, in <math>N</math>, the exponent of <math>251</math> is at least <math>3</math>, the exponent of <math>3</math> is at least <math>4</math>, and the exponent of <math>223</math> is at least <math>2</math>. | ||
+ | |||
+ | Thus, <math>N</math>, the minimum positive integer that satisfies the criteria, equals <math>2^3 \cdot 3^4 \cdot 251^3 \cdot 223^2</math>. Using [[modular arithmetic]] to find the remainder, | ||
+ | <cmath>N \equiv 8 \cdot 6 \cdot 1^3 \cdot (-2)^2 \pmod{25}</cmath> | ||
+ | <cmath>N \equiv 192 \pmod{25}</cmath> | ||
+ | <cmath>N \equiv 17 \pmod{25}</cmath> | ||
+ | The remainder when <math>N</math> is divided by <math>25</math> is <math>\boxed{\textbf{(R) } 17}</math>. | ||
+ | |||
+ | ==See Also== | ||
+ | {{iTest box|year=2007|num-b=23|num-a=25}} | ||
+ | |||
+ | [[Category:Intermediate Number Theory Problems]] |
Latest revision as of 18:49, 30 June 2018
Problem
Let be the smallest positive integer such that is a perfect square and is a perfect cube. Find the remainder when is divided by .
Solution
The prime factorization of is , and the prime factorization of is . Since is a perfect square and is a perfect cube, the exponents in the prime factorization of are even, and the exponents in the prime factorization of are a multiple of three.
To make the exponents of even, the exponent of in is at least , but since there are no powers of in , the exponent of in is at least . Similarly, in , the exponent of is at least , the exponent of is at least , and the exponent of is at least .
Thus, , the minimum positive integer that satisfies the criteria, equals . Using modular arithmetic to find the remainder, The remainder when is divided by is .
See Also
2007 iTest (Problems, Answer Key) | ||
Preceded by: Problem 23 |
Followed by: Problem 25 | |
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