Difference between revisions of "2003 AMC 12B Problems/Problem 18"
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== Solution == | == Solution == | ||
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+ | Substitute <math>a^cb^d</math> into <math>x</math>. We then have <math>7(a^{5c}b^{5d}) = 11y^{13}</math>. Divide both sides by <math>7</math>, and it follows that: | ||
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+ | <cmath>(a^{5c}b^{5d}) = \[\frac{11y^{13}}{7}.</cmath> | ||
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+ | Note that because <math>11</math> and <math>7</math> are prime, the minimum value of <math>x</math> must involve factors of <math>7</math> and <math>11</math> only. Thus, we try to look for the lowest power <math>p</math> of <math>11</math> such that <math>13p + 1 \equiv 0 \pmod{5}</math>, so that we can take <math>11^{13p + 1}</math> to the 5th root. Similarly, we want to look for the lowest power <math>n</math> of <math>7</math> such that <math>13n - 1 \equiv 0 \pmod{5}</math>. Again, this allows us to take the fifth root of <math>7^{13n - 1}</math>. With these conditions satisfied, we can simply multiply <math>11^{13p + 1}</math> and <math>7^{13n - 1}</math> and substitute this quantity into <math>y</math> to attain our answer. | ||
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+ | We can simply look for suitable values for <math>p</math> and <math>n</math>. We find that the lowest <math>p</math>, in this case, would be <math>3</math> because <math>13(3) + 1 \equiv 0 \pmod{5}</math>. Moreover, the lowest <math>q</math> should be <math>2</math> because <math>13(2) - 1 \equiv 0 \pmod{5}</math>. Hence, we can substitute the quantity <math>11^{3} \cdot 7^{2}</math> into <math>y</math>. Doing so gets us: | ||
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+ | <cmath>(a^{5c}b^{5d}) = \[\frac{11(11^{3} \cdot 7^{2})^{13}}{7} = 11^{40} \cdot 7^{25}.</cmath> | ||
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+ | Taking the fifth root of both sides, we are left with <math>a^cb^d = 11^{8} \cdot 7^{5}</math>. <math>a + b + c + d = 11 + 8 + 7 + 5 = \boxed{\textbf{(B)}\ 31}</math> | ||
== See Also== | == See Also== | ||
{{AMC12 box|year=2003|ab=B|num-b=17|num-a=19}} | {{AMC12 box|year=2003|ab=B|num-b=17|num-a=19}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 04:44, 9 June 2014
Problem
Let and be positive integers such that The minimum possible value of has a prime factorization What is
Solution
Substitute into . We then have . Divide both sides by , and it follows that:
(a^{5c}b^{5d}) = \[\frac{11y^{13}}{7}. (Error compiling LaTeX. Unknown error_msg)
Note that because and are prime, the minimum value of must involve factors of and only. Thus, we try to look for the lowest power of such that , so that we can take to the 5th root. Similarly, we want to look for the lowest power of such that . Again, this allows us to take the fifth root of . With these conditions satisfied, we can simply multiply and and substitute this quantity into to attain our answer.
We can simply look for suitable values for and . We find that the lowest , in this case, would be because . Moreover, the lowest should be because . Hence, we can substitute the quantity into . Doing so gets us:
(a^{5c}b^{5d}) = \[\frac{11(11^{3} \cdot 7^{2})^{13}}{7} = 11^{40} \cdot 7^{25}. (Error compiling LaTeX. Unknown error_msg)
Taking the fifth root of both sides, we are left with .
See Also
2003 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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 | |
All AMC 12 Problems and Solutions |
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