Difference between revisions of "2003 AMC 12B Problems/Problem 18"
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<math>\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 33 \qquad \textbf{(E)}\ 34</math> | <math>\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 33 \qquad \textbf{(E)}\ 34</math> | ||
− | == Solution == | + | == Solution 1== |
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: | 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: | ||
Line 18: | Line 18: | ||
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 + 7 + 8 + 5 = \boxed{\textbf{(B)}\ 31}</math> | 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 + 7 + 8 + 5 = \boxed{\textbf{(B)}\ 31}</math> | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | A simpler way to tackle this problem without all that modding is to keep the equation as: | ||
+ | |||
+ | <cmath>7*a^{5b}c^{5d} = 11y^13</cmath> | ||
+ | |||
+ | As stated above, <math>a</math> and <math>c</math> must be the factors 7 and 11 in order to keep <math>x</math> at a minimum. Moving all the non-y terms to the left hand side of the equation, we end up with: | ||
+ | |||
+ | <cmath>7^{5b+1}11^{5d-1}=y^13</cmath> | ||
+ | |||
+ | The above equation means that <math>y</math> must also contain only the factors 7 and 11 (again, in order to keep <math>x</math> at a minimum), so we end up with: | ||
+ | |||
+ | <cmath>7^{5b+1}11^{5d-1}=7^{13h}11^{13j}</cmath> | ||
+ | |||
+ | (<math>h</math> and <math>j</math> are arbitrary variables placed in order to show that <math>y</math> could have more than just one 7 or one 11 as factors) | ||
+ | |||
+ | Since 7 and 11 are prime, we know that <math>5b+1=13h</math> and <math>5d-1=13j</math>. The smallest positive combinations that would work are <math>b=5,h=2</math> and <math>d=8,j=3</math>. Therefore, <math>a+b+c+d=7+5+11+8=31</math>. <math>\boxed{\textbf{(B)}\ 31}</math> is correct. | ||
== 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 17:37, 12 August 2018
Contents
Problem
Let and be positive integers such that The minimum possible value of has a prime factorization What is
Solution 1
Substitute into . We then have . Divide both sides by , and it follows that:
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 fifth root. Similarly, we want to look for the lowest power of such that . Again, this allows us to take the fifth root of . Obviously, we want to add to and subtract from because and are multiplied by and divided by , respectively. 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:
Taking the fifth root of both sides, we are left with .
Solution 2
A simpler way to tackle this problem without all that modding is to keep the equation as:
As stated above, and must be the factors 7 and 11 in order to keep at a minimum. Moving all the non-y terms to the left hand side of the equation, we end up with:
The above equation means that must also contain only the factors 7 and 11 (again, in order to keep at a minimum), so we end up with:
( and are arbitrary variables placed in order to show that could have more than just one 7 or one 11 as factors)
Since 7 and 11 are prime, we know that and . The smallest positive combinations that would work are and . Therefore, . is correct.
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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