Difference between revisions of "2016 AMC 12B Problems/Problem 24"
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=Problem= | =Problem= | ||
− | There are exactly <math>77,000</math> ordered quadruplets <math>(a, b, c, d)</math> such that <math> | + | There are exactly <math>77,000</math> ordered quadruplets <math>(a, b, c, d)</math> such that <math>\gcd(a, b, c, d) = 77</math> and <math>\operatorname{lcm}(a, b, c, d) = n</math>. What is the smallest possible value for <math>n</math>? |
− | <math>\textbf{(A)}\ 13,860\qquad\textbf{(B)}\ 20,790\qquad\textbf{(C)}\ 21,560 \qquad\textbf{(D)} 27,720 \qquad\textbf{(E)}\ 41,580</math> | + | <math>\textbf{(A)}\ 13,860\qquad\textbf{(B)}\ 20,790\qquad\textbf{(C)}\ 21,560 \qquad\textbf{(D)}\ 27,720 \qquad\textbf{(E)}\ 41,580</math> |
=Solution= | =Solution= | ||
− | {{ | + | Let <math>A=a\div 77,\ B=b\div 77</math>, etc., so that <math>\gcd(A,B,C,D)=1</math>. Then for each prime power <math>p^k</math> in the prime factorization of <math>N=n\div 77</math>, at least one of the prime factorizations of <math>(A,B,C,D)</math> has <math>p^k</math>, at least one has <math>p^0</math>, and all must have <math>p^m</math> with <math>0\le m\le k</math>. |
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+ | Let <math>f(k)</math> be the number of ordered quadruplets of integers <math>(m_1,m_2,m_3,m_4)</math> such that <math>0\le m_i\le k</math> for all <math>i</math>, the largest is <math>k</math>, and the smallest is <math>0</math>. Then for the prime factorization <math>N=2^{k_2}3^{k_3}5^{k_5}\ldots</math> we must have <math>77000=f(k_2)f(k_3)f(k_5)\ldots</math> So let's take a look at the function <math>f(k)</math> by counting the quadruplets we just mentioned.. | ||
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+ | There are <math>14</math> quadruplets which consist only of <math>0</math> and <math>k</math>. Then there are <math>24(k-1)</math> quadruplets which include three different values, and <math>12(k-1)^2</math> with four. Thus <math>f(k)=14+12(2k-2+(k-1)^2)=14+12(k^2-1)</math> and the first few values from <math>k=1</math> onwards are <cmath>14,50,110,194,302,434,590,770,\ldots</cmath> Straight away we notice that <math>14\cdot 50\cdot 110=77000</math>, so the prime factorization of <math>N</math> can use the exponents <math>1,2,3</math>. To make it as small as possible, assign the larger exponents to smaller primes. The result is <math>N=2^33^25^1=360</math>, so <math>n=360\cdot 77=27720</math> which is answer <math>\textbf{(D)}</math>. | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2016|ab=B|num-a=25|num-b=23}} | {{AMC12 box|year=2016|ab=B|num-a=25|num-b=23}} |
Revision as of 09:19, 22 February 2016
Problem
There are exactly ordered quadruplets such that and . What is the smallest possible value for ?
Solution
Let , etc., so that . Then for each prime power in the prime factorization of , at least one of the prime factorizations of has , at least one has , and all must have with .
Let be the number of ordered quadruplets of integers such that for all , the largest is , and the smallest is . Then for the prime factorization we must have So let's take a look at the function by counting the quadruplets we just mentioned..
There are quadruplets which consist only of and . Then there are quadruplets which include three different values, and with four. Thus and the first few values from onwards are Straight away we notice that , so the prime factorization of can use the exponents . To make it as small as possible, assign the larger exponents to smaller primes. The result is , so which is answer .
See Also
2016 AMC 12B (Problems • Answer Key • Resources) | |
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 | |
All AMC 12 Problems and Solutions |