2016 AMC 12B Problems/Problem 24

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There are exactly $77,000$ ordered quadruplets $(a, b, c, d)$ such that $\gcd(a, b, c, d) = 77$ and $\operatorname{lcm}(a, b, c, d) = n$. What is the smallest possible value for $n$?

$\textbf{(A)}\ 13,860\qquad\textbf{(B)}\ 20,790\qquad\textbf{(C)}\ 21,560 \qquad\textbf{(D)}\ 27,720 \qquad\textbf{(E)}\ 41,580$


Let $A=a\div 77,\ B=b\div 77$, etc., so that $\gcd(A,B,C,D)=1$. Then for each prime power $p^k$ in the prime factorization of $N=n\div 77$, at least one of the prime factorizations of $(A,B,C,D)$ has $p^k$, at least one has $p^0$, and all must have $p^m$ with $0\le m\le k$.

Let $f(k)$ be the number of ordered quadruplets of integers $(m_1,m_2,m_3,m_4)$ such that $0\le m_i\le k$ for all $i$, the largest is $k$, and the smallest is $0$. Then for the prime factorization $N=2^{k_2}3^{k_3}5^{k_5}\ldots$ we must have $77000=f(k_2)f(k_3)f(k_5)\ldots$ So let's take a look at the function $f(k)$ by counting the quadruplets we just mentioned..

There are $14$ quadruplets which consist only of $0$ and $k$. Then there are $24(k-1)$ quadruplets which include three different values, and $12(k-1)^2$ with four. Thus $f(k)=14+12(2k-2+(k-1)^2)=14+12(k^2-1)$ and the first few values from $k=1$ onwards are \[14,50,110,194,302,434,590,770,\ldots\] Straight away we notice that $14\cdot 50\cdot 110=77000$, so the prime factorization of $N$ can use the exponents $1,2,3$. To make it as small as possible, assign the larger exponents to smaller primes. The result is $N=2^33^25^1=360$, so $n=360\cdot 77=27720$ which is answer $\textbf{(D)}$.

See Also

2016 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
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All AMC 12 Problems and Solutions
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