Revision as of 22:32, 22 December 2016

Problem

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 $n$ ?

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

Solution

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 $36(k-1)$ quadruplets which include three different values, and $12(k-1)(k-2)$ with four. Thus $f(k)=14+12(k-1)(3+k-2)=14+12(k^2-1)$ 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 $\textbf{(D)}$ .

@@ Line 9: / Line 9: @@
 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..
-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>.
+There are <math>14</math> quadruplets which consist only of <math>0</math> and <math>k</math>. Then there are <math>36(k-1)</math> quadruplets which include three different values, and <math>12(k-1)(k-2)</math> with four. Thus <math>f(k)=14+12(k-1)(3+k-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==
 {{AMC12 box|year=2016|ab=B|num-a=25|num-b=23}}

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