Revision as of 13:38, 12 August 2020

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 $A=\frac{a}{77},\ B=\frac{b}{77}$ , etc., so that $\gcd(A,B,C,D)=1$ . Then for each prime power $p^k$ in the prime factorization of $N=\frac{n}{77}$ , 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 $\boxed{\textbf{(D)}}$ .

Also, to get the above formula of $f(k)=12 k^2+2$ , we can also use the complementary counting by doing $(k+1)^4-k^4-k^4+(k-1)^4$ , while the first term $(k+1)^4$ is for the four integers to independently have k+1 choices each, with the second term indicating to subtract all the possibilities for the four integers to have values between 0 and k-1, and similarly the third term indicating to subtract all the possibilities for the four integers to have values between 1 and k, in the end the fourth term meaning the make up for the values between 1 and k-1.

@@ Line 5: / Line 5: @@
 =Solution=
-Let <math>A=\frac{a}{77},\ B=\frac{b}{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>.
+Let <math>A=\frac{a}{77},\ B=\frac{b}{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=\frac{n}{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>.
 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.

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

Page

Toolbox

Search

Difference between revisions of "2016 AMC 12B Problems/Problem 24"

Revision as of 13:38, 12 August 2020

Problem

Solution

See Also