Difference between revisions of "2016 AMC 12B Problems/Problem 24"
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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>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>\boxed{\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>\boxed{\textbf{(D)}}</math>. | ||
− | Also, to get the above formula of <math>f(k)=12 k^2+2</math>, we can also use the complementary counting by doing <math>(k+1)^4-k^4-k^4+(k-1)^4</math>, while the first term <math>(k+1)^4</math> 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. | + | Also, to get the above formula of <math>f(k)=12 k^2+2</math>, we can also use the complementary counting by doing <math>(k+1)^4-k^4-k^4+(k-1)^4</math>, while the first term <math>(k+1)^4</math> is for the four integers to independently have <math>k+1</math> choices each, with the second term indicating to subtract all the possibilities for the four integers to have values between <math>0</math> and <math>k-1</math>, and similarly the third term indicating to subtract all the possibilities for the four integers to have values between <math>1</math> and <math>k</math>, in the end the fourth term meaning the make up for the values between <math>1</math> and <math>k-1</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}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 20:55, 24 December 2020
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 .
Also, to get the above formula of , we can also use the complementary counting by doing , while the first term is for the four integers to independently have choices each, with the second term indicating to subtract all the possibilities for the four integers to have values between and , and similarly the third term indicating to subtract all the possibilities for the four integers to have values between and , in the end the fourth term meaning the make up for the values between and .
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 |
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