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>\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>. | ||
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+ | 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. | ||
==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 13:30, 6 July 2018
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 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.
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 |