Difference between revisions of "2017 AIME I Problems/Problem 7"
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+ | ==Solution 2 (Major Bash)== | ||
+ | Case 1: <math>a<b</math>. | ||
+ | |||
+ | Subcase 1: <math>a=0</math> | ||
+ | <cmath>\binom{6}{0}\binom{6}{1}\binom{6}{1}=36</cmath> | ||
+ | <cmath>\binom{6}{0}\binom{6}{2}\binom{6}{2}=225</cmath> | ||
+ | <cmath>\binom{6}{0}\binom{6}{3}\binom{6}{3}=400</cmath> | ||
+ | <cmath>\binom{6}{0}\binom{6}{4}\binom{6}{4}=225</cmath> | ||
+ | <cmath>\binom{6}{0}\binom{6}{5}\binom{6}{5}=36</cmath> | ||
+ | <cmath>\binom{6}{0}\binom{6}{6}\binom{6}{6}=1</cmath> | ||
+ | <cmath>36+225+400+225+36+1=923</cmath> | ||
+ | Subcase 2: <math>a=1</math> | ||
+ | <cmath>\binom{6}{1}\binom{6}{2}\binom{6}{3}=1800 \equiv 800 \pmod {1000}</cmath> | ||
+ | <cmath>\binom{6}{1}\binom{6}{3}\binom{6}{4}=1800 \equiv 800 \pmod {1000}</cmath> | ||
+ | <cmath>\binom{6}{1}\binom{6}{4}\binom{6}{5}=540</cmath> | ||
+ | <cmath>\binom{6}{1}\binom{6}{5}\binom{6}{6}=36</cmath> | ||
+ | <cmath>800+800+540+36=2176 \equiv 176 \pmod {1000}</cmath> | ||
+ | Subcase 3: <math>a=2</math> | ||
+ | <cmath>\binom{6}{2}\binom{6}{3}\binom{6}{5}=1800\equiv800\pmod{1000}</cmath> | ||
+ | <cmath>\binom{6}{2}\binom{6}{4}\binom{6}{6}=225</cmath> | ||
+ | <cmath>800+225=1025\equiv25\pmod{1000}</cmath> | ||
+ | |||
+ | |||
+ | <cmath>923+176+25=1124\equiv124\pmod{1000}</cmath> | ||
+ | |||
+ | Case 2: <math>b<a</math> | ||
+ | |||
+ | By just switching <math>a</math> and <math>b</math> in all of the above cases, we will get all of the cases such that <math>b>a</math> is true. Therefore, this case is also <math>124\pmod{1000}</math> | ||
+ | |||
+ | Case 3: <math>a=b</math> | ||
+ | <cmath>\binom{6}{0}\binom{6}{0}\binom{6}{0}=1</cmath> | ||
+ | <cmath>\binom{6}{1}\binom{6}{1}\binom{6}{2}=540</cmath> | ||
+ | <cmath>\binom{6}{2}\binom{6}{2}\binom{6}{4}=3375\equiv375\pmod{1000}</cmath> | ||
+ | <cmath>\binom{6}{3}\binom{6}{3}\binom{6}{6}=400</cmath> | ||
+ | <cmath>1+540+375+400=1316\equiv316\pmod{1000}</cmath> | ||
+ | |||
+ | |||
+ | <cmath>316+124+124=\boxed{564}</cmath> | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2017|n=I|num-b=6|num-a=8}} | {{AIME box|year=2017|n=I|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 21:19, 9 March 2017
Contents
[hide]Problem 7
For nonnegative integers and with , let . Let denote the sum of all , where and are nonnegative integers with . Find the remainder when is divided by .
Solution
Let , and note that . The problem thus asks for the sum over all such that . Consider an array of 18 dots, with 3 columns of 6 dots each. The desired expression counts the total number of ways to select 6 dots by considering each column separately. However, this must be equal to . Therefore, the answer is .
-rocketscience
Solution 2 (Major Bash)
Case 1: .
Subcase 1: Subcase 2: Subcase 3:
Case 2:
By just switching and in all of the above cases, we will get all of the cases such that is true. Therefore, this case is also
Case 3:
See Also
2017 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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