Difference between revisions of "2018 AIME I Problems/Problem 12"
(Adding solution) |
|||
| Line 56: | Line 56: | ||
P=1362/2^12=683/2^11 | P=1362/2^12=683/2^11 | ||
ANS=683 | ANS=683 | ||
| + | |||
| + | |||
| + | Solution 2: Consider the numbers {1,4,7,10,13,16}. Each of those are congruent to 1mod3. There is <math>{6 \choose 0}=1</math> way to choose zero numbers <math>{6 \choose 1}=6</math> ways to choose 1 and so on. There ends up being <math>{6 \choose 0}+{6 \choose 3}+{6 \choose 6}</math> possible subsets congruent to 0mod 3. There are <math>2^6=64</math> possible subsets of these numbers. | ||
Revision as of 20:14, 7 March 2018
Rewrite the set after mod3 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 All 0s can be omitted Case 1 No 1 No 2 1 Case 2 222 20 Case 3 222222 1 Case 4 12 6*6=36 Case 5 12222 6*15=90 Case 6 1122 15*15=225 Case 7 1122222 15*6=90 Case 8 111 20 Case 9 111222 20*20=400 Case 10 111222222 20 Case 11 11112 15*6=90 Case 12 11112222 15*15=225 Case 13 1111122 6*15=90 Case 14 1111122222 6*6=36 Case 15 111111 1 Case 16 111111222 20 Case 17 111111222222 1 Total 1+20+1+36+90+225+90+20+400+20+90+225+90+36+1+20+1=484+360+450+72=1366 P=1362/2^12=683/2^11 ANS=683
Solution 2: Consider the numbers {1,4,7,10,13,16}. Each of those are congruent to 1mod3. There is 
way to choose zero numbers 
ways to choose 1 and so on. There ends up being 
possible subsets congruent to 0mod 3. There are 
possible subsets of these numbers.
