Difference between revisions of "2018 AIME I Problems/Problem 11"
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If <math>3^n \cong 1 (\mod 121)</math>, one can see the sequence <math>1, 3, 9, 27, 81, 1, 3, 9...</math> so <math>n \mid 5</math>. | If <math>3^n \cong 1 (\mod 121)</math>, one can see the sequence <math>1, 3, 9, 27, 81, 1, 3, 9...</math> so <math>n \mid 5</math>. | ||
− | Now if <math>3^n \cong 1 (\mod 169)</math>, it is harder. But we do observe that <math>3^3 \cong 1 (\mod 13)</math>, therefore <math>3^3 = 13a + 1</math> for some integer <math>a</math>. So our goal is to find the first number <math>p_1</math> such that <math>(13a+1)^ | + | Now if <math>3^n \cong 1 (\mod 169)</math>, it is harder. But we do observe that <math>3^3 \cong 1 (\mod 13)</math>, therefore <math>3^3 = 13a + 1</math> for some integer <math>a</math>. So our goal is to find the first number <math>p_1</math> such that <math>(13a+1)^ {p_1} \cong 1 (\mod 169)</math>. In other words, the <math>a</math> coefficient must be <math>0 (\mod 169)</math>. It is not difficult to see that this first <math>p_1=13</math>, so ultimately <math>3^39 \cong 1 (\mod 169)</math>. Therefore, <math>n \mid 39</math>. |
The first <math>n</math> satisfying both criteria is <math>5*39=\boxed{195}</math>. | The first <math>n</math> satisfying both criteria is <math>5*39=\boxed{195}</math>. | ||
-expiLnCalc | -expiLnCalc |
Revision as of 17:44, 7 March 2018
Find the least positive integer such that when is written in base , its two right-most digits in base are .
Solutions
Modular Arithmetic Solution- Strange (MASS)
Note that . And . Because , and .
If , one can see the sequence so .
Now if , it is harder. But we do observe that , therefore for some integer . So our goal is to find the first number such that . In other words, the coefficient must be . It is not difficult to see that this first , so ultimately . Therefore, .
The first satisfying both criteria is .
-expiLnCalc