Difference between revisions of "2018 AIME I Problems/Problem 11"
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− | ==Solution== | + | ==Solution 2== |
Note that Euler's Totient Theorem would not necessarily lead to the smallest <math>n</math> and that in this case that <math>n</math> is greater than <math>1000</math>. | Note that Euler's Totient Theorem would not necessarily lead to the smallest <math>n</math> and that in this case that <math>n</math> is greater than <math>1000</math>. | ||
− | We wish to find the least <math>n</math> such that <math>3^n \equiv 1 | + | We wish to find the least <math>n</math> such that <math>3^n \equiv 1 \pmod{143^2}</math>. This factors as <math>143^2=11^{2}*13^{2}</math>. Because <math>gcd(121, 169) = 1</math>, we can simply find the least <math>n</math> such that <math>3^n \equiv 1 \pmod{121}</math> and <math>3^n \equiv 1 \pmod{169}</math>. |
− | Quick inspection yields <math>3^5 \equiv 1 | + | Quick inspection yields <math>3^5 \equiv 1 \pmod{121}</math> and <math>3^3 \equiv 1 \pmod{13}</math>. Now we must find the smallest <math>k</math> such that <math>3^{3k} \equiv 1 \pmod{13}</math>. Euler's gives <math>3^{156} \equiv 1 \pmod{169}</math>. So <math>3k</math> is a factor of <math>156</math>. This gives <math>k=1,2, 4, 13, 26, 52</math>. Some more inspection yields <math>k=13</math> is the smallest valid <math>k</math>. So <math>3^5 \equiv 1 \pmod{121}</math> and <math>3^{39} \equiv 1 \pmod{169}</math>. The least <math>n</math> satisfying both is <math>lcm(5, 39)=\boxed{195}</math>. (RegularHexagon) |
+ | |||
+ | ==Solution 3== | ||
+ | Listing out the powers of <math>3</math>, modulo <math>169</math> and modulo <math>121</math>, we have: | ||
+ | <cmath>\begin{array}{c|c|c} | ||
+ | n & 3^n\pmod{169} & 3^n\pmod{121}\\ \hline | ||
+ | 0 & 1 & 1\\ | ||
+ | 1 & 3 & 3\\ | ||
+ | 2 & 9 & 9\\ | ||
+ | 3 & 27 & 27\\ | ||
+ | 4 & 81 & 81\\ | ||
+ | 5 & 74 & 1\\ | ||
+ | 6 & 53\\ | ||
+ | 7 & 159\\ | ||
+ | 8 & 139\\ | ||
+ | 9 & 79\\ | ||
+ | 10 & 68\\ | ||
+ | 11 & 35\\ | ||
+ | 12 & 105\\ | ||
+ | 13 & 146\\ | ||
+ | 14 & 100\\ | ||
+ | 15 & 131\\ | ||
+ | 16 & 55\\ | ||
+ | 17 & 165\\ | ||
+ | 18 & 157\\ | ||
+ | 19 & 133\\ | ||
+ | 20 & 61\\ | ||
+ | 21 & 14\\ | ||
+ | 22 & 42\\ | ||
+ | 23 & 126\\ | ||
+ | 24 & 40\\ | ||
+ | 25 & 120\\ | ||
+ | 26 & 22\\ | ||
+ | 27 & 66\\ | ||
+ | 28 & 29\\ | ||
+ | 29 & 87\\ | ||
+ | 30 & 92\\ | ||
+ | 31 & 107\\ | ||
+ | 32 & 152\\ | ||
+ | 33 & 118\\ | ||
+ | 34 & 16\\ | ||
+ | 35 & 48\\ | ||
+ | 36 & 144\\ | ||
+ | 37 & 94\\ | ||
+ | 38 & 113\\ | ||
+ | 39 & 1\\ | ||
+ | \end{array}</cmath> | ||
+ | |||
+ | The powers of <math>3</math> repeat in cycles of <math>5</math> an <math>39</math> in modulo <math>121</math> and modulo <math>169</math>, respectively. The answer is <math>lcm(5, 39) = \boxed{195}</math> | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2018|n=I|num-b=10|num-a=12}} | {{AIME box|year=2018|n=I|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 18:04, 26 March 2018
Find the least positive integer such that when is written in base , its two right-most digits in base are .
Contents
Solutions
Modular Arithmetic Solution- Strange (MASS)
Note that and . Because , the desired condition is equivalent to 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 . It is not difficult to see that the smallest , so ultimately . Therefore, .
The first satisfying both criteria is thus .
-expiLnCalc
Solution 2
Note that Euler's Totient Theorem would not necessarily lead to the smallest and that in this case that is greater than .
We wish to find the least such that . This factors as . Because , we can simply find the least such that and .
Quick inspection yields and . Now we must find the smallest such that . Euler's gives . So is a factor of . This gives . Some more inspection yields is the smallest valid . So and . The least satisfying both is . (RegularHexagon)
Solution 3
Listing out the powers of , modulo and modulo , we have:
The powers of repeat in cycles of an in modulo and modulo , respectively. The answer is
See Also
2018 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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