2020 AIME I Problems/Problem 12
Contents
Problem
Let be the least positive integer for which is divisible by Find the number of positive integer divisors of
Solution 1
Lifting the Exponent shows that so thus, divides . It also shows that so thus, divides .
Now, multiplying by , we see and since and then meaning that we have that by LTE, divides .
Since , and all divide , the smallest value of working is their LCM, also . Thus the number of divisors is .
~kevinmathz
Solution 2 (For people who don't know like crazy stuff)
BY THE WAY, please feel free to correct my formatting. I don't know latex.
Note that for all n, 149^n - 2^n is divisible by 149-2 = 147 because that is a factor. That is 3*7^2, so now we can clearly see that the smallest n to make the expression divisible by 3^3 is just 3^2. Similarly, we can reason that the smallest n to make the expression divisible by 7^7 is just 7^5.
Finally, for 5^5, take mod (5) and mod (25) of each quantity (They happen to both be -1 and 2 respectively, so you only need to compute once). One knows from Fermat's theorem that the maximum possible minimum n for divisibility by 5 is 4, and other values are factors of 4. Testing all of them(just 1,2,4 using mods-not too bad), 4 is indeed the smallest value to make the expression divisible by 5, and this clearly is NOT divisible by 25. Therefore, the smallest n to make this expression divisible by 5^5 is 2^2 * 5^4.
Calculating the LCM of all these, one gets 2^2 * 3^2 * 5^4 * 7^5. Using the factor counting formula, the answer is 3*3*5*6 = . Can someone please format better for me?
-Solution by thanosaops
See Also
2020 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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All AIME Problems and Solutions |
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