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 (Simpler, just basic mods and Fermat's theorem)
BY THE WAY, please feel free to correct my formatting. I don't know latex.
Note that for all , is divisible by because that is a factor. That is , so now we can clearly see that the smallest to make the expression divisible by is just . Similarly, we can reason that the smallest to make the expression divisible by is just .
Finally, for , take and of each quantity (They happen to both be and respectively, so you only need to compute once). One knows from Fermat's theorem that the maximum possible minimum for divisibility by is , and other values are factors of . Testing all of them(just ,, using mods-not too bad), is indeed the smallest value to make the expression divisible by , and this clearly is NOT divisible by . Therefore, the smallest to make this expression divisible by is .
Calculating the LCM of all these, one gets . Using the factor counting formula, the answer is = . Can someone please format better for me?
-Solution by thanosaops -formatted by MY-2
See Also
2020 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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