What is the largest value of x for which $12^x$ leaves zero remainder when it divides $\frac{50!}{10!}$ ?

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Comments

ConqAug 25

Man, that exponent on x had me rereading the stem three times. Hard 700+ vibes for sure.

HassibAug 30

Same, I literally wrote 12x at first and was like wait that's way too easy for a 705+.

YugeshSep 21

I got 19 but I'm second-guessing myself. Did anyone else make a table for 2s and 3s separately? That felt like the only way to keep it straight.

MarcosSep 27

Yes, separate counting is the move. Don't try to count 12s directly.

JhaaduSep 29

I did the same but then realized I had to compare the two counts, not just add them.

ChinmayMay 7

Is there a trick here with the 10! denominator that I'm missing? I keep getting confused about whether to subtract or what.

GarvitaMay 12

Think about what dividing by 10! does to the factors you counted in 50!.

IyaanApr 18

Respect to whoever writes these. Every choice looks plausible if you mess up one step.

What this question tests

Your grasp of exponent rules — multiplying and dividing powers, power-of-a-power, and negative/zero exponents.

Adding exponents when multiplying bases that have the same exponent, or confusing (x^a)^b with x^(a·b).

What is the largest value of x for which 12x12^x12x leaves zero remainder when it divides 50!10!\frac{50!}{10!}10!50!​?