What is the largest exponent r such that the product of all multiples of 15 between 171 and 286, denoted as X, is divisible by ?
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This one took me a while. I knew I needed to count factors of 2 and 5, but the range 171 to 286 is tricky. Are the endpoints included? I assumed not, but want to be sure.
Usually 'between' excludes endpoints, but for multiples of 15 it might not matter since 171 and 286 aren't multiples of 15.
I think it's safe to assume exclusive. But even if inclusive, no multiples of 15 at the endpoints.
I got 6, but I'm not 100% sure. Can someone confirm? I counted 8 multiples of 15: 180, 195, 210, 225, 240, 255, 270, 285. Then factored out 5s and 2s. Seemed to work.
Yeah, I got 6 too. The 225 gives two 5s and 240 gives extra 2s. But I'm not sure if I missed any.
This is a classic trap. I initially thought the answer was 8 because there are 8 multiples, but that's wrong. You need to find the limiting factor, which is 5. So count the 5s carefully.
Why is the answer not 4? I must be missing something. I counted only four 5s? Let me re-evaluate.
Check 225: it's 15*15, so it has two 5s. And 210, 240, 270 each have one 5 from the multiple of 5 and one from the 3? Actually, each multiple of 15 has at least one 5. But some have more.
Also, 180, 195, 210, 225, 240, 255, 270, 285: all have at least one 5. But 225 has two, and 240, 270, etc. might have extra 2s but not extra 5s. So total 5s = 8 + 1 = 9? Wait, then r could be 9? But 2s are fewer? I'm confused.
What it tests
Your grasp of exponent rules — multiplying and dividing powers, power-of-a-power, and negative/zero exponents.
Common trap
Adding exponents when multiplying bases that have the same exponent, or confusing (x^a)^b with x^(a·b).