If ${x}$ is the product of all integer multiples of 5 from 1 to x, inclusive, what is the greatest prime number, y, such that $({36}+{41})/y$ is an integer?

Answer Choices

Correct answer marked below

5

11

19

23

41

Correct

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Comments

LeoMay 3

Wait, so {x} is the product of multiples of 5 from 1 to x? So {36} = 5*10*15*20*25*30*35? That's a lot of factors of 5. And {41} = 5*10*15*20*25*30*35*40? Then {36}+{41} is huge. How do you find the greatest prime factor? I'm lost.

SerenaMay 12

But careful: {36} itself has prime factors. The greatest prime among those could be larger than 41? Let's see: {36} = product of 5,10,...,35. The primes in there: 5, 2, 3, 7, 11, 13, 17, 19, 23, 29, 31. The largest prime is 31. So the sum has prime factors including 41 and 31. So the greatest is 41? But wait, the answer choices include 41. So maybe it's 41.

ArthurMay 14

This is tricky because you have to realize that {41} = {36} * 40. Then {36}+{41} = {36}*(1+40) = 41*{36}. Then the greatest prime factor of the sum is the greatest prime factor of 41*{36}. Since 41 is prime, and {36} contains primes up to 31, the greatest is 41. So answer E. But I initially thought it might be 31 because 31 is in {36}, but 41 is larger.

AkshayaMay 20

Also note that 40=5*8, so it adds more factors of 5 and 2, but no new primes beyond what's already in {36} except 5 and 2 which are already there. So the only new prime is 41 from the (1+40) factor.

HrishikeshAug 2

I got E. But I want to double-check: is there any prime factor in {36} that is greater than 41? The multiples of 5 up to 35: 5,10,15,20,25,30,35. Their prime factors: 5; 2,5; 3,5; 2,5; 5; 2,3,5; 5,7. So primes are 2,3,5,7. Wait, what about 10=2*5, 15=3*5, 20=2^2*5, 25=5^2, 30=2*3*5, 35=5*7. So actually {36} only has primes 2,3,5,7! I was wrong earlier. There's no 11,13, etc. because those are not factors of any multiple of 5 up to 35. So the greatest prime factor of {36} is 7. Then {36}+{41}=41*{36} has primes 41,7,5,3,2. So greatest is 41. So answer E. That makes it even simpler.

IsaacAug 14

So the answer is clearly E. The trap is thinking that numbers like 11,13, etc. are in {36}, but they are not multiples of 5. So good catch.

SofiaMay 23

So the answer is 41. But wait, the question asks for the greatest prime number y such that ({36}+{41})/y is an integer. That means y is a prime factor of the sum. Since the sum = 41*{36}, and {36} has prime factors 2,3,5,7, the prime factors of the sum are 2,3,5,7,41. The greatest is 41. So E. I think this is a 700-level question because you have to simplify the sum and then factor.

DanielMay 25

But also, you have to be careful not to assume that {36} contains large primes. It only contains primes that divide multiples of 5 up to 35, which are small.

What this question tests

What it tests

Your ability to evaluate and compose functions, including unusual defined operations.

Common trap

Applying function operations in the wrong order or ignoring domain restrictions.

If x{x}x is the product of all integer multiples of 5 from 1 to x, inclusive, what is the greatest prime number, y, such that (36+41)/y({36}+{41})/y(36+41)/y is an integer?