A number theory problem by Munem Shahriar

Number Theory Level 1

$1\times2\times3\times\cdots\times888$ is obviously divisible by 888.

But what is the smallest value of $N$ such that $1\times2\times3\times\cdots\times N$ is divisible by 888?

The answer is 37.

2 solutions

Stephen Mellor
Dec 16, 2017

Prime factorisation of $888 = 2^3 \times 3 \times 37$ . Therefore we must go up to at least 37 to get this factor. Using $N = 37$ , we cover the other required factors so this is the answer.

Your answer is incomplete.

If both the numbers 888 are replaced with 300, then would the answer be 5 or $5^2$ ?

Pi Han Goh - 3 years, 5 months ago

$300=2^2*5^2*3$ , so we need two numbers that are divisible by 5. 5 an 10, bingo. So all we need to do is to check if 300 divides 10!, which is true.

I still don't really get what you mean when you ask that, even though this solution is, of course, incomplete (of the "cover the other required factors" part)

Steven Jim - 3 years, 5 months ago

The way Stephen phrased it implies that the answer is simply the largest prime factor of 888, hence my rebuttal.

Pi Han Goh - 3 years, 5 months ago

For the use of the number 888, my answer is complete (although I could have been more explicit with the cover other factors part). As 37 is a prime factor, I showed that the lower bound for N is 37. I then said that if N = 37, we have covered the 2x2x2x3 in the numbers less than 37 (e.g. simply by the 24).

For 300, (as @Steven Jim said), we would do the prime factorisation, get the largest prime factor as the lower bound, however the lower bound is not possible. Therefore, we need 2 lots of the number 5, so 10 is the answer.

Stephen Mellor - 3 years, 5 months ago

Piero Sarti
Jan 2, 2018

Factoring out $888$ we get that $888 = 3 \times 8 \times 37$ .

Since $37$ is prime, it cannot be made by multiplying any other numbers in the multiplication up to $N$ . Therefore, $N = \boxed{37}$

A number theory problem by Munem Shahriar

The answer is 37.

2 solutions

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