# Integer that divides into 2π to yield an integer

Here's a seemingly simple riddle (and it's a riddle and not just a word problem) I thought up two days ago (although it's simple enough that I can easily imagine someone else has thought of it before me):

I am a multiple of 7 of the digits between 1 and 9, and surprisingly enough, if you divide me into 2π, you get one of the two digits that I am not a multiple of.

What number am I?

The answer is 72. I came up with that number yesterday, reasoning that neither 6 nor 8 could be the "left out" digits since any number divisible by those numbers are also divisible by 2 and 3, or 2 and 4, respectively, which would leave at least 3 of the digits 1 - 9 out.

So including those numbers and their factors we must include: 1, 2, 3, 4, 6, 8.

It made the most sense to me to add in 9, as it is a multiple of 3, and exclude the two remaining primes, 5 and 7.

The least common multiple of 1, 2, 3, 4, 6, 8, 9 is 8 * 9 = 72.

What I did not realize until today's hint is that instead of dividing 2 pi by 72 and looking at the first few decimal digits of the answer, which gave .087266, we should convert to degrees and divide 360 by 72, which gives 5. This is indeed one of the digits we excluded from the list 1 through 9 and solves the riddle.

• Excellent answer, although technically 8 could have been a "left out" digit, as 4 is divisible by 2. – Ben Hocking Jan 1 '15 at 16:31
• Nice answer @ScottAllen, and nice puzzle too! Unfortunately, I read "a multiple" as "the product" :-( I'd still respectfully suggest though that "I am a multiple of 7 of the digits" could be more pedantically written as "I am a multiple of each of 7 of the digits" or I am divisible by 7 of the digits". – h34 Jan 1 '15 at 16:47
• Yeah my logic may be backwards on the inclusion of 6 and 8 and even 9, but 360 was chosen because it is divisible by so many of the numbers 1 - 9. But it turns out, since you want the quotient to not be one of the original numbers, that 5 is the only number you can exclude. It is the only prime in the list besides 7 whose square is not a multiple of 360. – Scott Allen Jan 1 '15 at 17:11
• 6 was good, because if it was divisible by 2 and 3, it had to be divisible by 6, and of course, if only 2 numbers are being left out of 1-9, then 2 and 3 both have to be in the list because in the former case of (2, 4, and 8) being in those digits, and in the latter case of (3, 6, and 9) being in those digits. – Ben Hocking Jan 1 '15 at 17:16

This is a work in progress. Take each of the (31 distinct) multiples of seven different digits between 1 and 9 inclusive and divide them into 360 (getting from $2\pi$ to 360 by converting from radians to degrees). Only a single quotient has a decimal fraction that contains, after any initial zeroes, only a single repeating digit. That's 0.0(5), the result of dividing 360 by 6480. Unfortunately, 6480 = $1\times 2\times 3\times 4\times 5\times 6\times 9$, the left out digits being 7 and 8, neither of which is 5.

• I think "divide into 360" is supposed to mean 360/X, not X/360. That combined with what you already have should give you the answer. – Lopsy Jan 1 '15 at 15:58
• 360/X is what I did. If you do X/360 (i.e. dividing by rather than into) where X is each of the 31 multiples, you get results ranging from 14 to 504, none of which contains only a single digit, repeating or otherwise. If you round to the nearest integer, a single multiple does, namely 8064, which gives 8064/360 = 22.4 ~ 22, but 8064 = $1 \times 2\times 3\times 4\times 6\times 7\times 8$, the two left out digits being 5 and 9, neither of which is 2. – h34 Jan 1 '15 at 16:14
• Ah I see. My bad. In that case, my hint is that just because it's a multiple of 7 of the digits, doesn't mean you get it by multiplying together 7 of the digits. – Lopsy Jan 1 '15 at 16:55