For what value of N is the 5 digit # 7n,933 divisible by 33? (hah i messed up my question again!)?

Question

exxplain

Answer 1

5-five.

75933 / 33 = 2301

Answer 2

Oh, come on. There are only ten numbers to try. Surely you could have done that on a calculator yourself quicker than asking on Yahoo! and waiting for an answer.

But for division by 3, 7+n+9+3+3 must be a multiple of 3, so n must be 2, 5 or 8. For division by 11, (7+9+3) - (n+3) must be divisible by 11, so n must be 5. It's a good job the answer for 11 is one of the answers for 3, otherwise there wouldn't have been an answer.

By the way, (7+9+3) is the sum of alternate digits and (n+3) is the sum of the other alternate digits. It's a pity there are two 3's in the question number because it makes it less obvious what I was doing.

Answer 3

First, there is a thing called adding digits where you add up the digits in a number and if it comes out over 9 then you add those digits... over and over until you get a number less than 10...

if you add up all the digits of a number and they add up to 3, 6, or 9 then the number is divisible by 3... examples: 12, 15, 18, 1233, etc.

So, your number 7n933 has values for n for which it will be divisible by 3...

7+9+3+3 = 2+2 = 4... what do you need to add to this to make it add up to 3, 6, or 9?

now, which of those numbers would make it also divisible by 11?