Question

Problem:
The number 12345678910111213....979899 consists of all the integers from 1 through 99 written back to back. That's a very big number! Is it a multiple of 9? Explain.


Help?

Answer 1

It's a multiple of 9 if the sum of all the digits is a multiple of 9. So, add all the digits: there are 20 of each (if you include 0). So, the sum is 20*(1+2+3+4+5+6+7+8+9) = 20*45, which is a multiple of 9. So, your number is a multiple of 9.

Answer 2

Ok. Thanks.

Answer 3

Can you help me with this?
What is the smallest multiple of 18 of the form 2A945B, where A and B are digits?

Answer 4

So, it has to be divisible by both 9 and 2. For 2, B has to be even, and for 9, 2+A+9+4+5+B = 20+A+B has to be divisible by 9. You want to minimize the number, so make A as small as possible. It can't be 0, since then B would have to be 7, so make it 1, and B is 6. The number is 219456.

Answer 5

Thanks! One more please?
What is the sum of all positive 1-digit numbers that 4221462 is divisible by?
and
What's the largest 5-digit number that is a multiple of both 4 and 9?

Answer 6

After that I'll be your fan.

Answer 7

@srossd?

Answer 8

Sorry! '

Answer 9

I'm back

Answer 10

It's fine. Just was hoping you didn't bail on me.

Answer 11

The sum of its digits is 21, so it's divisible by 3 but not 9. Its last digit is even, so it's divisible by 2. Therefore, it's divisible by 6 as well. Its last two digits (62) don't form a number divisible by 4, so it's not divisible by 4 (or 8). It's fastest to just do division to see that it's not divisible by 7. So, the sum is: 1+2+3+6 = 12.

If something is a multiple of 4 and 9, it's a mulitple of 36. The largest 5-digit number is 99999. Dividing it by 36, we get 2777.75. So, the greatest 5 digit number divisible by 36 is 36*2777 = 99972.

Answer 12

For the first one it says that is incorrect.

Answer 13

?

Answer 14

So sorry! I messed up dividing, it actually is divisible by 7. So the answer is 19.

Answer 15

Oh, it's perfectly alright. I really appreciate your help.

Answer 16

Thanks, I'm glad I could help.