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January 31, 2015

January 31, 2015

Posted by **Amy** on Friday, August 31, 2012 at 7:41pm.

You can only use the digits 0-9. I need an odd number that is a multiple of five with no repeated digits. Half the digits are odd and the other half is even. The largest digit in the number is in the tens place (but it is not the largest digit). The digit in the hundreds place is half the digit in the tens place. The sum of the digits is greater than 20.

- 5th Grade Math -
**David Q/R**, Saturday, September 1, 2012 at 6:27amThe number is odd, and divisible by 5 - so the final digit must be 5.

The largest digit is in the tens place, but isn't the largest digit, and no digits are repeated - so it must be 6, 7 or 8. The digit in the hundreds place is half the digit in the tens place, so the digit in the tens place must be even - so it must be 6 or 8, and the digit in the hundreds place must correspondingly be 3 or 4.

The number must have an even number of digits because there are the same number of odd digits as even ones. Suppose it's got exactly four digits. Call it ABCD. We know that D=5, so it's ABC5. Since the total of the digits is greater than 20, we're probably looking for large digits - so let's assume that the digit in the tens place is 8, and therefore that the digit in the hundreds place is 4. That would make our number A485. A must now be odd, but it can't be 9 (because that would make it the largest digit), so it must be 1, 3 or 7. That would make the answer 1485, 3485 or 7485. But only the third of these totals to more than 20, so the answer must be 7485.

It's just as well that you can only use the digits 0-9 to solve this problem, since there aren't any others :)

- 5th Grade Math -
**James**, Thursday, September 6, 2012 at 5:06pm20.007 in expanded form

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