What is the only number that is equal to twice the sum of it's digits?

Answer 1

Well 18 works. [1+8 = 9]. It's not exactly the only one, though. Consider zero. The sum of the digits (0) is zero. And twice zero = zero.

Here are some ideas to see if there are others: For 1-digit numbers, the number is the sum, so the zero, above, is the only one that satisfies the conditions. For 2-digit numbers, the largest sum possible is 18 [99: 9 + 9]. Twice that sum is 36, so no numbers greater than 36 (with 2-digit numbers).

Now consider the unknown number with digits AB. So the sum = A + B, and the number = 10*A + B. Now the number is twice the sum, so:

10*A + B = 2*(A + B). Rearranging, we have 8*A = B, so you can substitute whole numbers for A, and calculate B. There is the trivial case of zero: 8*0 = 0, which was already covered. If A = 1, then B = 8, which gives us 18. With A = 2, then B= 16, so that's not a 2-digit number. So 18 is the only 2-digit number.

What about 3-digits: the largest 3-digit sum is 9+9+9=27. Twice that is 54, which is a 2-digit number. So there can be no 3-digit numbers, satisfying the conditions.