2007 iTest Problems/Problem TB1
Problem
The sum of the digits of an integer is equal to the sum of the digits of three times that integer. Prove that the integer is a multiple of 9.
Solution
A number is divisible by 3 or 9 if the sum of its digits is divisible by 3 or 9, respectively. A proof can be found here. Since the sum of the digits of 3 times the integer is a multiple of 3, the sum of the digits of the original integer is a multiple of 3, and hence the original integer is a multiple of 3. The sum of the digits of 3 times the integer is a multiple of 9. Hence the sum of the digits of the number itself is a multiple of 9. Hence the integer itself is a multiple of 9.
See also
2007 iTest (Problems) | ||
Preceded by: Problem 60 |
Followed by: Problem TB2 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 • 51 • 52 • 53 • 54 • 55 • 56 • 57 • 58 • 59 • 60 • TB1 • TB2 • TB3 • TB4 |