The first and third digit of the five-digit number \( h6h41 \) are the same. Given that the number is exactly divisible by 9, what is the sum of its five digits?

If a number is divisible by 3, then the sum of its digits is divisible by 3 .
If ( h6h41 ) is divisible by 9, then both it and 1/3 of it must be divisible by 3.
If ( h6h41 ) is divisible by 3, then (2h + 6 + 4 + 1) = (2h + 11) is divisible by 3.
If 2h+11 = 12, h = 1/2 If 2h+11 = 15, h = 2 If 2h+11 = 18, h = 3-1/2 If 2h+11 = 21, h = 5 If 2h+11 = 24, h = 6-1/2 If 2h+11 = 27, h = 8 If 2h+11 = 30, h = 9-1/2
But 'h' is a digit, so it must be 2, 5, or 8 .
-- If h6h41 is 26,241 then 1/3 of it is 8,747 . . . not divisible by 3
-- If h6h41 is 56,541 then 1/3 of it is 18,847 . . . not divisible by 3
-- If h6h41 is 86,841 then 1/3 of it is 28,947 . . . divisible by 3
So ' h ' = 8, and the sum of the digits in h6h41 is 27 .

Answered by AL2006 | 2024-06-10

The five-digit number h 6 h 41 is divisible by 9 when h = 8 . The sum of its digits is 27. Therefore, the answer is 27.
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Answered by AL2006 | 2024-12-26