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The first and third digit of the five digit number h6h41 are the same. Given that the number is exacley divisible by 9, what is it sum of it's five digit number? Please help !!☺️



Answer :

AL2006
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 .

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