<PRE><B>Question 168568</B>
let a = hundreds digit
let b = tens digit
let c = ones digit
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number is divisible by ten so c must = 0.
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hundreds digit is one less than tens digit so a = b-1
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let x = number
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number is 52 times the sum of its digits so x = 52 * (a + b + c)
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since x = the number, then
x = 100*a + 10*b + 1*c
since c = 0, and a = b-1, this becomes:
x = 100*(b-1) + 10*b
which becomes:
x = 100*b - 100 + 10*b
which becomes:
x = 110*b - 100
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since x = 52 * (a + b + c)
and c = 0, and a = b-1, this becomes:
x = 52 * ((b-1) + b)
which becomes:
x = 52 * (2b-1)
which becomes:
x = 104*b - 52
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since x = 110*b - 100, and
x = 104*b - 52, then
110*b - 100 = 104*b - 52
subtract 104*b from both sides of equation:
110*b - 104*b -100 = -52
add 100 to both sides of equation:
110*b - 104*b = 100 - 52
simplify:
6*b = 48
b = 8
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if b = 8, then a = b-1 = 7
since c = 0, number you are looking for is:
780
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to prove the number is correct:
780 / 10 = 78 which means number is divisible by 10.
780 = 52 * (a + b + c) = 52 * (7 + 8 + 0) = 52 * 15 = 780 which is a true statement.
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780 is the number you are looking for.
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