Question 659776
given:

a three-digit number, which is divisible by {{{10}}}, 

has a {{{hundreds}}} digit that is {{{one}}}{{{ less}}} than its {{{tens}}} digit 

the number also is {{{52}}}{{{ times}}} the {{{sum}}} of its digits 

solution:

Since it is divisible by {{{10}}}, let the number be 

{{{100a + 10b}}}............1 ...where {{{a = b - 1}}}

{{{100a + 10b = 52(a+b)}}}.......2
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Use substitution to solve for "{{{b}}}":

{{{100(b-1) + 10b = 52(b-1+b)}}}

{{{100b - 100 + 10b = 104b - 52}}}

{{{6b = 48}}}

{{{highlight(b = 8)}}}

now, solve for "{{{a}}}":

{{{a = 8-1}}}

{{{highlight(a = 7)}}}


so, the number is {{{100*7 + 8*10 = highlight(780)}}}