Question 1136134


Since the three-digit number is equal to {{{13}}} times the product of its digits, it is divisible by its {{{hundreds}}} digit, which is the largest of the three. 

If we increase both the tens digit and the units digit to the hundreds digit, the quotient will be {{{111}}}. If instead we decrease both of them to {{{0}}}, the quotient will be {{{100}}}.

The actual quotient is between {{{100}}} and {{{111}}}, and since it is divisible by {{{13}}}, it must be {{{104}}}.
 
The product of the tens digit and the units digit is {{{104/ 13=8}}}.

Since{{{ 8=2 * 4=1* 8}}}, the three-digit number is one of {{{624}}}, {{{642}}}, {{{918}}} and {{{981}}}.

The {{{only}}}{{{ one}}} divisible by {{{13}}} is {{{13*6* 2*4= 624}}}. 

answer:The number is {{{624}}}