Question 1109542
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First, a solution using formal algebra....<br>
The 3-digit number is ABC; its value is 100A+10B+C.
The number with the hundreds and tens digits switched is BAC; its value is 100B+10A+C.
The digit with the ones and hundreds digits switched is CBA; its value is 100C+10B+A.<br>
The problem tells us
{{{(100A+10B+C)-90 = 100B+10A+C}}}  (BAC is 90 less than ABC)
{{{90A = 90B+90}}}
(1) {{{A = B+1}}}<br>
It also tells us
{{{(100A+10B+C)+198 = 100C+10B+A}}}  (CBA is 198 more than ABC)
{{{99A+198 = 99C}}}
(2) {{{A+2 = C}}}<br>
So {{{A = B+1}}} and {{{C = A+2 = B+3}}}<br>
Then since the sum of the digits is 10,
{{{(B+1)+B+(B+3) = 10}}}
{{{3B+4 = 10}}}
{{{3B = 6}}}
{{{B = 2}}}<br>
So B is 2; A is B+1 = 3; and C is B+3 = 5.<br>
The original number ABC is 325.<br>
We can get to the answer with less work using only a little algebra and some logical reasoning.<br>
The derivation of equation (2) above shows that, whenever a 3-digit numbers has its digits reversed (i.e., the ones and hundreds digits are switched), and the value of the number decreases by 198, it means the ones digit is 2 more than the hundreds digit.<br>
We also know that the sum of all three digits is 10.<br>
The numbers with a digit sum of 10 and a ones digit 2 more than the hundreds digit are
163, 244, and 325<br>
Only one of these, 325, satisfies the other condition of the problem -- that switching the hundreds and tens digit decreases the value of the number by 90.<br>
So the original number is 325.