Question 1169861
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A solution using formal algebra....<br>
x = tens digit
y = units digit
10x+y = original number
10y+x = original number with digits reversed<br>
(1) {{{10x+y = 2(x+y)+26}}}  [the number exceeds twice the sum of its digits by 26]<br>
(2) {{{(10y+x)-(10x+y) = 18}}}  [the number with the digits reversed exceeds the given number by 18]<br>
Use (2) to find y in terms of x:<br>
{{{(10y+x)-(10x+y) = 18}}}
{{{9y-9x = 18}}}
{{{9(y-x) = 18}}}
{{{y-x = 2}}}
{{{y = x+2}}}<br>
Substitute "x+2" for "y" in (1):<br>
{{{10x+y = 2(x+y)+26}}}
{{{10x+x+2 = 2(x+x+2)+26}}}
{{{11x+2 = 4x+30}}}
{{{7x = 28}}}
{{{x = 4}}}
{{{y = x+2 = 6}}}<br>
ANSWER: the number is 10x+y = 46<br>
The first calculation above shows that, when a 2-digit number is subtracted from the same number with the digits reversed, the result is always a multiple of 9.  If a formal algebraic solution is not required (for example, in a timed math contest where a quick solution is wanted), you can use this fact along with trial and error to quickly find the answer.<br>
Since the difference in this case is 18, we know the difference between the digits is 2; and since the number with the digits reversed is greater than the given number, we know the units digit is 2 more than the tens digit.<br>
So the possible 2-digit numbers are 13, 24, 35, 46, 57, 68, and 79.<br>
Then simply find which of those satisfies the condition that the number exceeds twice the sum of its digits by 26.<br>