A three-digit number has two properties. The tens-digit and the ones-digit add up to 5. If the number is written with the digits in the reverse order, and then subtracted from the original number, the result is 792. ﬁnd all of three-digits?

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Answer 1 · 2016-06-07T09:32:56

Numbers are $941$ $941$ or $850$ $850$

Let the unit digit be $x$ $x$ , then ten's digit is $(5 - x)$ $(5-x)$ (note this means $x \leq 5$ $x<=5$ ). Let hundred's digit be $y$ $y$

Then the number is $100 y + 10 (5 - x) + x$ $100y+10(5-x)+x$ and on reversing the number becomes $100 x + 10 (5 - x) + y$ $100x+10(5-x)+y$ .

As second subtracted from first is $792$ $792$ , we have

$100 y + 10 (5 - x) + x - 100 x - 10 (5 - x) - y = 792$ $100y+10(5-x)+x-100x-10(5-x)-y=792$

Hence $99 y - 99 x = 792$ $99y-99x=792$ or $y - x = 8$ $y-x=8$ or $y = x + 8$ $y=x+8$

As $y$ $y$ can take values $8$ $8$ and $9$ $9$ only, $x$ $x$ takes values $0$ $0$ and $1$ $1$ only and hence

Numbers are $941$ $941$ or $850$ $850$