Hello!
The original number may be written as `bar(ab),` where `a` and `b` are one-digit natural numbers, `0lt=blt=9,` `1lt=alt=9.` The value of the number `bar(ab)` is obviously `b + 10a.`
When we interchange its digits, the resulting number is `bar(ba) = a + 10b.` The sum is `(b + 10a) + (a + 10b) = 11a + 11b = 110` (this is given). So
`a + b = 10.`
The second condition gives that
`(b...
Hello!
The original number may be written as `bar(ab),` where `a` and `b` are one-digit natural numbers, `0lt=blt=9,` `1lt=alt=9.` The value of the number `bar(ab)` is obviously `b + 10a.`
When we interchange its digits, the resulting number is `bar(ba) = a + 10b.` The sum is `(b + 10a) + (a + 10b) = 11a + 11b = 110` (this is given). So
`a + b = 10.`
The second condition gives that
`(b + 10a) - 20 = 4(a + b) + 4,` or `6a - 3b = 24, or 2a - b = 8.`
Thus we have a linear system for `a` and `b.` If we add up these two equations, we obtain `3a = 18,` or `a = 6.` From the first equation `b = 10 - a = 4.` The initial restrictions for `a` and `b` are also satisfied.
The answer: the original number is 64.
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