A 3 digit number - Math Central

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Question from Adrian, a student:

Consider a 3-digit number,abc,where a is not equal to c. Switch it so that it becomes cba. Subtract
the lesser from the greater to obtain rst. Switch this so that it is tsr. Prove that rst+tsr=1089

Hi Adrian.

Consider a>c to begin.

then abc > cba

and abc - cba = (a00 + b0 + c) - (c00 + b0 - a)
which requires a borrow of one in the tens, so the units
digit is t = 10 + c - a, the tens digit is s = 9 and the hundreds digit is r = a - 1 - c.

So rst + tsr makes a sum which is
ones digit: (10 + c - a) + (a - 1 - c) = 9.
tens digit: 8 (because 2 x 9 = 18, so carry the one)
hundreds digit: 1 + (10 + c - a) + (a - 1 - c) = 10.

Hence the sum is 1089.

Cheers,
Stephen La Rocque.

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