Digit Reversal Trick ExplainedDate: 03/23/2001 at 15:40:50 From: I. Kouraklis Subject: Algebra trick/puzzle Take a 3-digit number (e.g. 875) and subtract its reverse (578). Then, take the result (297) and add its reverse (792). The answer is always 1089, no matter what the initial numbers are. I would like to know why this happens. I noticed that this rule has some exceptions; it doesn't apply to numbers with the same first and third digits, since the subtraction gives zero. Thank you Date: 03/26/2001 at 14:42:06 From: Doctor Rob Subject: Re: Algebra trick/puzzle Thanks for writing to Ask Dr. Math. Let the original number be ABC, that is, 100*A + 10*B + C Here A, B and C are digits, whole numbers from 0 to 9. In order to be a three-digit number, presumably A is not zero. Then its reverse is: 100*C + 10*B + A Subtracting the smaller from the larger, you get: 99*A - 99*C or 99*C - 99*A Notice that the B's drop out. Suppose A > C, so it is the former result you get. (The same kind of argument works for A < C; A = C is one of the exceptions below.) Then: 99*(A-C) = 100*D + 10*E + F Notice that 0 < A-C <= 9, so A-C is a digit. Add this to its reverse: 100*F + 10*E + D and you get 101*(D+F) + 20*E Now F, the units digit of 99*(A-C) is 10-(A-C). Also E, the tens digit, is always 9, and F, the hundreds digit, is (A-C)-1. This all works because 9 >= A-C >= 1, and: 99*1 = 099 99*2 = 198 99*3 = 297 99*4 = 396 99*5 = 495 99*6 = 594 99*7 = 693 99*8 = 792 99*9 = 891 That means that the final sum is 101*(10-A+C+A-C-1) + 20*9 = 101*9 + 180 = 1089. Notice that the A's and C's drop out. The exception, of course, is when A = C, and then 99*(A-C) is zero. If a 3-digit number is chosen so that the last digit is 0, when you reverse it, you can drop the initial zero. Thus if you start with 730, you reverse it to 037 and drop the leading 0, giving 37, and subtract that from 730 to give 693. Then 693 + 396 = 1089, as before. Another condition is that when the difference is less than 100, you have to put on a leading zero to make it a 3-digit number before reversing and adding. For example, if you start with 392, reverse to 293, and subtract, you get 99. Make this 099 before you reverse it to get 990, and add 99, getting 1089 as before. For another answer to this question in our archives, see: Proving Something is Always So http://mathforum.org/dr.math/problems/emma4.1.98.html - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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