The Trick

I have used this number trick many times with beginning Algebra classes. I take a dictionary to class and before had write a word on a piece of paper then place it in an envelope. At the beginning of class I select a student at random and give them the envelope then have each student pick a 3 digit number without any zeroes or repeating numbers, such as 389. Next have them perform the following calculations:

389 reverse the digits in the number to get 983.
Take the larger and subtract the smaller.
983 – 389 = 594
Reverse the digits of 594 and get 495.
Add them, 594 + 495 = 1089

Next I give the dictionary to a random student and have them take the result of their calculation (I check to ensure no arithmetic errors), use the first 3 numbers for the page and the last number for the word on the page.
Ask them to say the word aloud then have the first student open the envelope and say the word. The class is always amazed and soon realize they al got the same result in their calculation and wonder why – which is the segway to discussing some algebra.

The Proof

Call initial number abc = a*100 + b*10 + c
after reversing cda = c*100 b*10 + a
we will assume a>c (does not matter which one is larger).

A = abc – cba = (a – c)*100 + (b – b)*10 + (c – a)

Since c<a we will need to borrow to do the subtraction (in our example above we had to take 3 – 9 for the units digits). to borrow we take 1 from the 10s unit and 1 from 100s unit and get (since middle term of both numbers are the same we need to borrow from 10s and 100s units:

A = abc – cba = (a – c – 1)*100 + (b – b – 1 + 10)*10 + (c – a) + 10
A = (a – c – 1)*100 + (9)*10 + (c – a) + 10

Now reverse A to get
B = (c – a + 10)*100 + 9*10 + (a – c- 1)

Next add them
A + B = (a – c – 1)*100 + (9)*10 + (c – a) + 10 + (c – a + 10)*100
+ 9*10 + (a – c- 1)
= (a – c – 1 + c – a + 10)*100 + (9 + 9)*10 + (c – a + 10) + (a – c – 1)
= 9*100 + (10 + 8)*10 + 9
= 9*100 + 100 + 8*10 + 9
= 10*100 + 8*10 + 9
= 1000 + 80 + 9
= 1089

This is the same answer regardless of values of a,b,c as long as we keep the digits all different and non zero.

Also check Kaprekar constant  for another set of calculation which will eventually end up with the same number.

An excerpt:

The 4-digit Kaprekar constant. That is, if the digits of a 4-digit number are not all equal, there is a certain number that we will end up with if we repeat the enumerated process above. Let’s have an example.

  1. Choose any 4-digit integer whose digits are not all equal: e.g. 4358
  2. Arrange the digits from in decreasing order: 8543
  3. Arrange the digits in increasing order: 3458
  4. Subtract: 8543 – 3458 = 5085

Repeat 2-4 certain number of times. For 4-digits the process will always reach 6174 and remain constant.

Mystery explained.

Advertisements