I am puzzled by the below exercise:

Step 1: Select any number having 3 digits (all different from one another). Ex. $125$.

Step 2: Now, write all possible combination of two digit number forming from selected digits. Here it is $12$,$21$,$15$,$51$,$25$,$52$. Add all of them.

here, $ 12+21+15+51+25+52=176$

Step 3: Divide the addition, (here $176$) by sum of all 3 digits selected. i.e.

$ \dfrac{176}{1+2+5} = 22$.

Always. Why, so?

I have tried many combination, it works. Can anyone give proof and explain the reason behind this?

  • 7
    $\begingroup$ What happens if you instead write all 6 possible three digit numbers from the selected digits and divide by the sum of the digits? Can you figure now, why? $\endgroup$ – N. S. Oct 30 '11 at 19:06
  • $\begingroup$ @ user9176: Great!! I will always get $222$..:) $\endgroup$ – orion Oct 30 '11 at 19:23
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    $\begingroup$ Now try to see if you can prove it... Seeing the proofs below should help you ;) $\endgroup$ – N. S. Oct 30 '11 at 19:41

If your original digits are $x, y, $ and $z$, then among your two-digit numbers are:
two with $x$ in the ones place,
two with $y$ in the ones place,
two with $z$ in the ones place,
two with $x$ in the tens place,
two with $y$ in the tens place, and
two with $z$ in the tens place.

Therefore they sum to

$2\cdot 10(x+y+z)+2\cdot(x+y+z)=22(x+y+z)$.

So when you divide you are left with $22$.


You can trace it backwards via $$ 22(a+b+c) = (10a+ b) + (10a + c) + (10b +a) + (10b + c) + (10c + a) + (10c + b) $$

  • 2
    $\begingroup$ Simplifying the right hand side gives identity. Bravo! $\endgroup$ – Hassan Muhammad Oct 30 '11 at 18:48

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