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## Puzzles Corner - Multiplication in Alphametics

Previously, we looked at alphametics involving addition and subtraction. The logic becomes a little more involved when the alphametic involves multiplication. When a two digit number is multiplied by another two digit number, the process of finding the result is composed of multiple levels. Today, let’s see how to handle those multiple levels.  The question involves quite a few steps and observations using number properties. Hence, you are unlikely to see such a question in actual GMAT but you might see a simpler version so it’s good to be prepared.

Question: The following alphametic shows multiplication of two numbers, IF and DR. The product you obtain is DORF.

What is the value of D + O + R + F?
(A) 17
(B) 20
(C) 22
(D) 23
(E) 30

Solution: The good thing is that we know D + O + R + F has a single value. This means there will be a logic to obtain the value of each of D, O, R and F.

As discussed last week, we first focus on the big picture, but we will have to go one level at a time.

(i) IF * R = OFF

(ii) IF * D = IF

(iii) OF + IF = DOR

A few interesting points to note from the above:

– From (ii), IF * D = IF

When you multiply IF by D, you get IF itself. This means that D must be 1. D can take no other value.

D = 1

– From (iii), F + F has unit’s digit of R.

Also O + I gives O as unit’s digit and 1 as tens digit (D of DORF obtained from above). How can this happen? Say, if O = 4, 4 + I = 14. This is possible only when I = 9 and there is a 1 carry over from the previous addition of F + F. This means that F must be 5 or greater to have a carryover of 1. It cannot be 5 because 5+5 will give you 10 making R = 0. This would mean that F*R would end in R (0). But in (i), F * R has unit’s digit of F, not R. So F cannot be 5.

D = 1, I  = 9

– Another interesting point: From (i), F * R has unit’s digit of F. This is possible only when F = 0 or F = 5 or R = 6 (Think of multiplication tables of numbers to convince yourself why this is so)

Since F has to be greater than 5 (as seen above), R must be 6.
If R = 6, then from (iii), F + F has unit’s digit of 6 and a carryover of 1 so F = 8. When you add 8 + 8, you will get 16 (units digit of 6 and a carryover)

D = 1, I = 9, R = 6, F = 8

– From (i), when we multiply IF by R, we get OFF. That is, when we multiply 98 by 6, we get 588. So O must be 5.

This gives us: D + O + R + F = 1 + 5 + 6 + 8 = 20