Puzzles Corner – Solving Alphametics

Today, let’s learn how to solve alphametics. An alphametic is a mathematical puzzle where every letter stands for a digit from 0 – 9. The mapping of letters to numbers is one-to-one; that is, the same letter always stands for the same digit, and the same digit is always represented by the same letter.

First focus on the big picture of the alphametic – such as, a two number is added to another two digit number to give a three digit number etc. Then look at the nitty gritty – for which digit can each letter stand?

Question 1: With # and & each representing different digits in the problem below, the difference between #&& and ## is 667. What is the value of &?

# & &

– # #

_______

6 6 7

_______

(A) 3

(B) 4

(C) 5

(D) 8

(E) 9

Solution: The big picture: A two digit number is subtracted from a three digit number to give 667. So the three digit number must be a bit larger than 667. This means that the hundreds digit of #&& must be either 6 or 7. It cannot be 8 because you cannot obtain 800+ by adding a two digit number to 667.

Let’s look at both cases:

# = 6: If you subtract 66 from 6&&, you will not get 667 – the largest value you can get is 699 – 66 = 633. So # cannot be 6.

# must be 7.

Now the question is very simple

7&& – 77 = 667

7&& = 667 + 77 = 744

Answer (B)

There are many other ways in which you can solve this question including plugging in the answer choices. We should now take a look at a DS question on alphametics.

Question 2:

M R

N R

+ P R

_______

S T V

_______

In the correctly worked addition problem above, M, N, R, S, T and V are distinct digits. Is R > 3?

Statement 1: M, N and P are positive even integers.

Statement 2: S = 2

Solution: This is certainly harder than the PS question but our process will remain the same.

First, let’s see what information we are given in the question – the units digits of all three numbers are the same. The two-digit numbers add up to give a three digit number. The hundreds digit, S, is either 1 or 2. Three two-digit numbers cannot add up to give a number 300 or more since 99 + 99 + 99 = 297. We have no information on what the value of R can be. All we know is that R cannot be 0 because 0+0+0 = 0 but V needs to be different from R.

Let’s look at the statements now.

Statement 1: M, N and P are positive even integers.

So M, N and P must take distinct values out of 2, 4, 6 and 8 and add up to give the units digit of T (again, distinct)

Every time you add three even numbers, you will get an even number. Let’s see which combinations we can get:

2 + 4 + 6 = 12

2 + 6 + 8 = 16

2 + 4 + 8 = 14

4 + 6 + 8 = 18

Note that in all four cases, the units digit is one of the numbers but T must be distinct. This means that there must have been a carryover from the previous addition. So when we added the three Rs, we must have got a carryover. Had R been 3 or less, we would not have got a carryover since 1+1+1 = 3, 2+2+2 = 6 and 3+3+3 = 9. So R must be greater than 3.

One such case would be

8 7

4 7

+ 6 7

_______

2 0 1

_______

This statement alone is sufficient.

Statement 2: S = 2

The result of addition gives us a number which is more than 200. We know from statement 1 that there must exist a case in which R is greater than 3. Now all we have to do is find a case in which R is less than 3. One of these cases is

8 1

9 1

+ 7 1

_______

2 4 3

_______

So this statement alone is not sufficient.

Answer (A)