Alphametics- Number Puzzles

Today, I want to show you a dream . A dream where you are sitting in a nice restaurant and waiting for your favorite dish to arrive, and you hear that little sweet voice (well, not that sweet anymore after hearing the same words repeated 5 times) “Mom, I am hungry. When will the food arrive”, and you reach out to a piece of paper and write down this puzzle for them. They would be working out their brain, figuring it and the food would arrive. Isn’t it nice, to have a quiet dinner once in a while. Thats the magic of this puzzle. So, let’s wake up from our dream and figure it out.

Alphabets and Arithmetics when combined together makes Alphametics. Alphametics are a type of mathematical puzzle consisting of a mathematical equation where the digits are unknown and replaced by letters. These letters form words or phrases and the goal is to find the value of each letter.

S E N D
 + M O R E
——————
M O N E Y

This classic puzzle was first introduced by the famous mathematician Henry Ernest Dudeney and is one of the most famous one. Generally, alphametic puzzles requires some guessing and checking and narrowing down of the possibilites. There are two rules to be followed to solve these puzzles

  • The leftmost letter can’t be zero in any word.
  • There is one to one mapping between the letters and the digits. Each letter can be assigned only one digit between 0 and 9.

So, lets give it a try.

Starting from the leftmost digit in the answer, we notice that there are 5 letters whereas the question has only 4 letters  which means M can not be 0 and it is a carry over from the thousands place.

The biggest possible 4 digit number is 9999. And sum of two largest 4 digit number can be 9999 + 9999 = 19998 . Since, our answer is a 5 digit number, the resulting sum should be < 20000 which means the word MONEY < 20000. Therefore, M = 1 

 Now we have:

  S E N D
+ 1 O R E
—————-
1 O N E Y

Now, in the thousand’s place there is a 1, so we can have either of the two values

S  + 1 = 10 + O   => S = 9 + O ( If there is no carry over from hundred’s place ) 

1 + S  + 1 = 10 + O  => S = 8 + O  ( If the carry over from hundred’s place is 1 ) 

Since, S is a single digit , O can be either 0 or 1 . But M = 1  so  O = 0

So, we get :

  S E N D
+ 1 0 R E
—————
1 0 N E Y

Now, in the hundred’s place, we have E + 0 = N . If there were no carry over from the ten’s place, E should be equal to N , but as per the rule E and N can’t be same, so there must be a carry over from the tens place. So, we would have 

1 + E = N ( There MUST be a carry over from ten’s place ) 

Also N <> 0 , so E must be less than 9 which means there is no carry over from hundred’s place to the thousand’s place. Looking back at the thousand’s place, it means we have 

S  + 1 = 10    => S = 9 

So, S = 9 is the only value that could cause a carry over to the 5th column .  So, we get:

9 E N D
+ 1 0 R E
—————-
1 0 N E Y

Looking back at the hundred’s place, we know 1 + E = N . And at the ten’s place we can have two possible values

N + R = E + 10  ( if there is no carry over from the one’s place ) 

 1 + N + R = E + 10  ( if there is a carry over from the one’s place ) 

Substituting N = 1 + E  in the first equation above , we get 

N + R = E + 10  

=>  1 + E + R  = E + 10  

=> 1 + R = 10

= >  R = 9  ( But E cannot be 9 since S is 9 ) 

Substituting N = 1 + E  in the second equation above , we get 

1 + N + R = E + 10  

= >    1  + 1  + E + R = E + 10 

=>  2 + R = 10 

=>  R = 8 

So, we get :

    9 E N D
+ 1 0 8 E
—————–
1 0 N E Y

Looking at the unit’s place, we have D + E = Y . Y needs to be greater than or equal to 10 because there is a carry over to the ten’s place. But Y cannot be 0 or 1 because they are already taken. So Y needs to be greater than or equal to 12. 

Only possible solution for Y >= 12 are when D and E are either 5, 6 or 7  ie 7+ 5 = 12 or 7+ 6 = 13. D and E cannot be 8 and 9 as they are  already taken. 

Let’s say E = 7, then it means N = 8 because 1 + E = N but that is not possible because N cannot be 8 as its already taken. So E cannot be 7 and therefor D = 7 and E is either 5 or 6.

If E = 6 then N = 7 because 1 + E = N but N cannot be 7 as D is already 7 so E = 5 and N = 6 . Now units place would be D + E = 7 + 5 = 12  which means Y = 2 

So, we get :

  9 5 6 7
+ 1 0 8 5
————
1 0 6 5 2

Was it fun and did it keep your kids engaged for a good enough time so that you could have a cup of tea in peace. If so, please Like my facebook page and subscribe to my blog to be the first one to read my post.

I am also doing a little give away, I would email a list of 20 Alphametrics to all the readers who like and share this post on their facebook page.

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