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Mathematical Games of Martin Gardner – Part 5

The line between entertaining math and serious math is a blurry one. (Martin Gardner, “A Quarter-Century of Recreational Mathematics,” Scientific American, August 1998: p. 68).

Penrose tile portrait of Martin Gardner by Bruce Torrence.

Martin Gardner – Puzzlemaster, by Bruce Torrence. August 12, 2010. Licensed under Flickr CC BY-NC 2.0.

As we come to the end of April and Mathematics and Statistics Awareness Month (MSAM) we are concluding our weekly nod to recreational mathematics inspired by “Mathematical Games,” a monthly column that appeared in Scientific American between 1956 and 1981. This column was authored by Martin Gardner, a tireless advocate for mathematics education and mathematical games.

Thank you for coming with us on this journey. The past month we have shown only a fraction of the types of math puzzles people can enjoy and we hope that these posts have given you the inspiration to find more.

Last week’s puzzle was a cryptarithm: GOOD + TIME = MATHS

To solve cryptarithms it does take patience and sometimes quite a bit of time. As stated last week, there are multiple solutions to this puzzle. 10254, 10259, 12603, and 16750. We’re going to work our way to 10254 together in the next few paragraphs so buckle in!

Before we begin, I will be referring to columns 1-5 where 1 is made up of D, E, and S; 2 is made up of O, M, and H; etc. All letters that are part of the puzzle will be bolded to differentiate them from words such as I and A.

First, we know that G, M, and T can not equal 0. We also know that M must equal 1 because of the carryover from column 4. In addition to this we have O = 3 thanks to the hint from last week. Periodically I will insert the original equation with the numbers we’ve figured out. To solve these it helps to write out the possibilities next to each letter and cross them out as you go. Good old fashioned pen/pencil and paper is a great way to keep track. Let’s highlight the numbers we have and cross out possibilities where we can:

G: 0 1 2 3 4 5 6 7 8 9
O: 0 1 2 3 4 5 6 7 8 9
D: 0 1 2 3 4 5 6 7 8 9
T: 0 1 2 3 4 5 6 7 8 9
I: 0 1 2 3 4 5 6 7 8 9
M: 0 1 2 3 4 5 6 7 8 9
E: 0 1 2 3 4 5 6 7 8 9
A: 0 1 2 3 4 5 6 7 8 9
H: 0 1 2 3 4 5 6 7 8 9
S: 0 1 2 3 4 5 6 7 8 9

Column 5 Column 4 Column 3 Column 2 Column 1
G 3 3 D
+ T I 1 E
1 A T H S

Because there is a carryover that creates column 5, we know that A is equal to or greater than 10. Now, there are two options: 1) column 4 has a carryover from column 3, which would mean the sum is greater than 10 OR 2) column 4 has no carryover and the sum of column 3 is less than 10. Let’s focus on option 1. This means A = G + T + 1, so G + T is less than or equal to 9 and because G and T can not equal 0 we can eliminate 9 as a possibility for either letter. Also with a carryover from column 3 we know that 3 + I is greater than or equal to 10 which makes I greater than or equal to 7. Let’s scratch out everything less than 7 as possibilities for I.

To find the value of T, we substitute I with 7, 8, and 9 to see what works. We know I can not be 7 because that would make T equal 10, where T would become 0. And because T can not equal 1, we know that I can not equal 8 so it must be 9 which makes T = 2. Because of the carryover to column 5, A must be 10 or greater. If A = G + 1 + T, we know that G must be 7. We know this because 9 has been taken, which leaves 7 or 8. But if G = 8, then A = 8 + 1 + 2, which would be 11. G must be 7 and A must be 10 where the 1 carries over. So far we have this:

Column 5 Column 4 Column 3 Column 2 Column 1
7 3 3 D
+ 2 9 1 E
1 0 2 H S

We know there isn’t a carryover to column 3 from column 2, but we don’t know if there’s a carryover from column 1 to column 2. So what we have left is: 1) H = 4 or 5; 2) D = 4 or 5 or 6 or 8; 3) E = 4 or 5 or 6 or 8; and 4) S = 4 or 5 or 6 or 8. Because S can not be 0, we know that D and E can not be a combination of 4 and 6, and with the remaining possibilities we know S is greater than 10 and there is a carryover to column 2: H = 3 + 1 + 1, or 5:

Column 5 Column 4 Column 3 Column 2 Column 1
7 3 3 D
+ 2 9 1 E
1 0 2 5 S

Now, since S can not be a combination of 4 and 6, we know that D + E is a combination of 6 + 8 which makes S 14 where the 1 carries over to column 2. In this instance D and E do not repeat anywhere else in the puzzle so D could be 6 or it could be 8 with E being the other:

Column 5 Column 4 Column 3 Column 2 Column 1
7 3 3 8
+ 2 9 1 6
1 0 2 5 4

Now see if you can solve this cryptarithm to get the other possible solutions: 10259, 12603, and 16750. Not each will have 3 replace O and keep in mind there will be 2 ways to get to each solution because D and E can be switched.

We loved learning about rep-tiles, contact numbers (kissing numbers), polyominos, and cryptarithms and hope we inspired you to continue to learn and play mathematical games. In addition to his monthly column in Scientific American, Gardner was a prolific author of full-length articles and a number of books. You can find these books using the Library of Congress online catalog and an author search using Gardner, Martin, 1914-2010. will return 173 distinct items. To find similar works of playful mathematics, click on the following links:

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