Cryptotastic #2

Algebra Level 5

A B C D + E F G H \begin{array} { l l l } & A & B \\ & C & D \\ + & E & F \\ \hline & G & H \\ \end{array}

In the above cryptogram, all the letters represent distinct digits from 1 to 9.

If S S is the sum of all possible values of G H \overline{GH} , find S S .


As a warmup try Cryptotastic .


The answer is 534.

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1 solution

The possible values for G H \overline{GH} are

  • 14 + 25 + 39 = 78 14 + 25 + 39 = 78 ,

  • 15 + 26 + 38 = 79 15 + 26 + 38 = 79 ,

  • 13 + 25 + 49 = 87 13 + 25 + 49 = 87 ,

  • 12 + 36 + 47 = 95 12 + 36 + 47 = 95 ,

  • 13 + 26 + 58 = 97 13 + 26 + 58 = 97 .

I'm fairly certain that 78 , 79 , 87 , 95 , 97 78, 79, 87, 95, 97 are the only options, but if you know of more please let me know so that I can get the correct answer posted.

Edit: As Pi Han Goh and Siva Bathula (in the reports section) have pointed out, there is a sixth option, namely 47 + 36 + 15 = 98 47 + 36 + 15 = 98 , so the correct answer should be 534 534 .

You missed one:

47+36+15=98

Pi Han Goh - 4 years, 1 month ago

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Oh darn. Thanks for finding that! I'll edit the question for now to ask for the sum minus 98 and ask staff to change the answer to 534.

Brian Charlesworth - 4 years, 1 month ago

Thanks. I've updated the answer to 534.

In future, if you notice such errors, you can report the problem directly by selecting that option in the menu.

Brilliant Mathematics Staff - 4 years, 1 month ago

You missed three: 14+25+36=75, 13+25+46=84, 12+35+46=84

Debanik Samanta - 4 years, 1 month ago

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All the digits must be distinct.

Pi Han Goh - 4 years, 1 month ago

Please stop calling these cryptograms. Cryptograms are enciphered messages of ordinary text. These are cryptarithms.

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Hi Michael, thanks for the response.

Yes, "cryptarithm" is certainly a more accurate term. However, the word "cryptogram" is still correct because the message (in this case, the numbers) is written in codes (or replaced with letters). We chose the latter term because it's a more recognisable and familiar term to most people.

Brilliant Mathematics Staff - 4 years ago

You've come up with 1 as being the A term for all your solutions, but what if instead of 14+25+39=78 you had 25+14+39=78? Would that not be another solution, or would there be the case where every different GH solution would then have 6 times as many possible arrangements making S=3204 then?

alex phillips - 2 years, 11 months ago

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While you are correct that there are other combinations of summands for a given value of G H \overline{GH} , they do not produce any different values of G H \overline{GH} . So since the question is to find the sum of all possible values of G H \overline{GH} we don't need to count repeated instances of the same sum, and so I believe that the posted answer remains valid.

Brian Charlesworth - 2 years, 11 months ago

0 pending reports

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