Cryptarithms

Can you crack these cryptarithms?

Problem

Cryptarithms printable worksheet

A cryptarithm is a mathematical puzzle where the digits in a sum have been replaced by letters.

In each of the puzzles below, each letter stands for a different digit. 

Bearing in mind that none of the numbers below have 0 as a leading digit, can you find a solution to all of these cryptharithms?

Do any of them have more than one solution?

1.

 
Image
A+A+A=BA, where each letter represents one digit.
2.

 
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BB+A=ACC, where each letter represents a digit.
3.

 
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AB+A=BCC, where each letter represents a digit.
4.

 
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AB+A=CDC, where each letter represents a digit.
5.

 
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AB+BC=BCB, where each letter represents a digit.
6.

 
Image
AB+CB=BA, where each letter represents a digit.
7.

 
Image
AB+CB=BBA, where each letter represents a digit.
8.

 
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AB+AB+AB+AB=CA, where each letter represents a digit.



 

9.

 
Image
AA+BB=CBC, where each letter represents a digit.
10.

 
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AB+AB=CBB, where each letter represents a digit.
11.

 
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AB+AB=CA, where each letter represents a digit.
12.

 
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AB+AB=BC, where each letter represents a digit.
13.

 
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AAA+BB+A=CAB, where each letter represents a digit.
14.

 
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ABC+ACB=CBA, where each letter represents a digit.
15.

 
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ABC+ABC=CDDB, where each letter represents a digit.
16.

 
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ABC+CBC=CDEB, where each letter represents a digit.
17.

 
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ABC+ABC+ABC=CCC, where each letter represents a digit.
18.

 
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ABC+ABC+ABC=BBB, where each letter represents a digit.
19.

 
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AB+BC+CA=ABC, where each letter represents a digit.
20.

 
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A+BB+CCC=BAB, where each letter represents a digit.
21.

 
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A+BB+CCC=BCB, where each letter represents a digit.
22.

 
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BAA+BAA+BAA+BAA=CAAD, where each letter represents a digit.
23.

 
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ABA+ABA+ABA+ABA+ABA=CDBA, where each letter represents a digit.
  

 

Final Challenge

Is it possible for all of the digits 1 to 9 to appear exactly once in the addition below?

   # # #

+ # # #

   # # #

Using each digit from 1 to 9 once, what is the largest sum you can obtain in the addition above?

 

If you enjoyed this problem, you may also like to take a look at Two and Two.

 

With thanks to Don Steward, whose ideas formed the basis of this problem.