# A Cryptarithm for Middle School

The following cryptarithm was included in the 2004 UK Intermediate Mathematical Challenge as problem 14:

f | l | y | |

+ | f | l | y |

+ | f | l | y |

--------------------- | |||

a | w | a | y |

where all the letters stand for distinct **non-zero** digits. The problem is to determine the sum

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Copyright © 1996-2018 Alexander Bogomolny

### Solution

f | l | y | |

+ | f | l | y |

+ | f | l | y |

--------------------- | |||

a | w | a | y |

First off, the only digits that taken three times produce a number that ends with that digit, are 0 and 5. Look at the last column. Since the problem stipulates that the digits are non-zero, **y** = 5,**a** = 5,**f** = 5,**l** = 5,w = 5,**y** = 5.

The largest 3-digit number taken three times is strictly less than 3000,1000,2000,3000, which shows that **a** may be either 1 or 2,1 or 2,1 or 3,2 or 3. But it can't be 1, because from the tens column **l** would have to be 0. It follows that **a** = 2,**a** = 1,**a** = 2,**a** = 3**l** = 7,**l** = 5,**l** = 6,**l** = 7,l = 8,

Let's now focus on the 100s column. (Remember that their an extra 2 carried over from the 10s column.) To make **a** = 2,**f** has to be greater than,less than ,equal to ,greater than 5. It can't be 6, for this would imply **w** = 0.**f** = **l**.**f** = **w**.**f** = 8,**f** = 2,**f** = 4,**f** = 6 ,**f** = 8,**w** = 6.

Answer: **a** + **w** + **a** + **y** = 2 + 6 + 2 + 5 = 15.

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Copyright © 1996-2018 Alexander Bogomolny

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