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 a + w + a + y.

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 and, therefore, l = 7,l = 5,l = 6,l = 7,l = 8, as the only digit whose product by 3 ends with 1.

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. It can't be 7, for then f = l. It can't be 9, for then f = w. The only remaining value is f = 8,f = 2,f = 4,f = 6 ,f = 8, implying w = 6.

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

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