The substitution of numbers for letters
is called cryptography, and a cryptarithm
is a mathematical problem in
which letters are substituted for numbers.
Here is a sample:
The problem is to find the number ABC which has been squared. Here is how to go about solving it: Start with C which is the last digit of the number and its square.
There are only four numbers, which, when multiplied by themselves, will have their last digit the same as the number.
They are: 0 (0 x 0=0),1 (1 x 1=1),5 (5 x 5 = 25) and 6 (6 X 6 = 36) . C cannot be equal to 0, for when you multiply a number by zero, you get zero, but in this problem we multiply C by Band get E.
Neither can C equal 6.
Note in the center column of the addition, we have D + C + C = D. If C equals 6, it would not be possible to add 6 + 6 + any number and have the sum equal the missing number. C cannot be equal to 1, since C X ABC would equal ABC, but in this problem it equals DBC.
Therefore, C must be equal to 5. We also know the number of another letter: A . We see in the multiplication that A X ABC = ABC. Therefore, A must equal 1.
We now have two digits: A = 1 and C = 5.
Write the problem over substituting the known numbers:
The problem is to find the number ABC which has been squared. Here is how to go about solving it: Start with C which is the last digit of the number and its square.
There are only four numbers, which, when multiplied by themselves, will have their last digit the same as the number.
They are: 0 (0 x 0=0),1 (1 x 1=1),5 (5 x 5 = 25) and 6 (6 X 6 = 36) . C cannot be equal to 0, for when you multiply a number by zero, you get zero, but in this problem we multiply C by Band get E.
Neither can C equal 6.
Note in the center column of the addition, we have D + C + C = D. If C equals 6, it would not be possible to add 6 + 6 + any number and have the sum equal the missing number. C cannot be equal to 1, since C X ABC would equal ABC, but in this problem it equals DBC.
Therefore, C must be equal to 5. We also know the number of another letter: A . We see in the multiplication that A X ABC = ABC. Therefore, A must equal 1.
We now have two digits: A = 1 and C = 5.
Write the problem over substituting the known numbers:
Look at the center column where
D + 5 + 5 is used in the problem. We
now know that 10 + D equals D and
we carry 1. Therefore, we know that
1 + B + B = 5. The only number that
B can stand for is 2 since 1 + 2 +
2 =5.
The problem is now solved: ABC=
125.
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