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March 2018 Puzzle Periodical - a, b, c - Variable Alphabet Equation

By Wendell W., NSA Mathematician

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Can you solve this alphabetical equation?

Puzzle Challenge Difficulty Rating: Medium

Consider the following equations:

a 2 × b × c 2 × g = 5,100

a × b 2 × e × f 2 = 33,462

a × c 2 × d 3 = 17,150

a 3 × b 3 × c × d × e 2 = 914,760

Find positive integers a, b, c, d, e, f, and g, all greater than 1, that satisfy all the equations.

Click to see the answer!

Did you find the only solution to all of these equations where a, b, c, d, e, f, and g are all integers greater than 1?

Solution

Did you find the only solution to all of these equations where a, b, c, d, e, f, and g are all integers greater than 1?

If all four equations are satisfied, then multiplying all four equations together will produce another valid equation. We obtain:

a 7 × b 6 × c 5 × d 4 × e 3 × f 2 × g = 2,677,277,333,530,800,000.

The number on the right factors into prime powers as 2,677,277,333,530,800,000 = 27 × 36 × 55 × 74 × 113 × 132 × 17, as can be determined by trial division by the small primes 2, 3, 5, 7, 11, 13, and 17, for instance. So we have the new equation:

a 7 × b 6 × c 5 × d 4 × e 3 × f 2 × g = 2 7 × 3 6 × 5 5 × 7 4 × 11 3 × 13 2 × 17.

If p is any prime dividing a, then p7 divides the right hand side. Since only 2 appears to the seventh power, the only value p can take is 2. Since a must be greater than 1, we see that 2 must divide a. Because 2 appears to exactly the seventh power on the right, we see that a = 2. Canceling a7 on the left with 27 on the right leaves:

b 6 × c 5 × d 4 × e 3 × f 2 × g = 3 6 × 5 5 × 7 4 × 11 3 × 13 2 × 17.

Repeating the same argument shows that b = 3, c = 5, d = 7, e = 11, f = 13, g = 17 is the unique solution to the new equation. However, it is still necessary that these values actually satisfy the original four equations, which in fact they do as can be checked easily. (To see why this last step is necessary, try using the same reasoning on the equations: x = 10, y2 = 9, z3 = 25, which clearly has no solutions in integers.)

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