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Another positive integer puzzle (Posted on 2006-11-29) Difficulty: 2 of 5
Given that a, b, c, and d are positive integers, determine analytically if it is possible that 6(6a+ 3b + c) = 5d.

  Submitted by K Sengupta    
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Solution: (Hide)
We can assume that gcd(a,b,c,d) =1. Otherwise, the equation can be simplified whenever gcd(a,b,c,d) is greater than 1.

Clearly, d is divisible by 6. So, substituting d = 6m, we obtain:

6a2+ 3b2 + c2 = 30m2

Replacing c=3n in the above equation, we obtain:
2a2+ b2 + 3n2 = 10m2......(#)

Now, we know that:

2a2 Mod 8 = 0, 2
2b2 Mod 8 = 0, 2
3n2 Mod 8 = 0, 3, 4
10m2 Mod 8 = 0, 2

By trying for each of the above quadruplet combinations, we observe that (#) is satisfied iff:

(a2 Mod 8, b2 Mod 8, n2 Mod 8, m2 Mod 8)
= (0,0,0,0); (2,2,0,0); (2,0,0,2); (0,2,0,2); (2,2, 4,0)

Accordingly, each of a, b, n and m are even.

This contradicts our original assumption that gcd(a,b,c,d) = 1

Consequently, no positive integer solution exists for the given problem.

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PROBLEM SOURCE: Mathematical Olympiad Treasures by Titu Andreescu and Bogdan Enescu.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
Some ThoughtsA start, perhaps?tomarken2006-11-29 21:57:53
D2?Richard2006-11-29 21:32:47
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