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Going Arithmetic With Alphametic (Posted on 2008-06-28) Difficulty: 2 of 5
Φ, Γ and Λ (in this order) with Φ < Γ < Λ are three positive integers in arithmetic sequence that satisfy this set of alphametic equations:

Λ3 - Γ3 = TWO, and:

Γ3 - Φ3 = TOW

where each of the letters T, O and W represent a different decimal (base 10) digit from 0 to 9.

What are the possible value(s) of the triplet (Φ, Γ, Λ)?

Note: T is not zero.

** Can you do this in a short time, using pen and paper, and eventually a hand calculator, but no computer programs?

See The Solution Submitted by K Sengupta    
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Solution Solution | Comment 1 of 2
Λ = a+d
Γ = a
Φ = a - d
100T+10W+O=3ad+3ad+d -- (1)
100T+10O+W=3ad-3ad+d -- (2)
Subtract (2) from (1)
=> 9(W-O)=6ad
=> 3(W-O)=2ad
As W,O are from 0 to 9, W-O can take values 2,4,6,8
=> ad  can take values 3,6,9,12
ad=3 => a=3, d=1
Given that T can not be 0.
This does not satisfy the given condition as 64-27 < 100
ad=6 => a=6,d=1.
This does not satisfy the given condition as 216-125 < 100
ad=9 => a=9,d=1
1000-729=271, 729-512=217
So,(8,9,10) is a solution
ad=12 => a=12,d=1 or a=3,d=2
(i):13-12=469 W<O. So, not a solution
(ii):27-1<100.So, not a solution

The only solution is (8,9,10)

  Posted by Praneeth on 2008-06-28 13:36:29
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