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Question

Let a, b, c, d, e be natural numbers in an arithmetic progression such that a+b+c+d+e is the cube of an integer and b+c+d is square of an integer. The least possible value of the number of digits of c is?

A
2
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B
3
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C
4
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D
5
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Solution

The correct option is C 3
Let the numbers be
a=C2Db=CDc=Cd=C+De=C+2D
Then,
a+b+c+d+e=5C=α3....(i)b+c+d=3C=β2......(ii)
From (i) and (ii)
α35=β23
For this least possibility is
α=5×3,β=5×3×3α=15,β=45C=α35=1535=675c=C=675
So option B is correct.

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