getgyan wrote:

A no when divided by D gives remainder 7 and when divided by 3D gives remainder 20.

What will be the remainder when twice the number is divided by 3D?

a.1

b.40.

c.27

d.13

e.can' be determined

Denote our number by N. Then we can write:

N = DQ + 7, where Q is a non-negative integer. Necessarily 7 < D.

N = 3Dq + 20, where q is also a non-negative integer, and 20 < 3D.

From the above two equations, we can deduce that DQ + 7 = 3Dq + 20, or D(Q - 3q) = 13.

It follows that D must be a factor of 13, and since D > 7, then D must be equal to 13.

From the second equality, multiplying by 2, we obtain 2N = 6 * 13q + 40 = (3 * 13)*(2q) + 3 * 13 + 1, which means that the remainder when 2N is divided by 3D = 3 * 13 is 1.

Answer A.

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