A number when divided by D leaves a remainder of 8 and when
divided by 3D leaves a remainder of 21 . What is the remainder
left, when twice the number is divided by 3D?
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Answer / sunil
let the no be
N = k1 * D + 8 = k2 * d +21
thus, D(k1 - 3K2)=13
as D cant be 1 and 13 is a prime no; D is 13
so, the smallest favorable no is 60
so, the remainder when twice the no is divided by 3D is 3
Answer is 3.
| Is This Answer Correct ? | 3 Yes | 0 No |
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