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:
3
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Step-by-step explanation:
Let n be the number
n divided by d leaves remainder 8
=> n = da + 8, for some a
n divided by 3d leaves remainder 21
=> n = 3db + 21, for some b
So
da + 8 = 3db + 21 => d ( a - 3b ) = 21 - 8 = 13
Since 13 is prime, this means that d = 13 ( supposing that d has to be positive ).
So
n = 3db + 21 = 39b + 21
=> 2n = 39(2b) + 42 = 39(2b) + 39 + 3 = 39 ( 2b + 1 ) + 3
=> 2n divided by 39 leaves remainder 3
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