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A number when successively divide by 3, 5 and

A number when successively divide by 3, 5 and 8 leaves remainders 1, 4 and 7 respectively. Find the respective remainders if the order of divisors be reversed.

  • Option 1)

    5, 4, 2

  • Option 2)

    6, 2, 4

  • Option 3)

    3, 1,4

  • Option 4)

    None of these

  • Option 5)

    1, 1, 3

Answers (1)
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Let us say that a number N, when successively divided by a, b, and c leaves a remainder of p, q, and r
=> Before the last division by c, the number must have been of the format of ck + r. Here k is a natural number.
Same logic can be extended to give the value of N.
\Rightarrow N= a\left [ b\left ( ck+r \right ) +q\right ]+p
In this case:
N = 3[5(8k + 7) + 4] + 1
=> N = 3[40k + 35 + 4] + 1
=> N = 3[40k + 39] + 1
=> N = 120k + 118
Now, we need to calculate the remainders when the number is successively divided by 8, 3 and 5. 
The question will be simpler if I just assume some value of 'k'
Let \: us \: put\: k = 0
=> \: N = 118
=> \frac{118}{8} = 14 + remainder \: of \: 6
=> \frac{14}{3} = 4 + remainder\: of \: 2
=> \frac{4}{5}= 0 + remainder\: of \: 4

So, the\: remainders\: are\: 6, 2, and 4

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