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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)

• Option 2)

• Option 3)

• Option 4)

None of these

• Option 5)

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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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