Question 103

If r is the remainder when each of 4749, 5601 and 7092 is divided by the greatest possible number d(>1), then the value of (d + r) will be:

Solution

Here, $$d$$ = H.C.F. of (5601-4749), (7092-5601) and (7092-4749)

= H.C.F. (852, 1491, 2343)

Prime factorisation of :

852 = $$2^2\times3\times71$$

1491 = $$3\times7\times71$$

2343 = $$3\times11\times71$$

=> H.C.F. = $$d=3\times71=213$$

Now, when any number say 4749 is divided by 213, remainder, => $$4749=213\times22+63$$

=> $$r=63$$

$$\therefore$$ $$d+r=213+63=276$$

=> Ans - (A)

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