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Q.

Find the largest number which divides 438 and 606 and leaves remainder 6 in each case.


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a

24

b

26

c

30

d

48 

answer is A.

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Detailed Solution

Given are the numbers 438 and 606.
We have to find the largest possible number which when divides both the given numbers leaves the remainder 6, for both the cases.
This implies that for 438 and 606 to be exactly divisible by the largest number, the remainder must be zero.
So, we will subtract the remainder 6 from each of the two numbers.
438−6=432 606−6=600
Now, this largest number will be the HCF of 432 and 600.
So, we proceed with the prime factorization of 432 and 600, to get their HCF as follows,
432=2×2×2×2×3×3×3 600=2×2×2×3×5×5
So, we can observe that the product of the common factors among the two numbers is 2×2×2×3 .
HCF (432, 600) = 2×2×2×3
= 24
So, the largest number which divides 438 and 606, leaving remainder 6 in each case is 24.
Hence, the correct option is (1).
 
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