If A is the smallest three digit number divisible by both 6 and 7 and B is the largest four digit number divisible by both 6 and 7, then what is the value of B – A? (SSC MTS 2018)

a)
9912

b)
9870

c)
9996

d)
9954

Correct answer is option 'B'. Can you explain this answer?

Question

If A is the smallest three digit number divisible by both 6 and 7 and B is the largest four digit number divisible by both 6 and 7, then what is the value of B – A? (SSC MTS 2018)

a)
9912
b)
9870
c)
9996
d)
9954

Correct answer is option 'B'. Can you explain this answer?

Answer

To be divisible by both 6 and 7, a number must be divisible by their least common multiple, which is 42.

The smallest three digit multiple of 42 is $2\cdot 42 = 84$, and the largest four digit multiple of 42 is $238\cdot 42 = 9996$.

Therefore, $A=84$ and $B=9996$.

The value of B is $\boxed{9996}$.

Answer

Answer

First of all inorder for that number to be divisible by 6&7 the number must also be divisible by 42 which is 7×6
42×3 gives 84 so 42×3=126 is the smallest three digit number divisible by 6 &7
and now we have to find out the largest four digit number divisible by 42
for that we can use divisibility
largest 4 digit number is 9999 but that number is not divisible by 6&7
inorder for a number to be divisible by 6 its 1 digit must be an even number now we divide 9998 (because it is the next largest number)
by 42 we get remainder as 2
now subtract 2 from 9998 which is 9996 that is the largest 4 digit number divisible by 42
now back to the question
the answer is 9996-126= 9870

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