Question

The largest 4 digit number divisible by 7 is
[a] 9993
[b] 9996
[c] 9992
[d]9990.

Answer 1

Hint: The least number to be subtracted from a number “b” such that it comes to be a multiple of another number “a” is the remainder of b obtained on division by a. If we subtract the largest 4 digit number by the smallest number such that the difference is divisible by 7, the resulting number will be the largest 4 digit number divisible by 7. The largest 4 digit number is 9999. Hence we find the smallest number which should be subtracted from 9999 so that the result is divisible by 7.

Complete step by step solution:
We know the least number to be subtracted from a number “b” such that it leaves it comes to be a multiple of another number “a” is the remainder of b obtained on division by a.
We know that the largest 4 digit number is 9999
Taking b = 9999 and a = 7, we get
We know that $9999=7\times 1428+3$
Hence by Euclid’s Division lemma, the remainder obtained on dividing 9999 by 7 is 3
Hence the remainder obtained on dividing 9999 by 7 is 3.
Hence the least number to be subtracted from =9999 so that it comes to be a multiple of 7 is 3
Hence the largest 4 digit number which is a multiple of 7 is $9999-3=9996$
Hence option [b] is correct

Note: If b = aq+ r where $0\le r < a$then the least non-negative number to be subtracted from b so that it comes to be a multiple of a is r.
Alternatively, we can check each of the options and check which of the options is correct.
Option [a]:
We have
$9993=7\times 1427+4$
Hence 9993 is not divisible by 7
Option [b]:
We have
$9996=7\times 1428$
Hence 9996 is divisible by 7
Option [c]:
We have
$9992=7\times 1427+3$
Hence, 9992 is not divisible by 7
Option [d]
We have $9990=7\times 1427+1$
Hence, 9990 is not divisible by 7
Hence option [b] is the only correct answer.