Find the least number which when divided by 5, 6, 7, 8, leaves a remainder 3, and is also a multiple of 9.

Correct option is A

Given:
The least number which when divided by 5, 6, 7 and 8 leaves a remainder 3
The least number when divided by 9 remainder = 0
Concept Used:
For the least number take LCM and also check that all conditions are satisfying or not
For same remainder in each case add the remainder in the LCM

​LCM of 5, 6, 7, and 8 + remainder condition ​
Solution:
LCM of 5, 6, 7, 8 is 23 × 31 × 51 × 71 = 840
For same remainder we have to add 3 to the LCM
=> 840 + 3
Now check weather it is divisible by 9 or not
843/9 = Not divisible by 9
Now we have to take next multiple of LCM and then add 3
840 × 2 + 3
=> 1683
Divide it by 9
1683/9 = 187
=> 1683 is divisible by 9
=> 1683 is the least number when divided by 5, 6, 7, 8 gives remainder 3
Alternate Method:
The least number must be divisible by 9
For divisibility of 9 sum of digits of the number must be divisible by 9
Checking it by options:
Option a) 1677
1 + 6 + 7 + 7 = 21
=> 1677 is not divisible by 9
Option b) 1683
1 + 6 + 8 + 3 = 18
=> 1683 is divisible by 9
Option c) 2523
2 + 5 + 2 + 3 = 12
=> 2523 is not divisible by 9
Option d) 3363
3 + 3 + 6 + 3 = 15
=> 3363 is not divisible by 9
∴ The least number is 1683 which when divided by 5, 6, 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder.