Let xbe the least number, which when divided by 5,6, 7 and 8 leaves a remainder 3 in each case but when divided by 9 leaves no remainder. The sum of digits of x is
Let xbe the least number, which when divided by 5,6, 7 and 8 leaves a remainder 3 in each case but when divided by 9 leaves no remainder. The sum of digits of x is 1). 21 2). 22 3). 18 4). 24
LCM of 5, 6, 7 and 8 = 840
LCM = 2 × 5 × 3 × 7 × 4 = 840
\ Required number = 840x + 3
which is divisible by 9 for a certain
least value of x.
Now,
840x + 3 = 93x × 9 + 3x + 3
3x + 3, is divisible by 9 for x = 2
Required number = 840 × 2
+ 3
= 1680 + 3 = 1683
Sum of digits = 1 + 6 + 8 + 3= 18