Here (18 - 7) = 11, (21 - 10) = 11, (24 - 13) = 11
LCM of 18, 21 and 24 is 504
Let the required number be 504k - 11
Least value of k for which (504k - 11) is divisible by 23 is k = 6
∴ Required number = 504 × 6 - 11 = 3024 - 11 = 3013
LCM of 6, 9, 12, 15, 18 is = 180
If 180 is divided by these given number remainder will be 0
⇒ To leave the same remainder 2
⇒ The number will be = 180 + 2 = 182
Greatest number of three digits is 999. LCM of 6, 9 and 12 is 36
On dividing 999 by 36, the remainder obtained is 27.
So, required number = (999 - 27) + 3 = 975
For least or minimum number of cans we should have maximum capacity cans for required quantity
⇒ For this we take HCF of given quantities.
HCF (21, 42, 63) = 21
∴ Maximum capacity of a can = 21 litres
∴ Number of cans of cow milk