Problems on H.C.F and L.C.M - 02

Explanation:-

LCM of 12, 16, 18, 21 = 1008
Next number = 1008 × 2 = 2016
Divisible by all
∴ 16 is added
Sum of digits = 1 + 6 = 7

Explanation:-

Clearly, (4 - 1) = 3, (5 - 2) = 3, (6 - 3) = 3, (7 - 4) = 3 and (8 - 5) = 3
LCM of 4, 5, 6, 7, 8 = 840
Greatest number of 4 digits = 9999
On dividing 9999 by 840, the remainder is 759
So, required number = (9999 - 759) - 3 = 9237

Explanation:-

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

Explanation:-

Here (12 - 7) = 5, (15 - 10) = 5 and (16 - 11) = 5
Required number = (LCM of 12, 15, 16) - 5
= 240 - 5
= 235

Explanation:-

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

Explanation:-

LCM of (6, 7, 8, 9, 12)
LCM = 3 × 2 × 7 × 4 × 3 = 504
They will toll after every 504 seconds

Explanation:-

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

Explanation:-

= (LCM of 24, 32, 36, 54) - 5
= 864 - 5 = 859

Explanation:-

HCF (GCD) of a, b number is 12
and a > b > 12 (given)
∴ Smallest value of a & b are (36, 24)

Explanation:-

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
$${\text{ = }}\frac{{21}}{{21}}{\text{ = 1}}$$ ∴ Number of cans of toned milk $${\text{ = }}\frac{{42}}{{21}} = 2$$ ∴ Number of cans of double toned milk $${\text{ = }}\frac{{63}}{{21}} = 3$$ ∴ Total number of cans = 1 + 2 + 3 = 6