model 2 find lcm of numbers Section-Wise Topic Notes With Detailed Explanation And Example Questions

Answer: (c)

LCM of 24, 36 and 54 seconds

= 216 seconds

= 3 minutes 36 seconds

Answer: (d)

Using Rule 5,

When a number is divided by P, Q or R leaving remainders X, Y or Z respectively

Such that the difference between divisor and remainder in each case is the same

i.e., (P – X) = (Q – Y) = (R – Z) = T

(say) then that (least) number must be in the form of (K – T),

where K is LCM of P, Q and R

Here 4 – 1 = 3, 5 – 2 = 3, 6 – 3 = 3

∴ The required number = LCM of (4, 5, 6) – 3

= 60 – 3 = 57

Answer: (c)

LCM of 4, 6, 10, 15 = 60

Least number of 6 digits = 100000

The least number of 6 digits which is exactly divisible by 60 = 100000 + (60 – 40) = 100020

∴ Required number (N) = 100020 + 2 = 100022

Hence, the sum of digits = 1 + 0 + 0 + 0 + 2 + 2 = 5

Answer: (c)

The LCM of 5, 6, 7 and 8 = 840

∴ Required number = 840 k + 3 which is exactly divisible by 9 for some value of k.

Now, 840 k + 3 = 93 × 9 k + (3k + 3)

When k = 2, 3k + 3 = 9, which is divisible by 9.

∴ Required number = 840 × 2 + 3 = 1683

Answer: (b)

Using Rule 5,

When a number is divided by P, Q or R leaving remainders X, Y or Z respectively such that the difference between divisor and remainder in each case is same i.e., (P – X) = (Q – Y) = (R – Z) = T (say) then that (least) number must be in the form of (K – T), where K is LCM of P, Q and R

The smallest number divisible by 12 or 10 or 8

= LCM of 12, 10 and 8 = 120

⇒ Required number =120 + 6 = 126

Answer: (c)

The smallest number divisible by 16, 20 and 24

= LCM of 16, 20 and 24

∴LCM = 2×2×2×2×5×3

= $2^2$ × $2^2$ × 5 × 3

∴Required complete square number =$2^2$ × $2^2$ × $5^2$ × $3^2$ = 3600