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

Answer: (a)

We find LCM of 5, 6 and 8

5=5

6=3×2

8=$2^3$

= $2^3$ ×3 × 5 = 8 × 15 = 120

Required number = 120K + 3

∴ when K = 2, 120 × 2 + 3 = 243 required no.

It is completely divisible by 9

Answer: (c)

Using Rule 9,

The n digit largest number which when divided by a, b, c, leaves remainder ‘x’ will be

Required number = [n – digit largest number – R] + x

where, R is the remainder obtained when n – digit largest number is divided by the L.C.M of a, b, c.

We will find the LCM of 16, 24, 30 and 36

∴ LCM = 2 × 2 × 2 × 3 × 2 × 5 × 3 = 720

The largest number of five digits = 99999

On dividing 99999 by 720, the remainder = 639

∴ The largest five-digit number divisible by 720

= 99999 – 639 = 99360

∴ Required number = 99360 + 10

= 99370

Answer: (c)

∴ LCM = 2 × 2 × 5 × 7 × 8 = 1120

∴ Required number

= 5834 – 1120 = 4714

Answer: (d)

Required time = LCM of 200, 300, 360 and 450 seconds

= 1800 seconds

Answer: (a)

Using Rule 4,

When a number is divided by a, b or c leaving same remainder ‘r’ in each case then that number must be k + r where k is LCM of a, b and c.

LCM of 15, 12, 20, 54 = 540

Then number = 540 + 4 = 544 [4 being remainder]

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,12 – 2 = 10; 16 – 6 = 10; 24 – 14 = 10

Now, LCM of 12, 16 and 24 = 48

∴ The greatest 4–digit number exactly divisible by 48 = 9984

∴ Required number = 9984 – 10 = 9974