model 2 find lcm of numbers Practice Questions Answers Test with Solutions & More Shortcuts

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)

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

= 1800 seconds

Answer: (c)

15 = 3 × 5

18 = $3^2$ × 2

21 = 3 × 7

24 = $2^3$ × 3

LCM = 8 × 9 × 5 × 7 = 2520

The largest number of four digits = 9999

3$2439/2520$ = 9999

Required number = 9999 – 2439 – 4 = 7556

(Because 15 - 11 = 4

18 - 14 = 4

21 – 17 = 4

24 – 20 = 4)

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