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

Answer: (c)

Using Rule 5,

When a number is divided by a, b or c, leaving remainders p, q or r respectively such that the difference between divisor and remainder in each case is the same

i.e., (a – P) = (b – q) = (c – r) = t (say) then that (least) number must be in the form of (k – t),

Where k is LCM of a, b and c

Here, 12 – 5 = 7, 16 – 9 = 7

∴ Required number = (L.C.M. of 12 and 16) – 7

= 48 – 7 = 41

Answer: (b)

LCM of 9, 10 and 15 = 90

⇒ The multiple of 90 are also divisible by 9, 10 or 15.

∴ 21 × 90 = 1890 will be divisible by them.

∴ Now, 1897 will be the number that will give remainder 7.

1936 – 1897

Required number = 1936 – 1897 = 39

Answer: (b)

Using Rule 4,

i.e. When a number is divided by x, y or z leaving same remainder ‘R’ in each case then that number must be K + R where k is LCM of x, y and z.

L.C.M. of 4, 6, 8, 12 and 16 = 48

∴ Required number = 48 + 2 = 50

Answer: (b)

Required distance = LCM of 63, 70 and 77 cm.

= 6930 cm.

Illustration :

∴ LCM = 7 × 9 × 10 × 11

= 6930

Answer: (d)

The difference between the divisor and the corresponding remainder is same in each case ie. 18 – 5 = 13, 27 – 14 = 13, 36 – 23 = 13

∴ Required number = (LCM of 18, 27, and 36 ) – 13

= 108 – 13 = 95

Answer: (c)

LCM of 5, 10, 12, 15

∴ LCM = 2 × 3 × 5 × 2 = 60

∴ Number = 60k + 2

Now, the required number should be divisible by 7.

Now, 60k + 2 = 7 × 8k + 4k + 2 If we put k = 3, (4k + 2) is equal to 14 which is exactly divisible by 7.

∴ Required number = 60 × 3 + 2 = 182