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

Answer: (c)

Required number of students = LCM of 6, 8, 10 = 120

Answer: (d)

The difference between divisor and the corresponding remainder is equal.

LCM of 3, 5, 7 and 9 = 315 Largest 4-digit number = 9999

31$234/315$ = 9999

∴ Number divisible by 315 = 9999 – 234 = 9765

Required number = 9765 – 2 = 9763

Answer: (a)

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.

The greatest number of five digits is 99999.

LCM of 3, 5, 8 and 12

∴ LCM = 2 × 2 × 3 × 5 × 2 = 120 After dividing 99999 by 120, we get 39 as remainder 99999 – 39 = 99960 = (833 × 120)

99960 is the greatest five digit number divisible by the given divisors.

In order to get 2 as remainder in each case we will simply add 2 to 99960.

∴ Greatest number = 99960 + 2 = 99962

Answer: (a)

We find LCM of = 10, 16, 24

∴ LCM = $2^2$ × $2^2$ × 3 × 5

∴ Required number

= 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5

= 3600

Answer: (c)

LCM of 4, 6, 8, 9

∴ LCM = 2 × 2 × 3 × 2 × 3 = 72

∴ Required number = 72, because it is exactly divisible by 4, 6, 8 and 9 and it leaves remainder 7 when divided by 13.

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