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

Answer: (b)

Using Rule 4,

LCM of 4, 5, 6, 7 and 8

=

= 2 × 2 × 2 × 3 × 5 × 7 = 840.

let required number be 840 K + 2 which is multiple of 13.

Least value of K for which (840 K + 2) is divisible by 13 is K = 3

∴ Required number

= 840 × 3 + 2

= 2520 + 2 = 2522

Answer: (c)

Using Rule 8,

Greatest n digit number which when divided by three numbers A, B, C leaves no remainder will be

Required Number = (n – digit greatest number) – R

R is the remainder obtained on dividing greatest n digit number by L.C.M of A, B, C.

The largest number of 4-digits is 9999. L.C.M. of divisors

LCM = 2 × 2 × 3 × 3 × 3 × 5 = 540

Divide 9999 by 540, now we get 279 as remainder.

9999 – 279 = 9720

Hence, 9720 is the largest 4-digit number exactly divisible by each of 12, 15, 18 and 27.

Answer: (a)

Required time = LCM of 252, 308 and 198 seconds

∴ LCM = 2 × 2 × 7 × 9 × 11 = 2772 seconds

= 46 minutes 12 seconds

Answer: (d)

Required time = LCM of 48, 72 and 108 seconds

∴ LCM = 2 × 2 × 2 × 2 × 3 × 3 × 3 = 432 seconds

= 7 minutes 12 second

∴ Required time = 10 : 07 : 12 hours

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: (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