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51. The least perfect square which is divisible by each of 21, 36 and 66 is = ?
= 21 × 12 × 11
= 7 × 3 × 4 × 3 × 11
= 7 × 3 × 2 × 2 × 3 × 11
For perfect square multiply by 7 × 11
So that pairs of number from perfect square
∴ 7 × 7 × 3 × 3 × 2 × 2 × 11 × 11
Required result is
= 213444 (which is a perfect square)
Solution:
L.C.M. of (21, 36, 66)= 21 × 12 × 11
= 7 × 3 × 4 × 3 × 11
= 7 × 3 × 2 × 2 × 3 × 11
For perfect square multiply by 7 × 11
So that pairs of number from perfect square
∴ 7 × 7 × 3 × 3 × 2 × 2 × 11 × 11
Required result is
= 213444 (which is a perfect square)
55. Let the least number of six digits which when divided by 4, 6, 10, 15 leaves in each case same remainder 2 be N. The sum of digits in N is = ?
LCM = 2 × 2 × 3 × 5 = 60
⇒ Least number of six digit = 100000
⇒ Divide 100000 by 60 we get remainder 40
⇒ Least six digit number which is divisible by (4, 6, 10, 15)given number is
[100000 + (60 - 40)] = 100020
∴ N ⇒ 100020 + 2 = 100022
∴ Sum of digits = 1 + 0 + 0 + 0 + 2 + 2 = 5
Solution:
LCM (4, 6, 10, 15)LCM = 2 × 2 × 3 × 5 = 60
⇒ Least number of six digit = 100000
⇒ Divide 100000 by 60 we get remainder 40
⇒ Least six digit number which is divisible by (4, 6, 10, 15)given number is
[100000 + (60 - 40)] = 100020
∴ N ⇒ 100020 + 2 = 100022
∴ Sum of digits = 1 + 0 + 0 + 0 + 2 + 2 = 5
56. The smallest square number divisible by 10, 16 and 24 is ?
= 5 × 2 × 8 × 3 = 240
⇒ for square number split the LCM into its factors
= 5 × 2 × 2 × 2 × 2 × 3
= 5 × 5 × 2 × 2 × 2 × 2 × 3 × 3
= 3600
Solution:
LCM (10, 16, 24)= 5 × 2 × 8 × 3 = 240
⇒ for square number split the LCM into its factors
= 5 × 2 × 2 × 2 × 2 × 3
= 5 × 5 × 2 × 2 × 2 × 2 × 3 × 3
= 3600
57. The G.C.D. (greatest common divisor) of 1.08, 0.36, 0.9 is = ?
HCF of 108, 36 and 90 is 18
∴ HCF of given numbers = 0.18
Solution:
Given numbers are 1.08, 0.36 and 0.90HCF of 108, 36 and 90 is 18
∴ HCF of given numbers = 0.18
58. HCF of 3240, 3600, and a third number is 36 and their LCM is 24 × 35 × 52 × 72 . The third number is = ?
3600 = 24 × 32 × 52
HCF = 36 = 22 × 32
Since H.C.F. is the product of lowest powers of common factors, so the third number must have (22 × 32) as its factors.
Since L.C.M. is the product of highest powers of common prime factors, so the third number must have 35 and 72 as its factors.
∴ Third number = 22 × 35 × 72
Solution:
3240 = 23 × 34 × 53600 = 24 × 32 × 52
HCF = 36 = 22 × 32
Since H.C.F. is the product of lowest powers of common factors, so the third number must have (22 × 32) as its factors.
Since L.C.M. is the product of highest powers of common prime factors, so the third number must have 35 and 72 as its factors.
∴ Third number = 22 × 35 × 72
59. From a point on a circular track 5km long A,B and C started running in the same direction at the same time with speed of km per hour, 3 km per hour and 2 km per hour respectively. Then on the starting point all three will meet again after ?
They will meet again after 10 hours
Solution:
They will meet again after 10 hours
60. Three numbers which are co-prime to one another are such that product of the first two is 551 and that of the last two is 1073. The sum of the three numbers is = ?
Solution: