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Problems on H.C.F and L.C.M

Problems on H.C.F and L.C.M
he maximum number of students among them 1001 pens and 910 pencils can be distributed in such a way that each student gets the same number of pens and same number of pencils is:

1911
910
1001
91

ANSWER DOWNLOAD EXAMIANS APP

Problems on H.C.F and L.C.M
Find largest number of 4 digits divisible by 12, 15, 18, 27 ?

9720
12765
11340
10478

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

The largest 4 digit number is 9999.We know that , if LCM of given numbers divide a number N , then N is exactly divisible by all the given numbers .LCM of 12,15,18,27 is 540.On dividing 9999 by 540 we get 279 as remainder.Therefore number =9999 - 279 =9720.

Problems on H.C.F and L.C.M
The least number which when increased by 5 is divisible by each one of 24, 32, 36 and 54, is

427
859
869
4320

ANSWER DOWNLOAD EXAMIANS APP

Problems on H.C.F and L.C.M
The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is:

24
23
21
22

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

L.C.M. of 5, 6, 4 and 3 = 60. On dividing 2497 by 60, the remainder is 37.Number to be added = (60 - 37) = 23.

Problems on H.C.F and L.C.M
If LCM of two number is 693, HCF of two numbers is 11 and one number is 99, then find other

66
88
55
77

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

Product of two numbers = Product of their HCF and LCMSo Other number =(693×11/99)                              = 77

Problems on H.C.F and L.C.M
Which of the following has the most number of divisors?

99
101
182
176

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

Option A, 176 = 1 x 2 x 2 x 2 x 2 x 11.Option B, 182 = 1 x 2 x 7 x 13. Option C, 99 = 1 x 3 x 3 x 11. Option D, 101 = 1 x 101.Divisors of 99 are 1, 3, 9, 11, 33, 99.Divisors of 176 are 1, 2, 4, 8, 11, 16, 22, 44, 88 and 176. Divisors of 182 are 1, 2, 7, 13, 14, 26, 91 and 182.Hence, 176 have the most number of divisors.

MORE MCQ ON Problems on H.C.F and L.C.M

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