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

Problems on H.C.F and L.C.M
Let the least number of six digits, which when divided by 4, 6, 10 and 15, leaves in each case the same remainder of 2, be N. The sum of the digits in N is:

5
3
4
6

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Problems on H.C.F and L.C.M
Two numbers 4242 and 2903 when divided by a certain number of three digits, leave the same remainder. Find the number?

107
113
103
101

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Problems on H.C.F and L.C.M
The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is:

107
111
101
185

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

 Let the numbers be 37a and 37b.Then, 37a x 37b = 4107 ab = 3.Now, co-primes with product 3 are (1, 3).So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111). Greater number = 111.

Problems on H.C.F and L.C.M
Reduce 391/667 to lowest terms .

23/37
17/29
11/13
31/45

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

H.C.F. of 391 and 667 is 23.On dividing the numerator and denominator by 23, we get :391 = 391 ¸23 = 17667    667¸ 23    29

Problems on H.C.F and L.C.M
The least number of four digits which is divisible by 4, 6, 8 and 10 is?

1080
1095
1085
1075

ANSWER DOWNLOAD EXAMIANS APP

Problems on H.C.F and L.C.M
A drink vendor has 80 liters of Maaza, 144 liters of Pepsi and 368 liters of Sprite. He wants to pack them in cans, so that each can contains the same number of liters of a drink, and doesn't want to

37
30
42
35

ANSWER DOWNLOAD EXAMIANS APP
MORE MCQ ON Problems on H.C.F and L.C.M

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