Question

What is the least multiple of 13 which when divided by 3, 4, 5, 8 and 10 leaves remainder 2 in each case?​

Answer 1

Answer:

962

Step-by-step explanation:

LCM of 3,4,5,8 and 10 is 120.

Now, (120k+2)%13 = 0

So, do hit and trial for value of k and check if it is divisible by 13.

for k = 8;

120*8+2 = 962 and

962%13 = 0.

Answer 2

Given : least multiple of 13 which when divided by 3, 4, 5, 8 and 10 leaves remainder 2 in each case

To find : Number

Solution:

Let say least multiple of 13 is 13N  

13N = 3A  + 2  => 13N - 2 =  3A

13N = 4B + 2   => 13N - 2 =   4B

13N = 5C  + 2   => 13N - 2 =  5C

13N = 8D  + 2   => 13N - 2 =  8D

13N = 10E + 2   => 13N - 2 = 10E

13N-2   multiple of LCM of  3 , 4 , 5 , 8 & 10

LCM = 120

13N -2= 120P

=> 13N  = 120P + 2

=>  13N = 117P  + 3P  +2

=> 13(N  - 9P)  = 3P + 2

P = 8

13N  = 120P + 2

= 120 * 8 + 2

= 962

962 is  the least multiple of 13 which when divided by 3, 4, 5, 8 and 10 leaves remainder 2 in each case

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