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