Question

What is the least number which when divided by the numbers 3, 5, 6, 8, 10 and 12 leaves in each case remainder 2 but when divided by 13 leaves no remainder
A.312
B.962
C.1562
D.1586

Answer 1

Hint: Hint- To find the least number we have to calculate the the LCM of the given numbers first then we will set the number as 120k+2 as in the question given that remainder is 2. Then check for what value of k the number is divisible by 13. At last put the value of k and find the required least number.

Complete step-by-step answer:
Complete step by step solution:
The given numbers are 3, 5, 6, 8, 10 and 12
Take LCM of them we get,
 $ \Rightarrow $ LCM of 3, 5, 6, 8, 10 and 12 \[ = 2 \times 2 \times 2 \times 3 \times 5 = 120\]
As question says that the least number leaves remainder 2 when divided by the numbers 3, 5, 6, 8, 10 and 12
So the required number we can write as
∴Required number \[ = \left( {120k + 2} \right)\]
Question also says that the required number is also divisible by the 13
So it can be possible when the value of k is equal to 8
(120k+2) is divisible by 13 then k =8
On putting the value of k in the (120k+2)
We get,
 $ \Rightarrow $ Required number \[ = 120 \times 8 + 2 = 962\]
Hence the least number which leaves remainder 2 when divided by 3, 5, 6, 8, 10 and 12 is 962.
So, the correct answer is “Option B”.

Note: Question asked that the least number there can be more than one number that can be divisible by numbers 3, 5, 6, 8, 10 and 12 and leaves the remainder 2. So students should be aware of that.