Question

What is the least number which When divided by the numbers, 3, 5, 6, 10. and 12 leaves a remainder 2 in each case?​

Answer 1

Explanation:

The given numbers are 3, 5, 6, 8, 10 and 12

Take LCM of them we get,

⇒ LCM of 3, 5, 6, 8, 10 and 12 =2×2×2×3×5=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 =(120k+2)

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,

⇒ Required number =120×8+2=962

Hence the least number which leaves remainder 2 when divided by 3, 5, 6, 8, 10 and 12 is 962.

Answer 2

Answer:

Required number =120×8+2=962

Explanation: