​​Find the least positive number, which when divided by 5, 6, 8, 9 and 12, gives 1 as a remainder in each case and is completely divisible by 13.

Correct option is A

Given:

Smallest number giving remainder 1 when divided by 5,6,8,9 and 12

But no remainder when divided by 13

Formula Used:

LCM is the smallest number when divided by numbers gives remainder zero

Solution:

The LCM of (5,6,8,9,12 ) is 360

The number when divided by 5,6,8,9 and 12 leaves a remainder.

Hence the number is 360+1 = 361

But the number should by divisible by 13 hence the number will be LCM of 361 and 13

Using Euclid’s division lemma(a = bq +r) we get:

x = 360q +1   ….(1)

We substitute the value of q and check till we get number divisible by 13

Putting q = 1 in eq(1)

x = 360(1) +1 = 361

361 is not divisible by 13

Putting q = 2 in eq(1)

x= 360(2) +1 = 721

721 is not divisible by 13

Likewise when we put q = 10

x = 360(10) + 1  = 3601

Now 3601 is divisible by 13

Alternative Method:

The above method is lengthy hence we use trial method using the options:\

Option are 3614, 3640,3627 and 3601

So the number should be divisible by 13 and leaves remainder when divided by 5,6,8,9 and 12

Number 3614 is divisible by 13 but leaves remainder 4 when divided by 5 hence it does not satisfies both condition

Number 3640 is also divisible by 13 but also divisible by 5 leaving no remainder

Hence 3640 does not satisfies the condition

Number 3627 is also divisible by 13 but leaves remainder 2 when divided by 5

Hence it does not satisfies the condition

Number 3601 is divisible by 13 and leaves remainder 1 when divided by 5,6,8,9 and 12 hence

3601 is the correct answer