Question

Find the least number which when divided by 6,7,8,9 and 12 leaves the remainder 2 in each case

Answer 1

Given numbers,
6 , 7 , 8, 9 and 12
Prime factorization of,
6=2×3
7=1×7
8=2³
9=3²
12=2²×3
LCM=product of each prime factor of highest power
LCM=2³×3²×7=504
504 is the least number exactly divisible by 6,7,8,9 and 12.


Hence,
least number which when divided by 6,7,8,9 and 12 leaves remainder 2 in each case=504+2=506.

Answer 2

Answer:

Step-by-step explanation:

Given:

$6,7,8,9$, and $12 .$

To find the least number.

Step 1

The prime factors are:

$6=2*3$

$7=1*7$

$8=2*2*2$

$9=3*3$

$12=2*2*3$

Step 2

The least common multiple (LCM) of a set of numbers exists as the smallest number that leaves 0 as the remainder each time when divided by any of those digits.

$LCM}(6,7,8,912)=504$

Step 3

So, $504$ exists the least number that leaves 0 as the remainder each time when divided by any of $6,7,8,9$ and $12.$

So, the required number $=(504+2)=506$

Therefore $506$ will be the least number which when divided by $6,7,8,9$, $12$ leaves the remainder $2$ each time.

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