Question

The least number which when divided by $2,3,4,5$ and $6$ leaves the remainder$1$ in each case. If the same number is divided by $\mathbf{7}$ it leaves no remainder. The number is

Answer 1

Given

The numbers $2,3,4,5$ and $6$

Find out

We have to determine the least number which when divided by $2,3,4,5$ and $6$ leaves $1$ as a remainder and no remainder when divided by $7$

Solution

Let us determine the $LCM$ of the numbers $2,3,4,5$ and $6$

The value of $LCM$$(2,3,4,5,6)$ obtained will be the smallest number that is exactly divisible by $2,3,4,5$ and $6$

Multiples of $2=2,4,6,8,10$——— $54,56,58,60$ and so on
Multiples of $3=3,6,9,12,15$ ——— $48,51,54,57,60$ and so on
Multiples of $4=4,8,12,16,20$——— $44,48,52,56,60$ and so on
Multiples of $5=5,10,15,20,25,30$——— $45,50,55,60$ and so on
Multiples of $6=6,12,18,24,30$——— $48,54,60$ and so on

$60$ is the least common multiple for the numbers $2,3,4,5$ and $6$
Hence, the LCM of $2,3,4,5$ and $6$ is $60$.

Since the required number leaves the remainder $1$ when it is divisible by $2,3,4,5,6$ and no remainder when it is divisible by $7.$

Therefore the required number is of the form $60x+1$

$60x+1$ is a multiple of $7$

Now, let’s verify for $60x+1$ is divisible by $7$by substituting the natural numbers sequentially,

$60\left(1\right)+1=61=(7\times 8)+3$ is not divisible by $7.$

$60\left(2\right)+1=121=(7\times 17)+2$ is not divisible by $7.$

$60\left(3\right)+1=181=(7\times 25)+6$ is not divisible by $7.$

$60\left(4\right)+1=241=(7\times 34)+3$ is not divisible by $7.$

$60\left(5\right)+1=301=(7\times 43)+0$ is divisible by $7.$

$301$is divisible by $7$

Hence, The least number is $301$ when divided by $2,3,4,5,6$ leaves a remainder $1$, but when divided by $7,$ there will be no remainder.