Question

A number lying in between 1000 and 2000 is such that in the division by 2, 3, 4, 5, 6, 7, and 8 leaves remainders 1, 2, 3, 4, 5, 6, and 7 respectively. Find that number?
(a) 518
(b) 416
(c) 364
(d) 1679

Answer 1

Hint: We start solving the problem by assuming the required number as ‘x’. We convert ‘x’ to the form $dividend=\left( divisor\times quotient \right)+remainder$ for the dividends 2, 3, 4, 5, 6, 7 and 8. On doing this we can see x+1 is divisible by each number. We use L.C.M to the find the value of x+1, which makes us eventually to fins the value of x.

Complete step-by-step solution:
Given that we have a number lying in between 1000 and 2000. It is also given that if the number is divided by 2, 3, 4, 5, 6, 7 and 8 we get remainders 1, 2, 3, 4, 5, 6 and 7 respectively.
Let us assume the number be x. We know that $dividend=\left( divisor\times quotient \right)+remainder$.
We have given that if the number ‘x’ is divided by 2, it leaves remainder 1. Let us assume the quotient be p.
So, $x=2p+1$.
$\Rightarrow $ $x=2p+2-1$.
$\Rightarrow $ $x+1=2\left( p+1 \right)$ ---(1).
We have given that if the number ‘x’ is divided by 3, it leaves remainder 2. Let us assume the quotient be q.
$\Rightarrow $ $x=3q+2$.
$\Rightarrow $ $x=3q+3-1$.
$\Rightarrow $ $x+1=3\left( q+1 \right)$ ---(2).
We have given that if the number ‘x’ is divided by 4, it leaves remainder 3. Let us assume the quotient be r.
$\Rightarrow $ $x=4r+3$.
$\Rightarrow $ $x=4r+4-1$.
$\Rightarrow $ $x+1=4\left( r+1 \right)$ ---(3).
We have given that if the number ‘x’ is divided by 5, it leaves remainder 4. Let us assume the quotient be s.
$\Rightarrow $ $x=5s+4$.
$\Rightarrow $ $x=5s+5-1$.
$\Rightarrow $ $x+1=5\left( s+1 \right)$ ---(4).
We have given that if the number ‘x’ is divided by 6, it leaves remainder 5. Let us assume the quotient be t.
$\Rightarrow $ $x=6t+5$.
$\Rightarrow $ $x=6t+6-1$.
$\Rightarrow $ $x+1=6\left( t+1 \right)$ ---(5).
We have given that if the number ‘x’ is divided by 7, it leaves remainder 6. Let us assume the quotient be u.
$\Rightarrow $ $x=7u+6$.
$\Rightarrow $ $x=7u+7-1$.
$\Rightarrow $ $x+1=7\left( u+1 \right)$ ---(6).
We have given that if the number ‘x’ is divided by 8, it leaves remainder 7. Let us assume the quotient be v.
$\Rightarrow $ $x=8v+7$.
$\Rightarrow $ $x=8v+8-1$.
$\Rightarrow $ $x+1=8\left( v+1 \right)$ ---(7).
From equations (1), (2), (3), (4), (5), (6) and (7), we can see that the number ‘x+1’ is divided by all the numbers 2, 3, 4, 5, 6, 7 and 8.
We know that the number divided by the numbers a, b, c, d, e is equal to the L.C.M of all the numbers.
So, let us find the L.C.M of the numbers 2, 3, 4, 5, 6, 7 and 8.
\[\begin{align}
  & 2\left| \!{\underline {\,
  \begin{matrix}
   2 & 3 & 4 & 5 & 6 & 7 & 8 \\
\end{matrix} \,}} \right. \\
 & 3\left| \!{\underline {\,
  \begin{matrix}
   1 & 3 & 2 & 5 & 3 & 7 & 4 \\
\end{matrix} \,}} \right. \\
 & 2\left| \!{\underline {\,
  \begin{matrix}
   1 & 1 & 2 & 5 & 1 & 7 & 4 \\
\end{matrix} \,}} \right. \\
 & \left| \!{\underline {\,
  \begin{matrix}
   1 & 1 & 1 & 5 & 1 & 7 & 2 \\
\end{matrix} \,}} \right. \\
\end{align}\].
So, L.C.M of the numbers 2, 3, 4, 5, 6, 7 and 8 is $2\times 3\times 2\times 5\times 7\times 2$.
L.C.M of the numbers 2, 3, 4, 5, 6, 7 and 8 is 840.
Since the given number ‘x’ lies in between 1000 and 2000 and we don’t have the number 840 present in between 1000 and 2000.
So, ‘x+1’ can be one the multiples of 840. We can see that the 1680 is multiple of 840 and is obtained when 840 is multiplies with 2.
We get the value of ‘x+1’ is 1680.
But we need the value of x which is 1680-1.
We found the value of x is 1679.
∴ The value of x is 1679.
The correct option for the given problem is (d).

Note: Whenever we get the problems involving division and remainders with many numbers, we use $dividend=\left( divisor\times quotient \right)+remainder$ to find whether we get any relation with the number and the dividends. We can also expect problems to get all values of ‘x’ that lies between 1000 and 5000, 1000 and 10000.