Question

Divide 136 into two parts one of which when divided by 5 leaves remainder 2 and the other divided by 8 leaves remainder 3.

Answer 1

Hint:
We can take the two parts as x and 132 – x. Then we can take the quotient of the 1st part as p and the quotient of the \[{2^{nd}}\] part as q when divided with 5 and 8 respectively. Then we can form equations for each part with quotient and remainder. Then we can eliminate x by subtracting the 2 equations. Then we can solve for p and q using the hit and trial method. Then we can substitute these values to find the required parts.

Complete step by step solution:
We can take the 2 parts of 136 as x and 132 – x.
It is given that \[{1^{st}}\] part when divided by 5 leaves remainder 2. Let p be the quotient when divided with 5. So, we can write the \[{1^{st}}\] part as,
$x = 5p + 2$ … (1)
It is given that ${2^{nd}}$ part when divided by 8 leaves remainder 3. Let p be the quotient when divided with 5. So, we can write the ${2^{nd}}$ part as,
$136 - x = 8q + 3$ … (2)
Now, we can add equations (1) and (2) to eliminate x.
$
  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 5p + 2 \\
  \underline {\,\left( + \right)136 - x = 8q + 3} \\
  \,\,\,\,\,\,\,\,\,136 + 0 = 5p + 8q + 5 \\
 $
On adding 5 on both sides, we get
$ \Rightarrow 136 - 5 = 5p + 8q$
On simplification, we get
$ \Rightarrow 131 = 5p + 8q$
On rearranging, we get
$ \Rightarrow 5p = 131 - 8q$
Now we can find the value of p and q by hit and trial method.
We can see that RHS is a multiple of 5. So, it must end with either 0 or 5. But we know that multiple of 8 is always even. So, we can take the values of q such that 8q ends in 6.
Let q = 2,
$ \Rightarrow 5p = 131 - 8 \times 2$
On simplification, we get
$ \Rightarrow 5p = 131 - 16$
On dividing the equation by 5, we get
$ \Rightarrow p = \dfrac{{115}}{5}$
Hence, we have
$ \Rightarrow p = 23$
 On substituting the value of p in equation (1), we get
$ \Rightarrow x = 5 \times 23 + 2$
On simplification, we get
$ \Rightarrow x = 115 + 2$
Hence, we have
$ \Rightarrow x = 117$
Now the ${1^{st}}$ part is 117. The \[{2^{nd}}\] part is given by,
$136 - 117 = 19$

Therefore, the required parts of the number 136 are 117 and 19.

Note:
We must eliminate x and then solve for p and q by hit and trial method. Even though it is not stated, we must know that p and q are natural numbers by the theorem of division algorithm. The values of p and q may not be unique. Another value of p and q is given by,
We have the equation $5p = 131 - 8q$
Now we can find the value of p and q by hit and trial method.
We can see that RHS is a multiple of 5. So, it must end with either 0 or 5. But we know that multiple of 8 is always even. So, we can take the values of q such that 8q ends in 6
Let q = 7,
$ \Rightarrow 5p = 131 - 8 \times 7$
On simplification, we get
$ \Rightarrow 5p = 131 - 56$
On dividing the equation by 5, we get
$ \Rightarrow p = \dfrac{{75}}{5}$
So, we have
$ \Rightarrow p = 15$
 On substituting the value of p in equation (1), we get
$ \Rightarrow x = 5 \times 15 + 2$
On simplification, we get
$ \Rightarrow x = 75 + 2$
So, we have
$ \Rightarrow x = 77$
Now the \[{1^{st}}\] part is 77. The \[{2^{nd}}\] part is given by,
$136 - 77 = 59$
Now, the 2 parts are 77 and 59.