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.
$$\begin{gathered} x=5p+2 \\ \underset{―}{(+)136-x=8q+3} \\ 136+0=5p+8q+5 \end{gathered}$$
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
$\Rightarrow p={\displaystyle \frac{115}{5}}$
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
$\Rightarrow p={\displaystyle \frac{75}{5}}$
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.