If 4 is divided into the positive integer x, it leaves a remainder of 3. If 9 is divided into x, it leaves a remainder of 4. If y is a positive integer such that x + y is divisible by 36, what is the smallest possible value of y?
(A) 4
(B) 5
(C) 7
(D) 33
(E) 36
OA is B
pls, How could B be right? I need a profound explanation from an Expert
Hi Roland2rule,
Let's take a look at your question.
If 4 is divided into the positive integer x, it leaves a remainder of 3.
If p represents the quotient, then,
$$4p+3=x...(i)$$
If 9 is divided into x, it leaves a remainder of 4.
If q represents the quotient, then,
$$9q+4=x...(ii)$$
Using equation(i) and (ii),
$$4p+3=9q+4$$
$$4p=9q+4-3$$
$$4p=9q+1$$
$$p=\frac{9}{4}q+\frac{1}{4}$$
On test and trial basis, we can find the value of p.
Since p and q are the quotients so these are integers, lets put q = 1 to find if the value of P comes out to be an integer.
$$p=\frac{9}{4}\left(1\right)+\frac{1}{4}=\frac{9}{4}+\frac{1}{4}=\frac{10}{4}=\frac{5}{2}$$
Which is not an integer, so q can not be equal to 1.
Let's try q = 2,
$$p=\frac{9}{4}\left(2\right)+\frac{1}{4}=\frac{18}{4}+\frac{1}{4}=\frac{19}{4}$$
Which is also not an integer, hence q can not be 2.
Let's try q = 3,
$$p=\frac{9}{4}\left(3\right)+\frac{1}{4}=\frac{27}{4}+\frac{1}{4}=\frac{28}{4}=7$$
Which is an integer, hence,
$$q=3,\ p=7$$
We can now find x using eq(i),
$$4p+3=x$$
$$4(7)+3=x$$
$$28+3=x$$
$$x=31$$
The question states, " If y is a positive integer such that x + y is divisible by 36, what is the smallest possible value of y".
since x = 31, therefore, we need to find 31+y is divisible by 36.
The smallest possible value of y should be 5, so it will be divisible by 36.
therefore, Option
B is correct.
Hope it helps.
I am available if you'd like any follow up.
