M30-17

Official Solution:


When a positive integer \(x\) is divided by a positive integer \(y\), the quotient is 3 and the remainder is \(z\). When \(z\) is divided by \(y\), the remainder is 2. Which of the following could be the value of \(x\)?

I. 5

II. 8

III. 32


A. I only
B. II only
C. III only
D. I and II only
E. I, II and III


We are given that a positive integer \(x\) divided by a positive integer \(y\) gives a quotient of 3 and a remainder of \(z\). This can be expressed as \(x = 3y + z\), where \(0 \leq z < y\) (since the remainder is always less than the divisor).

The problem also states that when \(z\) is divided by \(y\), the remainder is 2. Since the remainder is always less than the divisor, it follows that \(2 < y\). Furthermore, we know that when the divisor (\(y\) in this case) is greater than the dividend (\(z\) in this case), the remainder must equal the dividend. Therefore, we can deduce that \(z = 2\).

Substituting \(z = 2\) in \(x = 3y + z\), we get \(x = 3y + 2\). As \(y\) is a positive integer and \(2 < y\), the smallest possible value for \(y\) is 3. Thus, the minimum value of \(x\) is \(x = 3*3 + 2 = 11\). Therefore, \(x\) cannot be 5 or 8.

To check if \(x\) could be 32, we can set \(y = 10\), which satisfies the condition that \(2 < y\). Substituting \(y = 10\) in \(x = 3y + 2\), we get \(x = 3*10 + 2 = 32\). Therefore, \(x\) could be 32, and option III is a possible value for \(x\).


Answer: C
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Alternative approach.

Solving problems of such a type we have to remember 2 fundamental rules:
1. Divident is always equal or greater than remainder.
2. Divisor is always greater than remainder.

We have 2 equations:
1. x/y = 3 + z. Means: x ≥ z, y > z, taking into account the quotient, x ≥ y * 3 + z.
2. z/y = ? + 2. Meаns: z ≥ 2, y > 2.

Make conclusion from both 1 and 2:
I. x ≥ z ≥ 2.
II. y > z ≥ 2; y can be at least 3, because y is integer and y > z = 2.
III. x ≥ 11. Because 3 * 3 + 2 = 11.

Therefore y can't be 5 nor 8.

My approach towards this question was abit different, I found it easy, I guess I might have overlooked something then.
Is there a number "y" that gives us quotinent 3 if you divide 5/y or 8/y? Answer is no.
5/1 = 5, 5/2 = 2 + 1, 5/3 = 1 + 2, 5/4 = 1 + 1 ...
8/1 = 8, 8/2 = 4, 8/3 = 2 + 2, 8/4 = 2, 8/5 = 1 + 3, 8 /6 = 1 + 2 ...

thus the only option is III coz we got no "no options possible"

When z is divided by y, the remainder is 2 does not mean that z=y+2. It means that z =yq + 2.

Check the solution:
When \(z\) is divided by \(y\), the remainder is 2: when divisor (\(y\) in our case) is more than dividend (\(z\) in our case), then the reminder equals to the dividend (for example, 2 divided by 5 gives the remainder of 2). Therefore, \(z=2\) and \(2 < y\).

So, we have that \(x=3y+2\) and \(2 < y\). This implies that the least value of \(x\) is \(x=3*3+2=11\): \(x\) cannot be 5 or 8.

Could \(x\) be 32? Yes. If \(y=10\), then \(x=3*10+2=32\).
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I think this question is good and helpful.

I think this is a high-quality question and I agree with explanation.

I think this is a high-quality question and I agree with explanation.

I think this is a high-quality question and I agree with explanation. Good Question...Quant questions in Club Tests always have a surprise element
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I think this is a high-quality question and I agree with explanation.

I think this is a high-quality question and I agree with explanation. very good question

Excellent Question!!

I think this is a high-quality question and I agree with explanation. Beautiful question

Wasted a lot of time to realise that x = 3y + 2
Once I figured that out, then was able to solve it by substituting values of x in the equation.

Posted from my mobile device

very nice question..
thank you for posting this. decent explanation.

Got a li'l doubt. same line of thinking as i came to y>2 bt substituted z in the first equation and tried to solve.
I got the equation x=3y+z
substituting z=y+2
comes to x=4y+2. (and y>2)
I'm not getting 32 either

Hi Banuel, I am not making the connection of how we definitively know that the z = 2. i infer think Y>Z, however and Y >2 but why couldn't Z = 0 or 1 since Z less than or equal to 2? Thank you!

We are told that when z is divided by y, the remainder is 2.

If z = 0, then 0 divided by positive integer y, yields the remainder of 0, not 2. 0 is divisible by every integer (except 0 itself): 0/y = 0.
If z = 1, then 1 divided by positive integer y, where y > z, yields the remainder of 1, not 2.

Hope it's clear.
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I think this is a high-quality question and I agree with explanation.

I think this is a high-quality question and I agree with explanation.

I think this is a high-quality question and I agree with explanation. very good concept based question !