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Bunuel
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Bunuel
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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"
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Bunuel
When the positive integer is \(x\) divided by the 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

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.
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I think this is a high-quality question and I agree with explanation.
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I think this is a high-quality question and I agree with explanation.
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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.
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I think this is a high-quality question and I agree with explanation. very good question
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Excellent Question!!
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I think this is a high-quality question and I agree with explanation. Beautiful question
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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
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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
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venkivety
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

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