Challenge #2: If x and y are different positive integers, wh

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Tue Oct 30, 2018 1:24 pm

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If x and y are different positive integers, which of the following COULD be true:

i) When x is divided by y, the remainder is 2x
ii) When x is divided by 2y, the remainder is y
iii) When (2x + y) is divided by (x + y), the remainder is y

A) i only
B) ii only
C) iii only
D) i & ii only
E) ii & iii only

ASIDE: Many Integer Properties questions can be solved by identifying values that satisfy some given conditions. This question is intended to strengthen that skill.

Answer: B
Difficulty level: 650 - 700
Source: www.gmatprepnow.com

Related question: https://www.beatthegmat.com/challenge-i ... 04939.html

Last edited by Brent@GMATPrepNow on Sat Nov 03, 2018 6:09 am, edited 1 time in total.

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Source: Beat The GMAT — Problem Solving | Tue Oct 30, 2018 1:24 pm

deloitte247: Legendary Member; Posts: 2214; Joined: Fri Mar 02, 2018 2:22 pmFollowed by: 5 members

edited

by deloitte247 » Fri Nov 02, 2018 7:29 am

i) When x is divided by y remainder = 2x
Remainder cannot be greater than dividend or remainder (x) hence (i) could not be true

ii) When x is divided by 2y the remainder = y
when the divisor is greater than the dividend possible values of dividend = 3y, 5y, 7y, 9y.................... e.t.c
$$if\ x=5y$$
$$\frac{x}{2y}=\frac{5y}{2y}=\frac{\left(y+y+y+y+y\right)}{y+y}$$
$$remained\ \ 3y,\ hence\ statement\ \left(ii\right)\ could\ be\ TRUE$$

iii) When (2x+y) is divided by (x+y) The remainder is y
$$\frac{\left(2x+y\right)}{x+y}=\frac{\left(x+x+y\right)}{x+y}=x$$
$$remainder\ =\ x,\ hence\ \frac{\left(2x+y\right)}{x+y}\ne y$$
$$answer\ =option\ B$$

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by Brent@GMATPrepNow » Sat Nov 03, 2018 7:56 am

Brent@GMATPrepNow wrote:If x and y are different positive integers, which of the following COULD be true:

i) When x is divided by y, the remainder is 2x
ii) When x is divided by 2y, the remainder is y
iii) When (2x + y) is divided by (x + y), the remainder is y

A) i only
B) ii only
C) iii only
D) i & ii only
E) ii & iii only

i) When x is divided by y, the remainder is 2x
The remainder cannot be greater than the dividend (the number we're dividing)
For example, it CANNOT be the case that 17 divided by 5 leaves a remainder of 34
Statement i can never be true
Check the answer choices. . . . ELIMINATE A and D

ii) When x is divided by 2y, the remainder is y
Nice rule: If N divided by D leaves remainder R, then the possible values of N are R, R+D, R+2D, R+3D,. . . etc.
For example, if k divided by 5 leaves a remainder of 1, then the possible values of k are: 1, 1+5, 1+(2)(5), 1+(3)(5), 1+(4)(5), . . . etc.

So, some possible values of x are: y, y + 2y, y + 4y, y + 6y, . . . etc
Let's TEST the 1st option: x = y. No good.
The question tells us that x and y are different

So, let's TEST the 2nd option: x = y + 2y = 3y
So, how about x = 15 and y = 5
When we plug those values into statement ii, we get: When 15 is divided by 10, the remainder is 5
PERFECT!
Statement ii CAN be true.
Check the answer choices. . . . ELIMINATE C

iii) When (2x + y) is divided by (x + y), the remainder is y
When we apply the above rule, we get....
Some possible values of (2x + y) are: y, y + (x + y), y + 2(x + y), y + 3(x + y), . . . etc

Let's TEST the 1st option: (2x + y) = y.
Solve to get x = 0
No good. We're told x is POSITIVE

Let's TEST the 2nd option: (2x + y) = y + (x + y)
Solve to get: x = y
No good. The question tells us that x and y are different

Let's TEST the 3rd option: (2x + y) = y + 2(x + y)
Solve to get: y = 0
No good. We're told y is POSITIVE

Let's TEST the 4th option: (2x + y) = y + 3(x + y)
Solve to get: x = -3y
No good. If y is POSITIVE, then that means x is NEGATIVE, but we're told x is POSITIVE

Let's TEST the 5th option: (2x + y) = y + 4(x + y)
Solve to get: 2x = -4y
No good. If y is POSITIVE, then that means x is NEGATIVE, but we're told x is POSITIVE

At this point, we should recognize that, if we keep going, we'll keep running into the same problem where either x or y is NEGATIVE (which contradicts the given information.
So, statement iii can never be true

Answer: B

Cheers,
Brent

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by Scott@TargetTestPrep » Sat Nov 03, 2018 5:05 pm

Brent@GMATPrepNow wrote:If x and y are different positive integers, which of the following COULD be true:

i) When x is divided by y, the remainder is 2x
ii) When x is divided by 2y, the remainder is y
iii) When (2x + y) is divided by (x + y), the remainder is y

A) i only
B) ii only
C) iii only
D) i & ii only
E) ii & iii only

Since the remainder can't be greater than the dividend or the divisor, we see that statement (i) is not true.

If x = 9 and y = 3, we see that the remainder is 3 when 9 is divided by 6. So statement (ii) could be true.

Since (2x + y)/(x + y) = (x + y + x)/(x + y) = (x + y)/(x + y) + x/(x + y) = 1 + x/(x + y), we see that the remainder must be x. However, since x and y are different positive integers, then the remainder can't be y. We see that statement (iii) is not true.

Answer: B

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