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If y is a positive integer, is 3y odd? 1) y+319 is even 2) y

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If y is a positive integer, is 3y odd? 1) y+319 is even 2) y [#permalink] New post 24 Sep 2005, 13:51
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If y is a positive integer, is 3y odd?
1) y+319 is even
2) y cannot be evenly divided by any odd number other than 1

Question 2:
If the integer x is positive, is x>17?
1) When x is divided by 4 or 5, the remainder is 3
2) x is evenly divisible by 21

Please let me know your answers

Thanks.
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 [#permalink] New post 24 Sep 2005, 15:55
Question 1:
A)

statement 1: y+319=even. Thus y has to be odd. 3 * odd number = odd number.

statement 2: y cannot be divided by any odd number other than 1. y could be even, or prime. eg., 2 or 5. insufficient.

Question 2:
C)

statement 1: this gives us two equations, x=4y+3 and x=5z+3 or 4y=5z. which is insufficient to calculate x or figure out whether x>17. Here x could be 3, and y and z would be zero.

statement 2: x is a multiple of 21. x could be 0, so this is insufficient.

together, we know x is greater than zero, and by statement 2, we know x is a multiple of 21. thus we can be assured that x>17, for example x could be 63. Thus sufficient together.
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 [#permalink] New post 24 Sep 2005, 16:07
Be careful with the first one chet. Statement 2 says that y cannot evenly be divided by any other odd number except 1. Prime numbers like 5, 7, 11 etc, can also be divided by themselves. Therefore, statement 2 must be an even number. Either answer choice is sufficient. D

Also, in question 2, the information in the second statement is unnecessary. A

OA please
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 [#permalink] New post 24 Sep 2005, 17:21
I agreed with Chet on both questions.
But the answers given were D for both the questions
I dont understand why they dont consider the number 1 in the first question and a multiple of 0 in the second question.
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 [#permalink] New post 25 Sep 2005, 00:29
Answer to the first one is D, as GMATT73 explaned.

Answer to the second one is also D

1 tells you that when divided by 4 or 5, x leaves a remainder of 3. So x is the common multiple of 4 and 5 increased by 3. The least common multiple of 4 and 5 is 20, so x is at least 23. Sufficient

2 the question tells you that x is POSITIVE, so x cannot be 0. Thus 2 is sufficient.

Answer is D
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Re: DS - 2 integer questions [#permalink] New post 25 Sep 2005, 06:06
good questions.
asaratchandra wrote:
If y is a positive integer, is 3y odd?
1) y+319 is even
2) y cannot be evenly divided by any odd number other than 1


from i, it is clear that y is odd.
from ii, y is 1, not any other values.

so both are sufficient......

asaratchandra wrote:
If the integer x is positive, is x>17?
1) When x is divided by 4 or 5, the remainder is 3
2) x is evenly divisible by 21


from i, x is > than 17 because the minimum value that satisfies to have 3 as reminder when divided by eithr 4 or 5 is 23.
from ii, x is > 17 because the lowest value that is evenly divided by 21 is 21.

so i and ii, both are suff.....
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 [#permalink] New post 25 Sep 2005, 09:16
Got D for both :lol:
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 [#permalink] New post 25 Sep 2005, 17:19
For the 1st question:
from the second statement the value can be 1,2,4 etc. Since the set contains the number 1 the statement is insufficient

for the 2nd question:
x is positive, so x=3 is valid and that satisfies the first statement, thus this statement is insufficient.

any thoughts??
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Re: DS - 2 integer questions [#permalink] New post 25 Sep 2005, 18:47
I agree with asaratchandra
Actually for the first question, I dont see how D is ok
I agree that statement 1 is suff.
However, concerning statement 2, Y can also be 2 or 4. No ???

HIMALAYA wrote:
good questions.
asaratchandra wrote:
If y is a positive integer, is 3y odd?
1) y+319 is even
2) y cannot be evenly divided by any odd number other than 1


from i, it is clear that y is odd.
from ii, y is 1, not any other values.

so both are sufficient......

asaratchandra wrote:
If the integer x is positive, is x>17?
1) When x is divided by 4 or 5, the remainder is 3
2) x is evenly divisible by 21


from i, x is > than 17 because the minimum value that satisfies to have 3 as reminder when divided by eithr 4 or 5 is 23.
from ii, x is > 17 because the lowest value that is evenly divided by 21 is 21.

so i and ii, both are suff.....
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 [#permalink] New post 26 Sep 2005, 08:15
Question 1: A

Question 2: D

Explain:

Question 1:
(1) y must be odd. Enough to anwer the question
(2) y can be 1 or 2^n where n is 1,2,3,...
=> y can be even or odd => Not sufficient to answer the question.

Question 2:

(1) Just "try and error" with some numbers:
4+3
8+3
12+3
16+3
20+3 (Here it is)

So the least number that devide 4 or 5 with a remainder 3 would be 23 that is greater than 21. Enough to answer the question
(2) x must be an multiple of 21. Enough to answer the question too


Hope this clear
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Last edited by tibeo - vietnam on 27 Sep 2005, 08:54, edited 1 time in total.
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 [#permalink] New post 26 Sep 2005, 08:36
I say D for 1st

A for 2nd

II in the 2nd problem:

x is evenly divisible by 21...

well x can be 42/21=2

X can be 0/21=0 which is even...

42 > than 17

0 < than 17...
(II) is insuff...
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Re: DS - 2 integer questions [#permalink] New post 26 Sep 2005, 18:31
oh that seems correct. y can be divided by 2, 4, 8 and any power of 2.

Antmavel wrote:
I agree with asaratchandra
Actually for the first question, I dont see how D is ok
I agree that statement 1 is suff.
However, concerning statement 2, Y can also be 2 or 4. No ???

HIMALAYA wrote:
good questions.
asaratchandra wrote:
If y is a positive integer, is 3y odd?
1) y+319 is even
2) y cannot be evenly divided by any odd number other than 1


from i, it is clear that y is odd.
from ii, y is 1, not any other values.

so both are sufficient......

asaratchandra wrote:
If the integer x is positive, is x>17?
1) When x is divided by 4 or 5, the remainder is 3
2) x is evenly divisible by 21


from i, x is > than 17 because the minimum value that satisfies to have 3 as reminder when divided by eithr 4 or 5 is 23.
from ii, x is > 17 because the lowest value that is evenly divided by 21 is 21.

so i and ii, both are suff.....
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 [#permalink] New post 27 Sep 2005, 08:56
See my revised answer above.
@ fresinha: y is positive integer therefore y cannot be 0.

Sincerely
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 [#permalink] New post 27 Sep 2005, 12:22
yes you are right, I did not read the question properly...
D it is for both...

tibeo - vietnam wrote:
See my revised answer above.
@ fresinha: y is positive integer therefore y cannot be 0.

Sincerely
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 [#permalink] New post 27 Sep 2005, 13:20
asaratchandra wrote:
For the 1st question:
from the second statement the value can be 1,2,4 etc. Since the set contains the number 1 the statement is insufficient

for the 2nd question:
x is positive, so x=3 is valid and that satisfies the first statement, thus this statement is insufficient.

any thoughts??


I agree with asaratchandra and I am not very fond of the two Ds...

Regarding the first problem: statement 1 is straightforward and obviously sufficient. Any non-negative power of 2 will satisfy the second condition. This means 1, 2, 4, 8, etc. In that case 3y can be either positive or negative.

Regarding the second problem: I agree that statement 2 is sufficient. But what about statement 1? y=3 does seem like a value, which satisfies the condition about the remainder...

I think the official answers could be wrong in this case.
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 [#permalink] New post 27 Sep 2005, 16:10
tibeo - vietnam wrote:
Question 1: A

Question 2: D

Explain:

Question 1:
(1) y must be odd. Enough to anwer the question
(2) y can be 1 or 2^n where n is 1,2,3,...
=> y can be even or odd => Not sufficient to answer the question.

Question 2:

(1) Just "try and error" with some numbers:
4+3
8+3
12+3
16+3
20+3 (Here it is)

So the least number that devide 4 or 5 with a remainder 3 would be 23 that is greater than 21. Enough to answer the question
(2) x must be an multiple of 21. Enough to answer the question too


Hope this clear


If i have to type, i would type same explanation as did tibeo.
Can anyone point how 1st one results into D???

y cannot be evenly divided by any odd number other than 1

dont forget y can be "1" which is odd.
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Re: DS - 2 integer questions [#permalink] New post 28 Sep 2005, 01:46
Q1 Answer is D. y + 319 is even then y is odd, hence 3y is odd.
y cannot be evenly divided by any odd number other than 1, hence y is odd so 3y is also odd.

Q2 Answer is D, if x is divisible by 4 or 5 then x should be divisible by 20, hence x should be > 17.
X is evenly divisible by 21, then x should be greater than 17.

So for both the questions my answer is D.
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 [#permalink] New post 28 Sep 2005, 02:55
Q1 answer is A.

(2) is insufficient:

y cannot be evenly divided by any odd number other than 1:

y=1, 1 is a positive integer that cannot be evenly divided by
any odd number other than 1. 3y=3 odd

y=2, 2 s a positive integer that cannot be evenly divided by
any odd number other than 1. 3y=6 even

Is there anything wrong in this reasoning?
  [#permalink] 28 Sep 2005, 02:55
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