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If y is a positive integer, is 3y odd? 1) y+319 is even 2) y [#permalink]
 24 Sep 2005, 14:51
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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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Sarat
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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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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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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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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]
25 Sep 2005, 07:06
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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Got D for both 
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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]
25 Sep 2005, 19:47
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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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, 09:54, edited 1 time in total.
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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]
26 Sep 2005, 19:31
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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See my revised answer above.
@ fresinha: y is positive integer therefore y cannot be 0.
Sincerely
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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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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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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]
28 Sep 2005, 02:46
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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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?
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