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If y is the sum of x consecutive positive integers, and x>3,

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If y is the sum of x consecutive positive integers, and x>3,  [#permalink]

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New post 24 Mar 2014, 15:34
2
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A
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C
D
E

Difficulty:

  95% (hard)

Question Stats:

36% (01:45) correct 64% (01:26) wrong based on 218 sessions

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If y is the sum of x consecutive positive integers, and x>3, is y odd?

(1) x is odd.
(2) The largest number among the x consecutive integers is odd.

OE:
(1): plug in x=5 and y=1+2+3+4+5=15 -> y = odd
But, we can also plug in y=2+3+4+5+6=20 -> y = even
IS
(2): Make sure the largest number is odd
Plug in y=1+2+3+4+5=15 -> y = odd
But, we can also plug in y=2+3+4+5 -> y = even
IS
Combined: Plug in x=5 and make sure the largest number is odd:
y=1+2+3+4+5=15 -> y = odd
But, Plug in x=7 y=1+2+3+4+5+6+7=28 -> y = even
IS
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Re: If y is the sum of x consecutive positive integers, and x>3,  [#permalink]

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New post 25 Mar 2014, 03:41
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4
goodyear2013 wrote:
If y is the sum of x consecutive positive integers, and x>3, is y odd?

(1) x is odd.
(2) The largest number among the x consecutive integers is odd.

OE:
(1): plug in x=5 and y=1+2+3+4+5=15 -> y = odd
But, we can also plug in y=2+3+4+5+6=20 -> y = even
IS
(2): Make sure the largest number is odd
Plug in y=1+2+3+4+5=15 -> y = odd
But, we can also plug in y=2+3+4+5 -> y = even
IS
Combined: Plug in x=5 and make sure the largest number is odd:
y=1+2+3+4+5=15 -> y = odd
But, Plug in x=7 y=1+2+3+4+5+6+7=28 -> y = even
IS


If y is the sum of x consecutive positive integers, and x>3, is y odd?

The sum of x consecutive integers equals the mean multiplied by the number of terms, thus x*{mean}=y. Recall that in evenly spaced set mean=median, thus we have that x*{mean}=x*{median}=y.

The question asks whether y is odd.

(1) x is odd --> x*{median}=odd*{median}=y. If a set has odd number of terms the median is the middle term, when arranged in ascending/descending order, thus the median of the set must be an integer. For y to be odd the median must be odd, but the median of odd number of consecutive integers can be odd as well as even, for example, consider {1, 2, 3, 4, 5} and {1, 2, 3, 4, 5, 6, 7}. Not sufficient.

(2) The largest number among the x consecutive integers is odd. If the set is {0, 1, 2, 3}, then y is even but if the set is {1, 2, 3, 4, 5}, then y is odd. Not sufficient.

(1)+(2) The sets we considered to prove insufficiency of the first statement are still valid because both of them have the largest term odd (to satisfy the second statement). Thus even taken together the statements are not sufficient.

Answer: E.
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Re: If y is the sum of x consecutive positive integers, and x>3,  [#permalink]

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New post 08 Sep 2014, 09:17
1
My Approach -

let Y = a + (a+1) + (a+2) + ......
Nth term is the Xth term

Y = X*a + (1+2+3+.......(X-1) )
Y = X*a + X(X-1)/2

Y(odd) = Odd + even
Or = even + odd

1) X is odd == > we doesn't know whether a is even or odd and also whether X(X-1)/2 is even or odd. Not Suff.

2) Same reason for 2 . Not suff.

1+2 --> not suff.

Please let me know if my approach is correct.

Thanks.
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Re: If y is the sum of x consecutive positive integers, and x>3,  [#permalink]

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New post 02 Nov 2015, 18:53
goodyear2013 wrote:
If y is the sum of x consecutive positive integers, and x>3, is y odd?

(1) x is odd.
(2) The largest number among the x consecutive integers is odd.

OE:
(1): plug in x=5 and y=1+2+3+4+5=15 -> y = odd
But, we can also plug in y=2+3+4+5+6=20 -> y = even
IS
(2): Make sure the largest number is odd
Plug in y=1+2+3+4+5=15 -> y = odd
But, we can also plug in y=2+3+4+5 -> y = even
IS
Combined: Plug in x=5 and make sure the largest number is odd:
y=1+2+3+4+5=15 -> y = odd
But, Plug in x=7 y=1+2+3+4+5+6+7=28 -> y = even
IS



Use the formula n(n+1)/2 . if x is 5 y= 15 and x is 7 y is 28 . Both stmt 1 and 2 states that x is odd. so Q solved E. Let me know your thoughts
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Re: If y is the sum of x consecutive positive integers, and x>3,  [#permalink]

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New post 08 Nov 2015, 05:59
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If y is the sum of x consecutive positive integers, and x>3, is y odd?

(1) x is odd.
(2) The largest number among the x consecutive integers is odd.

In the original condition, there are 2 variables (initial value and the number of integers) and 2 equations are given from the 2 conditions, so there is high chance (C) will be our answer.
Looking at the conditions together, the answer is 'yes' for 1,2,3,4,5 ==> 15, but 'no' for 1,2,3,4,5,6,7==> 28. So the conditions are insufficient and the answer becomes (E).

For cases where we need 2 more equation, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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Re: If y is the sum of x consecutive positive integers, and x>3,  [#permalink]

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New post 03 Oct 2017, 14:32
Based on the qx: it means, y = sum of x positive inte which are CONSECUTIVE.
Also, number of integers are more than 3. Can be 4,5,6....
But is Y (the total of x sequence) odd?

St 1: x (number of +ve consecutive inte) is ODD. So can be 5,7,9
If I total 1 to 5, it is odd. If I add 1 to 7 it is even. And from 1 to 9, it is odd again. Can try with other numbers too, but the result is not consistent. So NS.

St 2: largest number is odd. Makes no change. Just by above, I can say that answer changes.

St 1+ 2= no conclusion.

So Answer is E
Re: If y is the sum of x consecutive positive integers, and x>3, &nbs [#permalink] 03 Oct 2017, 14:32
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