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Updated on: 06 Mar 2012, 03:50
Originally posted by devinawilliam83 on 06 Mar 2012, 00:36.
Last edited by Bunuel on 06 Mar 2012, 03:50, edited 2 times in total.
Last edited by Bunuel on 06 Mar 2012, 03:50, edited 2 times in total.
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If a is the sum of x consecutive positive integers. b is the sum of y consecutive positive integers. For which of the following values of x and y is it impossible that a = b?
A. x = 2; y = 6
B. x = 3; y = 6
C. x = 7; y = 9
D. x = 10; y = 4
E. x = 10; y = 7
A. x = 2; y = 6
B. x = 3; y = 6
C. x = 7; y = 9
D. x = 10; y = 4
E. x = 10; y = 7
11 Apr 2014, 04:53
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devinawilliam83
Sum of 2 consecutive integers = Odd
Sum of 4 consecutive integers = Even
Sum of 6 consecutive integers = Odd
Sum of 8 consecutive integer = Even and so on is the pattern.
If we look at answer option D. Sum of 10 consecutive integer = Odd, Sum of 4 consecutive integers = Even.
Odd cannot be equal to Even.
Hence the answer is D.
06 Mar 2012, 03:48
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If a is the sum of x consecutive positive integers. b is the sum of y consecutive positive integers. For which of the following values of x and y is it impossible that a = b?
A. x = 2; y = 6
B. x = 3; y = 6
C. x = 7; y = 9
D. x = 10; y = 4
E. x = 10; y = 7
The sum n consecutive integers is give by: \(Sum=\frac{(2a_1+n-1)*n}{2}\) (check Number Theory chapter of Math Book for more: math-number-theory-88376.html);
Notice that:
If \(n=even=2*odd\), so when \(n\) (# of consecutive integers) is even but not a multiple 4 then \(Sum=\frac{(2a_1+n-1)*n}{2}=\frac{(even+even-odd)*(2*odd)}{2}=odd*odd=odd\);
If \(n=even=2*even\), so when \(n\) is a multiple 4 then \(Sum=\frac{(2a_1+n-1)*n}{2}=\frac{(even+even-odd)*(2*even)}{2}=odd*even=even\);
That's because a set of even number of consecutive integers has half even and half odd terms. The sum of even terms is obviously even. As for odd terms: their sum is even if their number is even (so total # of terms is multiple of 4) and their sum is odd if their number is odd (so total number of terms is even but not a multiple of 4);
So, the sum of 10 (not a multiple of 4) consecutive integers will be odd (the sum of 5 even and 5 odd integers) and the sum of 4 (multiple of 4) consecutive integers will be even (the sum of 2 even and 2 odd integers), so option D is not possible.
Answer: D.
Hope it's clear.
A. x = 2; y = 6
B. x = 3; y = 6
C. x = 7; y = 9
D. x = 10; y = 4
E. x = 10; y = 7
The sum n consecutive integers is give by: \(Sum=\frac{(2a_1+n-1)*n}{2}\) (check Number Theory chapter of Math Book for more: math-number-theory-88376.html);
Notice that:
If \(n=even=2*odd\), so when \(n\) (# of consecutive integers) is even but not a multiple 4 then \(Sum=\frac{(2a_1+n-1)*n}{2}=\frac{(even+even-odd)*(2*odd)}{2}=odd*odd=odd\);
If \(n=even=2*even\), so when \(n\) is a multiple 4 then \(Sum=\frac{(2a_1+n-1)*n}{2}=\frac{(even+even-odd)*(2*even)}{2}=odd*even=even\);
That's because a set of even number of consecutive integers has half even and half odd terms. The sum of even terms is obviously even. As for odd terms: their sum is even if their number is even (so total # of terms is multiple of 4) and their sum is odd if their number is odd (so total number of terms is even but not a multiple of 4);
So, the sum of 10 (not a multiple of 4) consecutive integers will be odd (the sum of 5 even and 5 odd integers) and the sum of 4 (multiple of 4) consecutive integers will be even (the sum of 2 even and 2 odd integers), so option D is not possible.
Answer: D.
Hope it's clear.
