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# a is the sum of x consecutive positive integers. b is the

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Director
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a is the sum of x consecutive positive integers. b is the [#permalink]  15 May 2006, 23:49
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

No OA
Senior Manager
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BG wrote:
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

No OA

D it is

Sum = average * number

the average of even number of consecutive integers is always decimal
while the average of odd number of con. int. is always integer

assume, the average of X a(x) and the average of Y a(y)

A) a(x)*2 = a(y)*6 => a(x) (decimal) = 3*a(y)(decimal) OK
B) a(x)*3 = a(y)*6 => a(x) (integer) = 2*a(y)(2*decimal = integer) OK
C) a(x)*7 (integer)= a(y)*9(integer) OK
D) a(x)*10 = a(y)*4 => 5*a(x)(decimal) = 2*a(y) ( 2 * decimal = integer ) DECIMAL != INTEGER
E) a(x)*10 ( 10 * decimal = integer ) = a(y)*7 (integer) OK
Director
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deowl wrote:
BG wrote:
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

No OA

D it is

Sum = average * number

the average of even number of consecutive integers is always decimal
while the average of odd number of con. int. is always integer

assume, the average of X a(x) and the average of Y a(y)

A) a(x)*2 = a(y)*6 => a(x) (decimal) = 3*a(y)(decimal) OK
B) a(x)*3 = a(y)*6 => a(x) (integer) = 2*a(y)(2*decimal = integer) OK
C) a(x)*7 (integer)= a(y)*9(integer) OK
D) a(x)*10 = a(y)*4 => 5*a(x)(decimal) = 2*a(y) ( 2 * decimal = integer ) DECIMAL != INTEGER
E) a(x)*10 ( 10 * decimal = integer ) = a(y)*7 (integer) OK

Hey, that is excellent !!
I was lost trying to figure out how to do this.
Manager
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For part a)
a=10+11=b=1+2+3+4+5+6=21
for part b)
a= 6+7+8=1+2+3+4+5+6=21
for part c)
a=6+7+8+9+10+11=3+4+5+6+7+8+9+10+11=63
for part d) impossible
for part e)
a=6+7+8+9+10+11+12+13+14+15=12+13+14+15+16+17+18=105

further explanation in support for d
a= n+(n+1)+(n+2)+(n+3)=4n+6
b=m+ (m+1)+ (m+2)+â€¦â€¦(m+9)= 10m+45

4n+6=10m+45
4n=10m+39
4n is even but 10m+39 is odd , so it is impossible
Intern
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deowl wrote:
BG wrote:
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

No OA

D it is

Sum = average * number

the average of even number of consecutive integers is always decimal
while the average of odd number of con. int. is always integer

assume, the average of X a(x) and the average of Y a(y)

A) a(x)*2 = a(y)*6 => a(x) (decimal) = 3*a(y)(decimal) OK
B) a(x)*3 = a(y)*6 => a(x) (integer) = 2*a(y)(2*decimal = integer) OK
C) a(x)*7 (integer)= a(y)*9(integer) OK
D) a(x)*10 = a(y)*4 => 5*a(x)(decimal) = 2*a(y) ( 2 * decimal = integer ) DECIMAL != INTEGER
E) a(x)*10 ( 10 * decimal = integer ) = a(y)*7 (integer) OK

Although (D) is correct the assertion below is not.
5*a(x)(decimal) = 2*a(y) Integer => DECIMAL = Integer

if a(x) = 0.4, a(y)=1 then 5*0.4 = 2 * 1, there are other examples too like a(x) = 7.2 and a(y) = 18

I solved it a slightly different way.

Sum of X cons integers = Kx + x(x+1)/2 (where K is lowest) - 1
Sum of Y cons integers = K1y + y(y+1)/2 (where K1 is lowest) - 2

If x = 10 and y = 4

SUM1 = 10K + 10*11/2 = 10K + 55
SUM2 = 4K1 + 4*5/2 = 4K1 + 10

SUM1 = ODD and SUM2 = EVEN, hence D is correct. The others do not provide such an inconsistency.
Intern
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saha wrote:
if a(x) = 0.4, a(y)=1 then 5*0.4 = 2 * 1, there are other examples too like a(x) = 7.2 and a(y) = 18

the sum of x consecutive positive integers. b is the sum of y consecutive positive integers

i think your example is not right.
Intern
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Saha, you are close. But i believe the formula would be

Sum of X cons integers = Kx + x(x-1)/2 (where K is lowest) - 1
Sum of Y cons integers = K1y + y(y-1)/2 (where K1 is lowest) - 2

If x = 10 and y = 4

SUM1 = 10K + 10*9/2 = 10K + 45
SUM2 = 4M + 4*3/2 = 4M + 6

SUM1 = ODD and SUM2 = EVEN, hence D is correct. The others do not provide such an inconsistency.
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kpoxa wrote:
saha wrote:
if a(x) = 0.4, a(y)=1 then 5*0.4 = 2 * 1, there are other examples too like a(x) = 7.2 and a(y) = 18

the sum of x consecutive positive integers. b is the sum of y consecutive positive integers

i think your example is not right.

a(x) is the average of X consecutive integers, a(y) is the averge of Y consecutive integers according to the example given above ? What's wrong ?
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mattflow wrote:
Saha, you are close. But i believe the formula would be

Sum of X cons integers = Kx + x(x-1)/2 (where K is lowest) - 1
Sum of Y cons integers = K1y + y(y-1)/2 (where K1 is lowest) - 2

If x = 10 and y = 4

SUM1 = 10K + 10*9/2 = 10K + 45
SUM2 = 4M + 4*3/2 = 4M + 6

SUM1 = ODD and SUM2 = EVEN, hence D is correct. The others do not provide such an inconsistency.

Yes, thanks for the correction. My formula would be correct if K+1 and K1+1 were the lowest, not K and K1.
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Re: sum of integers MNHTN [#permalink]  16 May 2006, 10:03
BG wrote:
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) take any 6 consecutive integers (+ve) for y = 10, 11, 12, 13, 14 and 15. their sum is 75. divide it by 2, we get 37.5. so take immidiate integers above and below 37.5. they are 37 and 38 and their sum is75.

(b) take any 6 consecutive integers for y = 10, 11, 12, 13, 14 and 15. their sum is 75. divide it by 3, we get 25. so put 25 in the middle. the 3 consecutive integers are 24, 25 and 26. their sum is 75.

(c) take any 9 consecutive integers for y = 3, 4, 5, 6, 7, 8, 9, 10, and 11. their sum is 63. divide it by 7, we get 9. so put 9 in the middle of 7 consecutive integers. they are 6, 7, 8, 9, 10, 11, and 12. their sum is 63.

(d) take any 10 consecutive integers for x = 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10. their sum is 55. similarly take any 4 consecutive integers and add them up, the sum is even number/integer.

here we have interesting pattren with 10 and 4 consecutive integers. any 10 such integers have sum ending at 5 where as sum of 4 consecutive integers end at even integer. so these two consecutive integers never be equal.

(e) take any 10 consecutive integers for x = 6, 7, 8, 9, 10, 11, 12, 13, 14 and 15. their sum is 105. divide it by 7, we get 15. so put 15 in the middle of 7 consecutive integers. they are 12, 13, 14, 15, 16, 17, and 18. their sum is 105.

it is D, however it is lengthy.
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saha wrote:

Although (D) is correct the assertion below is not.
5*a(x)(decimal) = 2*a(y) Integer => DECIMAL = Integer

if a(x) = 0.4, a(y)=1 then 5*0.4 = 2 * 1, there are other examples too like a(x) = 7.2 and a(y) = 18

The decimal average can never be 0.4 or anything different from
.5 in tenths. And it will never give an integer when multiplied by 5.
So my assertion is perfectly correct.
Intern
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deowl wrote:
saha wrote:

Although (D) is correct the assertion below is not.
5*a(x)(decimal) = 2*a(y) Integer => DECIMAL = Integer

if a(x) = 0.4, a(y)=1 then 5*0.4 = 2 * 1, there are other examples too like a(x) = 7.2 and a(y) = 18

The decimal average can never be 0.4 or anything different from
.5 in tenths. And it will never give an integer when multiplied by 5.
So my assertion is perfectly correct.

I realised that later but it wasn't mentioned in that form earlier. its a good method i think.
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mattflow wrote:
Saha, you are close. But i believe the formula would be

Sum of X cons integers = Kx + x(x-1)/2 (where K is lowest) - 1
Sum of Y cons integers = K1y + y(y-1)/2 (where K1 is lowest) - 2

If x = 10 and y = 4

SUM1 = 10K + 10*9/2 = 10K + 45
SUM2 = 4M + 4*3/2 = 4M + 6

SUM1 = ODD and SUM2 = EVEN, hence D is correct. The others do not provide such an inconsistency.

I did the same way, but i dont agree with the first solution of sum=average*number
CEO
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Instead of going into averages etc. Here is shorter method:

(A) x = 2; y = 6=> a is always ODD : b is always ODD
(B) x = 3; y = 6 => a may be ODD or EVEN : b is always ODD
(C) x = 7; y = 9 => a may be ODD or EVEN : b may be ODD or EVEN
(D) x = 10; y = 4=> a is always ODD : b is always EVEN
(E) x = 10; y = 7 => a may be ODD or EVEN : b is always ODD
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Last edited by ps_dahiya on 18 May 2006, 12:24, edited 1 time in total.
Manager
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ps_dahiya wrote:
Instead of going into averages etc. Here is shorter method:

(A) x = 2; y = 6=> a is always EVEN : b is always ODD
(B) x = 3; y = 6 => a may be ODD or EVEN : b is always ODD
(C) x = 7; y = 9 => a may be ODD or EVEN : b may be ODD or EVEN
(D) x = 10; y = 4=> a is always ODD : b is always EVEN
(E) x = 10; y = 7 => a may be ODD or EVEN : b is always ODD

Great shortcut! However, i think you made a typo in (A) --> a should be odd.
Senior Manager
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ps_dahiya wrote:
Instead of going into averages etc. Here is shorter method:

(A) x = 2; y = 6=> a is always EVEN : b is always ODD
(B) x = 3; y = 6 => a may be ODD or EVEN : b is always ODD
(C) x = 7; y = 9 => a may be ODD or EVEN : b may be ODD or EVEN
(D) x = 10; y = 4=> a is always ODD : b is always EVEN
(E) x = 10; y = 7 => a may be ODD or EVEN : b is always ODD

Yep I solved the same way!
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tl372 wrote:
ps_dahiya wrote:
Instead of going into averages etc. Here is shorter method:

(A) x = 2; y = 6=> a is always EVEN : b is always ODD
(B) x = 3; y = 6 => a may be ODD or EVEN : b is always ODD
(C) x = 7; y = 9 => a may be ODD or EVEN : b may be ODD or EVEN
(D) x = 10; y = 4=> a is always ODD : b is always EVEN
(E) x = 10; y = 7 => a may be ODD or EVEN : b is always ODD

Great shortcut! However, i think you made a typo in (A) --> a should be odd.

Yes that was a typo. Thanks buddy. I edited that.
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Manager
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ps_dahiya wrote:
Instead of going into averages etc. Here is shorter method:

(A) x = 2; y = 6=> a is always ODD : b is always ODD
(B) x = 3; y = 6 => a may be ODD or EVEN : b is always ODD
(C) x = 7; y = 9 => a may be ODD or EVEN : b may be ODD or EVEN
(D) x = 10; y = 4=> a is always ODD : b is always EVEN
(E) x = 10; y = 7 => a may be ODD or EVEN : b is always ODD

Shouldn't (E) be: a is always odd, b maybe odd or even?
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