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The sum of a three term sequence of consecutive positive odd integers

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Director
Director
avatar
G
Joined: 07 Dec 2014
Posts: 836

Kudos [?]: 266 [0], given: 15

The sum of a three term sequence of consecutive positive odd integers [#permalink]

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New post 04 Aug 2017, 15:04
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The sum of a three term sequence of consecutive positive odd integers is 1 less than the sum of a four term sequence of consecutive positive even integers.
If the sum of the odd sequence is less than 100, what is the number of such odd and even sequences?

A. 2
B. 4
C. 6
D. 8
E. 10

can anyone show me a faster way to do this?
thanks, gracie

sum of least odd sequence (1,3,5)=9
sum of next odd sequence (3,5,7)=15
difference=15-9=6
sum of least even sequence (2,4,6,8)=20
sum of next even sequence (4,6,8,10)=28
difference=28-20=8
equation1: 9+6x=20+8y-1➡3x-4y=5
least value of x=3
least value of y=1
substituting in equation1,
9+6*3=20+8*1-1
27=28-1
27-28 is least pair of odd-even sums with difference of 1
because 6 and 8 are respective differences between successive odd and even sums,
lowest common interval=24
let n+1=total odd sequences
equation2: 27+24n<100
24n<73
n<3.04
n=3
n+1=4
4*2=8=number of odd and even sequences
D
[Reveal] Spoiler: OA

Kudos [?]: 266 [0], given: 15

1 KUDOS received
Intern
Intern
avatar
B
Joined: 05 Nov 2016
Posts: 19

Kudos [?]: 6 [1], given: 58

Location: India
Concentration: Operations
Schools: Ivey '19
GMAT 1: 610 Q48 V26
GPA: 3.9
WE: Engineering (Consulting)
Re: The sum of a three term sequence of consecutive positive odd integers [#permalink]

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New post 04 Aug 2017, 23:18
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KUDOS
hi, I could not solve the sum correctly in my first attempt, but eventually figured out a way.

suppose,
Three consecutive positive odd numbers are: a, a+2 , a+4; where a is an odd integer, and its sum is 3a+6.
Four consecutive positive even numbers are: b, b+2 , b+4, b+6; where b is an even integer, and its sum is 4b+12.

Now,
4b+12-1=3a+6
4b-3a=6+1-12
3a-4b=5 (equation 1)


Now,
3a+6<100
a<94/3
a<31.333
so, a<=31, which is an odd integer.

similarly, 4b+12-1<100
b<89/4
so b<22.22
which is b<=22, which is an even integer.

Now the solution of 3a-4b=5 (equation 1) are.
(a,b) = (3,1), (7,4), (11,7), (15,10), (19,13), (23,16), (27,19), (31,22), (35,25) and so on....
but, a is <=31 & b<=22. So there are 8 set of solutions that satisfy this condition.
so the answer is 8.

correct answer: D

Kudos [?]: 6 [1], given: 58

Re: The sum of a three term sequence of consecutive positive odd integers   [#permalink] 04 Aug 2017, 23:18
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The sum of a three term sequence of consecutive positive odd integers

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