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04 Aug 2017, 15:04
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D
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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
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
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This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
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04 Aug 2017, 23:18
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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
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
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
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