Ask your own question, for FREE!
Mathematics 67 Online
OpenStudy (anonymous):

How many integers 1≤N≤1000 can be written both as the sum of 26 consecutive integers and as the sum of 13 consecutive integers? Details and assumptions The consecutive integers are allowed to be a mix of negative integers, 0 and positive integers (as long as they are consecutive).

OpenStudy (anonymous):

please give

OpenStudy (anonymous):

That's cheating, and if you don't know how to solve this one, wait till you see the really hard ones !

OpenStudy (anonymous):

pls tell

OpenStudy (anonymous):

since they are consecutive integers, the common difference in both series must be 1 d = 1 let the first term of the longer series be a let the first term of the shorter series be b sum(n) = (n/2)(2a + (n-1)d ) (26/2) (2a + 25(1) ) = (13/2)(2b + 12(1) ) 26(2a+25) = 13(2b+12) 52a + 650 = 26b + 156 52a + 494 = 26b b = (52a + 494)/26 = 2a + 19 if a=1 , b = 21 Long series: 1+2+3+...+26 = 351 short series: 21+22+..+33 = 351 if a=2 , b = 23 Long series: 2+3+..+27 = 377 short series: 23+24+...+35 = 377 ..... the last term in the sort series has to be ≤ 1000 , and clearly odd, making b = 987 and a = 484 short series: 987 + 988 + ... + 999 = 12909 long series : 484+485+...+509 = 12909 since a goes from 1 to 484 , there are 484 such series. I will leave it up to you to answer the question.

OpenStudy (anonymous):

i got itss 38

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!