Posted by **Anubhav** on Monday, May 27, 2013 at 6:09am.

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?

- heeeeeeeeeeeeeelp maths -
**Reiny**, Monday, May 27, 2013 at 8:12am
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.

- brilliant supervisor -
**brilliant supervisor**, Tuesday, May 28, 2013 at 9:02am
Good job reiny

visit on brilliant(dot)org

thank you

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