heeeeeeeeeeeeeelp maths
posted by Anubhav on .
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?

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 + (n1)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. 
Good job reiny
visit on brilliant(dot)org
thank you