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?

Answer 1

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.

Answer 2

Good job reiny

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