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 + (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.
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To find the integers that satisfy the given condition, we can set up two equations and solve for N.
Let's assume that the first integer in the sequence of 26 consecutive integers is x. Then the sum of these 26 integers can be expressed as:
x + (x+1) + (x+2) + ... + (x+25) = 26x + (1+2+...+25) = 26x + 325
Similarly, let's assume that the first integer in the sequence of 13 consecutive integers is y. Then the sum of these 13 integers can be expressed as:
y + (y+1) + (y+2) + ... + (y+12) = 13y + (1+2+...+12) = 13y + 78
Since N is the sum of both sequences, we have the following two equations:
26x + 325 = N
13y + 78 = N
Now we can solve these equations simultaneously. Rearranging the equations, we get:
26x = N - 325
13y = N - 78
Simplifying further, we have:
2x = (N - 325)/13
y = (N - 78)/13
From these equations, we can see that N must be divisible by both 13 and 325 for there to be a solution.
Now let's find the values of N that satisfy the condition:
For N to be divisible by both 13 and 325, it must be divisible by their least common multiple (LCM). The LCM of 13 and 325 is 4225.
Thus, N can be any multiple of 4225 that falls within the range 1 ≤ N ≤ 1000.
To find the number of such integers, we can divide the upper limit of N (1000) by the LCM (4225) and round down the result.
1000 / 4225 ≈ 0.2366
Therefore, there are no positive integers 1 ≤ N ≤ 1000 that satisfy the given condition.
To find the solution to the problem, we need to identify the range of integers between 1 and 1000 that can be written as both the sum of 26 consecutive integers and the sum of 13 consecutive integers.
To begin, let's consider the sum of 26 consecutive integers. The sum of an arithmetic series can be determined by using the formula Sn = (n/2)(a + l), where Sn represents the sum, n is the number of terms, a is the first term, and l is the last term.
In this case, we will find the sum of the series for 26 terms.
Sn = (26/2)(a + l)
Since we are looking for the sum of the integers between 1 and 1000, we have a = 1 and l = 1000.
Sn = (26/2)(1 + 1000)
Sn = 13(1001)
Sn = 13013
So, any integer that can be expressed as the sum of 26 consecutive integers will yield a sum of 13013.
Next, let's consider the sum of 13 consecutive integers. Again, using the formula Sn = (n/2)(a + l), we will find the sum of the series for 13 terms.
Sn = (13/2)(a + l)
Using the same reasoning, a = 1 and l = 1000.
Sn = (13/2)(1 + 1000)
Sn = 6.5(1001)
Sn = 6506.5
Therefore, any integer that can be expressed as the sum of 13 consecutive integers will yield a sum of 6506.5.
To find the integers that satisfy both conditions, we need to see if the sums 13013 and 6506.5 have any common integers within the range of 1 to 1000.
Since 6506.5 is not an integer, it is not possible to find any common integers within the given range. Hence, there are no integers between 1 and 1000 that can be written as the sum of both 26 and 13 consecutive integers.