Sums of Consecutive Integers

Suppose that $x$ can be expressed as the sum of 26 consecutive integers and 13 consecutive integers. Let the 13 integers be $a - 6, a - 5, ..., a + 5, a + 6$ . Then $x = 13a$ . Let the 26 consecutive integers be $b - 12, b - 11, ..., b + 12, b + 13$ . Then $x = 26b + 13$ . Thus, $x = 0 \pmod{13}$ , and $x = 13 \pmod{26}$ . Thus, the values of $x$ that work are $13(2n + 1)$ for nonnegative integers $n$ . It is well-known that $1001 = 7*11*13$ , so the largest value of $x$ that works is $13(77 - 2) = 13*75$ . Thus, the values of $x$ that work are $13*1, 13*3, ..., 13*75$ , so 38 numbers work, Q. E. D.

[Latex Edits - Calvin]

The sum:

$(-6)+(-5)+(-4)+ \dots +4+5+6= 0$

Is the sum of $13$ consecutive numbers. Any other sum of $13$ consecutive number can be obtained from this one adding $1$ to each number in the sum. Therefore we can express every sum of $13$ consecutive numbers with a number of the form $13a$ (where $a$ is an integer). The sum:

$(-12)+(-11)+(-10)+ \dots +11+12+13= 0$

Is the sum of $26$ consecutive numbers. Any other sum of $26$ consecutive number can be obtained from this one adding $1$ to each number in the sum. Therefore we can express every sum of $26$ consecutive numbers with a number of the form $26b+13$ (where $b$ is an integer). Clearly, all of these numbers can also be expressed as $13a$ for some integer $a$

therefore I have to find how many numbers between $1$ and $1000$ are multiples of $13$ but not of $26$ . These numbers are: $\left\lfloor \frac{1000}{13} \right\rfloor-\left\lfloor \frac{1000}{26} \right\rfloor=76-38=38$

Let the $13$ consecutive integers be $a, a+1, a + 2, \ldots a + 12$ for an integer $a$ . The sum of these integers is $(2a + 12)\frac{13}{2} = (a + 6)(13)$ . Thus, every integer that is divisible by $13$ can be written as the sum of $13$ consecutive integers.

Let the $26$ consecutive integers be $b, b+1, b + 2, \ldots b + 25$ for an integer $b$ . The sum of these integers is $(2b + 25)\frac{26}{2} = (2b + 25)(13)$ . Thus, every integer that is divisible by $13$ but not $26$ ( $2b + 25$ is odd) can be written as the sum of $26$ consecutive integers. Every integer of this form can also be written as the sum of $13$ consecutive integers (from above). Hence, there are $\lfloor \frac{1000}{13} \rfloor - \lfloor \frac{1000}{26} \rfloor = 38$ integers that can be written as the sum of $26$ and $13$ consecutive integers within the given range.

I split the problem up into two sets--the set of 26 consecutive integers and the set of 13 consecutive integers. If you examine any set of 26 integers (lets take one to 26 for example), you will find that you can split the set into 13 pairs of numbers adding up to twice the lowest integer + 25 (in the example, 27). This means that any set = 13 * {2X + 25}. 2X + 25 must be a positive integer, so X is greater than -25. 13 * {2X + 25} is less than one thousand, so 2X + 25 must be less than 76.9 (the same as 1000 divided by 13). X is an integer, so functionally, 2X+25 must be less than 75. So the domain of the set is all the odd numbers from 1 to 75. For the sum of the 13 consecutive integers, the sum=13 * {the middle number}. The domain of the middle number goes from 1 to 76. There are 38 numbers in the intersection of the domains, so the answer is 38

as a certain number has to be expressed as a sum of 13 consecutive integers, consider the arithmetic progression $A-6+A-5.........+A+A+1+.....A+6 = X$ (where $A$ is an integer and $X$ is a certain integer between 1 and 1000 inclusive of both 1 and 1000) we also notice that these are consecutive integers as the common difference between each integer is 1. the above equation when simplified gives $13A=X$ hence we know that $X$ is a multiple of 13.also since the maximum value of X is 1000 ,the maximum value that A can take is 76 (as 13 multiplied by 77 is 1001 which is greater than 1000)

now consider a similar progression $B-12+B-11..........+B+B+1......+B+13$ where $B$ is an integer the above sum should also equal the same $X$ as the previous expression as it is the same integer being expressed as the sum of 26 and 13 consecutive integers. we see that the above expession contains 26 integers each with common difference 1 therefore $B-12+B-11.......+B+B+1......+B+13=X$ but we already know that $X=13A$ plugging that in the above expression B-12+B-11.......+B+B+1......+B+13=13A we notice that this simplifies to 26B+13=13A which equals 26B=13(A-1) which futher simplifies to B= $\frac {A-1} {2}$ therefore for B to be an integer $\frac {A-1} {2}$ should be an integer.this we notice only happens when a is odd since a can take a maximum value of 76 and a minimum value of 1 there are 38 odd numbers between 1&76.therefore the ans is 38 NOTE-instead of the expression B-12+B-11.......+B+B+1......+B+13 we could also have written B-13+B-12+B-11.......+B+B+1....+B+12 which would also have 26 consecutive integers and the same simplification process would follow leading us to the equation B= $\frac {A+1} {2}$ here too B is only an integer when A is odd but the only difference between the 2 procedures is that in this equation A does not take the value 1 but instead take the value -1.hence the answer still remains 38

Pick an integer $n_1$ and sum it to its 25 consecutive integers forming the number $N_1$ :

$N_1 = \sum_{i=n_1}^{n_1 + 25}i = 26n_1 + \sum_{i=1}^{25}i = 26n_1 + 325$

Pick another integer $n_2$ and sum it to its 12 consecutive integers forming the number $N_2$ :

$N_2 = \sum_{i=n_2}^{n_2 + 12}i = 13n_2 + \sum_{i=1}^{12}i = 13n_2 + 78$

Asking that $N_1 = N_2 \equiv N$ , one gets:

$n_2 = 19 + 2n_1$

Let's see then the minimum and maximum values of $n_1$ such that $1 \le N \le 1000$ . It is not difficult to see that

$n_{1min} = -12$ for which $N_{min} = 13$

and

$n_{1max} = 25$ for which $N_{max} = 975$

Therefore, from $n_{1min} = -12$ to $n_{1max} = 25$ , we have a total of $\boxed{38}$ integers.

suppose the number is P, then $P=26a+13*25=13b+13*6$ this gives us $b-2a=19$ and there will be solution only if b is an odd number.......the first range gives us $-6<b<71$ .......there are 38 odd numbers in this range, so answer is 38! :D

Let the first number out of the thirteen consecutive numbers = $x$ . Then the final number = $x+13$ and the sum of all of these 13 numbers = $13x + \frac{12 \times 13}{2} = 13x + 78$ .

Using similar logic, we can express the sum of the twenty six consecutive numbers as $26y + 325$

Since these are two ways of expressing the same number, $26y + 325 = 13x + 78$

$2y + 19 = x$

$19 = x - 2y$ .

Now, we also know that the output number is going to be less than 1000 and greater than 1.

We can therefore form some inequalities:

$1000 > 26y + 235 > 1$

$1000 > 13x + 78 > 1$

Which simplify to give (rounded whole numbers)

$-9 < y < 29$ = 38 values of y. $-5 < x < 90$ = 95 values of x.

Since we are asked to find the number of integers, and not the number of solutions, we can see that the answer must be $\boxed{38}$