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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?

Question

Just solved this problem. Give it a try:
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?

parthkohli · Answer

It's very easy!

experimentx · Answer

use these conditions
13(2a+25) < 1000
6(2a+12) < 1000
13(2a+25) = 6(2a+12)

parthkohli · Answer

So what do you get as your answer?

experimentx · Answer

just plugin to W|A

experimentx · Answer

seems the result is negative

parthkohli · Answer

@FoolAroundMath got the correct answer (in the messages).

parthkohli · Answer

So I can disclose the solution.

parthkohli · Answer

FoolAroundMath
is the answer to your question 38?

Yup!

experimentx · Answer

ah .. aren't these you 13 consecutive numbers
 84    96   108   120   132   144   156   168   180   192   204   216   228   240

  Columns 15 through 28

   252   264   276   288   300   312   324   336   348   360   372   384   396   408

  Columns 29 through 42

   420   432   444   456   468   480   492   504   516   528   540   552   564   576

  Columns 43 through 56

   588   600   612   624   636   648   660   672   684   696   708   720   732   744

  Columns 57 through 70

   756   768   780   792   804   816   828   840   852   864   876   888   900   912

  Columns 71 through 84

   924   936   948   960   972   984   996    84    96   108   120   132   144   156

  Columns 85 through 98

   168   180   192   204   216   228   240   252   264   276   288   300   312   324

  Columns 99 through 112

   336   348   360   372   384   396   408   420   432   444   456   468   480   492

  Columns 113 through 126

   504   516   528   540   552   564   576   588   600   612   624   636   648   660

  Columns 127 through 140

   672   684   696   708   720   732   744   756   768   780   792   804   816   828

  Columns 141 through 154

   840   852   864   876   888   900   912   924   936   948   960   972   984   996

parthkohli · Answer

Huh

experimentx · Answer

looks like bug in my program.

shubhamsrg · Answer

@ParthKohli 
n+(n+1) ..(n+26) ---> these are 27 terms and not 26 ..!!

parthkohli · Answer

Actually $25$... haha sorry

parthkohli · Answer

But my answer is still correct. That's for sure :)

shubhamsrg · Answer

yes i get it,its still correct..

shubhamsrg · Answer

and not 25,they are, 27 terms only.. o.O

parthkohli · Answer

$$N = n + (n + 1) + (n + 2)\cdots (n + 25)  = 26(n + \text{something)}$$and$$N = k + ( k+1) + (k + 2)\cdots (k+12)=13(k + \text{something)}$$So, the number must be a multiple of 13 and 26. If a number is a multiple of 26, it is a multiple of 13. So we have to find the number of multiples of 26 less than or equal to 1000 which is $\left\lfloor1000/26\right\rfloor=38$.

parthkohli · Answer

@shubhamsrg I said that it'd be $n + (n + 1)\cdots(n + 25)$

shubhamsrg · Answer

ohh.. :P

parthkohli · Answer

:)

experimentx · Answer

here's a short code for doing that on mathematica
Length[Intersection[Table[Sum[j, {j, i, i + 12}], {i, 1, 70}], 
  Table[Sum[j, {j, i, i + 25}], {i, 1, 22}]]]