0
$\begingroup$

Here i have a question that

To find the number of consecutive odd integers whose sum can be expressed as $50^2-13^2$

Just i am unable to understand the question what is really it is asking. Please someone explain me.

$\endgroup$
  • $\begingroup$ I.e., how many (2n+1)+(2n+3).. etc is equal to (63*37) $\endgroup$ – Mann Jun 4 '15 at 9:01
  • $\begingroup$ I am somewhat getting it....i m trying.. $\endgroup$ – Pratyush Jun 4 '15 at 9:06
  • $\begingroup$ But i think there should be the reason of giving $50^2-13^2$ $\endgroup$ – Pratyush Jun 4 '15 at 9:09
4
$\begingroup$

We have $$1+ 3 +...+(2n-1) = n^2 $$ hence $$50^2 -13^2 =1+3 +...+99 -(1+3+...+25) =27 +29 +...+99$$

$\endgroup$
  • $\begingroup$ Pratyush, has just to count the numbers. $\endgroup$ – kmitov Jun 4 '15 at 9:20
  • $\begingroup$ Then answer should be 37 . Am i right? $\endgroup$ – Pratyush Jun 4 '15 at 9:24
  • $\begingroup$ Yes you're right. $\endgroup$ – MotylaNogaTomkaMazura Jun 4 '15 at 9:31
  • $\begingroup$ But this is not the only solution, because $50^2-13^2$ can be expressed in other ways as a difference of two squares. For instance, $50^2-13^2 = 63 \times 37 = 21 \times 111 = 66^2-45^2$. $\endgroup$ – TonyK Jun 4 '15 at 9:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.