[MOVED] What is the value of P+q ?

[nelsonling]

Hi I need your help to solve this question;
If (1+3+5+.....+p) + (1+3+5+.....+q) = (1+3+5+.....+25), what is the value of P+q ?

:roll:

 

[soroban]

Re: What is the value of P+q ?

Hello, nelsonling!

Edit: a really BAD blunder! . . . [Thanks for the heads-up, Subhotosh!]

If \(\displaystyle \,(1\,+\,3\,+\,5\,+\,\cdots\,+\,p)\:+\1\,+\,3\,+\,5\,+\,\cdots\,+\,q) \;= \;(1\,+\,3\,+\,5\,+\,\cdots\,+\,25)\),

what is the value of $\displaystyle p\,+\,q$ ?

It is well-known that the sum of the first $\displaystyle n$ odd numbers is $\displaystyle n^2$.


$\displaystyle 1\,+\,3\,+\,5\,+\,\cdots\,+\,p\,$ is the sum of the first $\displaystyle \frac{p+1}{2}$ odd numbers.
. . Hence, its sum is: $\displaystyle \,\left(\frac{p+1}{2}\right)^2$

$\displaystyle 1\,+\,3\,+\,5\,+\,\cdots\,+\,q\,$ is the sum of the first $\displaystyle \frac{q+1}{2}$ odd numbers.
. . Hence, its sum is: $\displaystyle \,\left(\frac{q+1}{2}\right)^2$

$\displaystyle 1\,+\,3\,+\,5\,+\,\cdots\,+\,25\,$ is the sum of the first $\displaystyle \frac{25+1}{2}\,=\,13$ odd numbers.
. . Hence, its sum is: $\displaystyle \,13^2$

So we have: $\displaystyle \:\left(\frac{p+1}{2}\right)^2\,+\,\left(\frac{q+1}{2}\right)^2\:=\:13^2$

We have: the sum of two squares equals a square.
. . This is the Pythagorean relationship!

The only Pythagorean triple is: $\displaystyle \:5^2\,+\,12^2\:=\:13^2$

So we have: $\displaystyle \:\frac{p+1}{2}\:=\:5\;\;\Rightarrow\;\;p \,=\,9$
. . . . . .and: $\displaystyle \:\frac{q+1}{2}\:=\:12\;\;\Rightarrow\;\;q\,=\,23$


Therefore: \(\displaystyle \\,+\,q\;=\;9\,+\,23\;=\;32\)

 

[Deleted member 4993]

Re: What is the value of P+q ?

Hi I need your help to solve this question;
If (1+3+5+.....+p) + (1+3+5+.....+q) = (1+3+5+.....+25), what is the value of P+q ?

:roll:

What grade problem are these?

 

[Deleted member 4993]

Re: What is the value of P+q ?

Hello, nelsonling!

Is there a typo? .I don't get integers answers.

If \(\displaystyle \,(1\,+\,3\,+\,5\,+\,\cdots\,+\,p)\:+\1\,+\,3\,+\,5\,+\,\cdots\,+\,q) \;= \;(1\,+\,3\,+\,5\,+\,\cdots\,+\,25)\),

what is the value of $\displaystyle p\,+\,q$ ?

It is well-known that the sum of the first $\displaystyle n$ odd numbers is $\displaystyle n^2$.


$\displaystyle 1\,+\,3\,+\,5\,+\,\cdots\,+\,p\,$ is the sum of the first $\displaystyle (2p-1)$ odd numbers.
. . Hence, its sum is: $\displaystyle \,(2p-1)^2$

small typo ...(1+3+5..+p) has (p+1)/2 terms - so the sum (p+1)^2/2

$\displaystyle 1\,+\,3\,+\,5\,+\,\cdots\,+\,q\,$ is the sum of the first $\displaystyle (2q-1)$ odd numbers.
. . Hence, its sum is: $\displaystyle \,(2q-1)^2$

$\displaystyle 1\,+\,3\,+\,5\,+\,\cdots\,+\,25\,$ is the sum of the first $\displaystyle 13$ odd numbers.
. . Hence, its sum is: $\displaystyle \,13^2$

So we have: \(\displaystyle \2p-1)^2\,+\,(2q-1)^2\:=\:13^2\)

We have: the sum of two squares equals a square.
. . This is the Pythagorean relationship!

The only Pythagorean triple is: $\displaystyle \:5^2\,+\,12^2\:=\:13^2$

But this makes: $\displaystyle \:2p\,-\,1\:=\:5\;\;\Rightarrow\;\;p \,=\,3$
. . . . . . . . and: $\displaystyle \:2q\,-\,1\:=\:12\;\;\Rightarrow\;\;q\,=\,\frac{13}{2}\;$ ??

 

[Denis]

p = 9, q = 23 : p + q = 32

p = 2a - 1 and q = 2b - 1 where a^2 + b^2 = 169

 

[Denis]

Re: What is the value of P+q ?

So we have: \(\displaystyle \2p-1)^2\,+\,(2q-1)^2\:=\:13^2\)

Methinks that should be [(p+1)/2]^2 + [(q+1)/2)^2 = 13^2

 

[Deleted member 4993]

Re: What is the value of P+q ?

Methinks that should be [(p+1)/2]^2 + [(q+1)/2)^2 = 13^2

Yep - then follow Soroban's method and you would find integer solution

 

[nelsonling]

Excellent ! thanks everyone for your help.