Odd + Even = Odd: Solving for the Largest Integer n

[physicsrox]

How do i solve this qns?
"Find the largest integer n such that n^6021=2007^2007" Pls tell me how to solve this. thanks.

 

[Integral]

Use logs...

 

[HallsofIvy]

How do i solve this qns?
"Find the largest integer n such that n^6021=2007^2007" Pls tell me how to solve this. thanks.

Surely that's not the exact statement of the problem. There is at most one integer such that n⁶⁰²¹= 2007²⁰⁰⁷. If the 6021 root of 2007²⁰⁰⁷ is an integer, then that is n. If it is not an
integer then there is no solution. If those large numbers bother you, then as Integral said, you can use logarithms:

log(n⁶⁰²¹= 6021 log(n)= 2007 log(2007)
log(n)= (2007 log(2007))/6021

Using natural logs, ln(n)= 2.5347987829321127138088652278561
so n= e^{2.5347987829321127138088652278561}= 12.61. Of course, we had to round off the logarithms at 31 decimal places so that is not exact. It is however, far enough from an integer that we can be sure that the exact value is not an integer: there is no such n.

Perhaps you meant "find the smallest integer,n, such that n⁶⁰²¹> 2007²⁰⁰⁷.

 

[Edgardo]

Hello physicsrox,

another method is to consider the numbers 6021 and 2007.
Do you see a relationship between those two numbers?

 

[HallsofIvy]

Oh, well, if you want to do it the intelligent way!

 

[threetheoreom]

i don't .please help me out here ..i see if 2007 turn was reversed( or partially) then it becomes 7020 which maybe related in some way to 6021..maybe gcd.. help or mods ..i can't figure this way.

 

[learningphysics]

i don't .please help me out here ..i see if 2007 turn was reversed( or partially) then it becomes 7020 which maybe related in some way to 6021..maybe gcd.. help or mods ..i can't figure this way.

You're making it too complicated. don't reverse it. There's a very simple relationship between 2007 and 6021. Do you see it? stare at the two numbers for a while if you don't see it...

 

[symbolipoint]

You're making it too complicated. don't reverse it. There's a very simple relationship between 2007 and 6021. Do you see it? stare at the two numbers for a while if you don't see it...

2007*3=6021

 

[threetheoreom]

2007*3=6021

thanks, i saw that but it said nothing to me ( please excuse my ignorance ) so i hesistated to mention. so can some one explain how todo it the intelligent way!

 

[learningphysics]

thanks, i saw that but it said nothing to me ( please excuse my ignorance ) so i hesistated to mention. Can someone please explain why this matters.

No problem. Edgardo's idea was that by factoring 6021, you can write $$n^{6021}$$ as $$(n^3)^{2007}$$

Now the left side and right side both have the same exponent 2007... when two quantites have the same exponent, you just need to compare the bases...

 

[HallsofIvy]

And, doing that, we find again that there is no such number n!

physicrox, I'll ask again: since you ask for "the largest integer n", is it possible that the problem is to find the largest integer n such that n⁶⁰²¹< 2007²⁰⁰⁷?

That should now be easy to solve.

 

[threetheoreom]

cool .. i tried something different (dont know if it makes any sense)

n^6021=2007^2007
n^6021 can be written as 6021^n for any n

so we have n^6021=2007^2007

so we have log_6021 n = log_2007 2007
which is quite easy to solve for n

but what does this mean . if it means anthing at all.

Edit: looking at it this way it produces an insane decimal, so ther is no integer n that is = .( disregard what i said about easy)

 

[VietDao29]

cool .. i tried something different (dont know if it makes any sense)

n^6021=2007^2007
n^6021 can be written as 6021^n for any n...

Well, nope, this is not correct at all. a^b is, definitely, not b^a.

Say, 1² = 1, but 2¹ = 2. So, they are not the same.

so we have n^6021=2007^2007

so we have log_6021 n = log_2007 2007

Oh, and these 2 lines are wrong, too. You are taking different bases of both sides.

 

[HallsofIvy]

cool .. i tried something different (dont know if it makes any sense)

n^6021=2007^2007
n^6021 can be written as 6021^n for any n

and how do you conclude that?

so we have n^6021=2007^2007

so we have log_6021 n = log_2007 2007
which is quite easy to solve for n

but what does this mean . if it means anthing at all.

Edit: looking at it this way it produces an insane decimal, so ther is no integer n that is = .( disregard what i said about easy)

 

[physicsrox]

erm threetheoreom, the qns is n^6021>2007^2007, not n^6021=2007^2007.
btw, thanks for the replies, especially learningphysics, using the simple law of indices. n^6021>2007^2007---(n^3)^2007>2007^2007. therefore, n^3>2007. n=12.

 

[d_leet]

erm threetheoreom, the qns is n^6021>2007^2007, not n^6021=2007^2007.
btw, thanks for the replies, especially learningphysics, using the simple law of indices. n^6021>2007^2007---(n^3)^2007>2007^2007. therefore, n^3>2007. n=12.

12^3=1728<2007, so n does not equal 12.

 

[physicsrox]

whoops, i typed wrongly. It should be n^6021<2007^2007. sorry. Here is another qns...

Which of the following numbers is odd for any integer values of k?
(A) 2007+k^3
(B) 2007+7k
(C) 2007+2k^2
(D) 2007+2007k
(E) 2007k

 

[Dick]

Odd+even=odd, right? Which of those choices could be this case?