A little math riddle to keep you on your toes...

[vypon]

I'm thinking of a 6-digit number. The sum of the digits is 43. And only two of the following three statements about the number are true: (1) it's a square number. (2) it's a cube number, and (3) the number is under 500000.

If you have the answer e-mail me at chrisdavis09@verizon.net

This stumped me (still does) :bmage: You know when the greatest mind the "Earth" doesn't get it, it's no use...

[Hatake Kakashi]

Hrm. A homework assignment, eh?

No thanks. But good luck to ya. Something in google might help you out.

[Krylo]

I'm too lazy to solve it, but here's how you do so: start by finding cubes of numbers in numerical order (start at at a biggish number) that come out to 99,000 but under 500,000, then start checking which of them are also squares by checking for square roots. THEN see which of the numbers also adds up to 43.

That should be a pretty easy way, I think... so long as you have a calculator.

[capnchicken]

707^2 = 499 849 is the highest square before 500 000 and adds up to 43. So that is AN answer, I don't know if its the one and only though.

FYI: there is only one number that is a six digits, is above 500,000, and is a perfect square and cube. 9^6 = 531 441, since only numbers that are powers of 6 are perfect squares and cubes. It doesn't add up to 43 though.

[vypon]

No, it's not a homework assignment actually! It's more of a... Hey, solve this and I'll give you20 bucks! I mean, who could turn it down?

[Elmer J.]

You never included that it had to be positive.
It could be negative and still technically under 500 000.
Or irrational.
Or imaginary.

[vypon]

First off if it is square it could not be negative, imaginary or irrational. Second, you must not have read it all the way through not to know that!

[Kikuichimonji]

HAH! You just divulged that it was a square ^_^

[RedScar]

Actually he said only two out of three were true. And if it was a cube then it would be negitive. example: -1 X -1 X -1 = -1 Therefore it could be negitive. and I think I'm going to try this.

[Kikuichimonji]

But what I said was that he just showed that it was definitely a square. So that makes it 1 of the next two is wrong.

[RedScar]

Well how so we know it's a square? Huh? I mean he said it but how do we know that's right? Me go get cacultor, me find.

Edit: The first 6 digt square is 317. I'm going up from there.

[Bosolai]

Well, It can't be a square and a cube. That would mean its something to the 6th power. And only 7^6, 8^6, and 9^6 have 6 digits, while none of those add up to 43. That means its a square OR a cube (but he gave away its a square, so it can't be a cube), AND under 500000. I gotta go now, so I can't bother to figure out exactly which one it is, but it won't be too difficult now.

[RedScar]

Actually it can be a square and a cube.
Square root 16 you get 4.
Cube root 16 you get 2.
Anyways and number is a cube or square, it's just some numbers are perfect squares and cube roots.

Bythe way is there a time limit to when you have to have the awanser?

[Bosolai]

Um... No.

Cubes are things to the 3rd power, not 4th. If its not already a square before you cube it, it won't be a square after you cube it. Its the same as saying that if an integer isn't even before you multiply by 3, its most definitely not gonna be even after you multiply by 3. capnchicken has it down.

[sasuto]

Squares are if they make a perfect square, like the polygon, cubed are if it makes a perfect cube polygon. you can't have a 2D square that's also and 3D cube. I really doesnt work

[vypon]

SPOILER!!! The answer is:... (Drum Roll Please...)

499899 !!!
Yup, the cubing was false!

Yes, I know I'm a genious... And nice signature Red... ;)

[RedScar]

Ya I fiugre oput what I said was wrong. But Then I for got the change it. Opps. Ya

Thanks for the coment on the sig. :)

[Meister]

Double posts are bad, m'kay?

[RedScar]

Wow I actually found a place with this question. Thanks for the awanser.

[vypon]

Sorry 'bout the double post, meh computer does that when it gets laggy... I'm not completely sure why...

[vypon]

O, this one's easier actually. What is the number of prime numbers between one and a million that become a perfect square if you add 1 to them?

I'm finding the answer as I type this! :D

:bmage: **Think... THINK!!!**

[Meister]

Merged threads. I think it's better when we have one single math riddle thread.

[RedScar]

Hey where did you find this question? Just asking.

[vypon]

A friend of mine supplies me with a couple... He gave me another last night, but I can't figure it out... He never tells me the answer!

[capnchicken]

Since its easier to deal with perfect squares then primes, there are 19 powers of 2 up to 1,000,000 (2^20 = 1,048,576) compared with some 78,000 primes between 1 and 1,000,000. We just add up the ones that are prime if you subtract one from them. These are known as Mersenne primes.


2^1 - 1 = 1
2^2 - 1 = 3
2^3 - 1 = 7
2^5 - 1 = 31
2^7 - 1 = 127
2^13 - 1 = 8191
2^17 - 1 = 131071
2^19 - 1 = 524287

There are 8 Mersenne primes between one and one million. And your previous answer was incorrect vypon, look at my previous response. Your digits add up to 48 not 43.