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[TheQueen'sOwn]

Ok so I probably shouldn't be doing math this late.. I can't focus at all. But if anyone could help me with these it would be greatly appreciated. Show me the light!

Proofs
(1) Prove that the square of an odd integer is always of the form 8k + 1, where k is an integer.

(2) Observe that the last two digits of 7^2 are 49, the last two digits of 7^3 are 43, the last two digits of 7^4 are 01, and the last two digits of 7^5 are 07. Prove that the last two digits of 7^201 are 07.

Thanks a lot! I know you can do it GAF!

 

[Xellos]

It's been a long time since I did anything like this so please feel free to tell me if I've made some major error.

Part 1:

Odd integers can be expressed in a series 1 + 2x, so the first is 1, the next is 3, the next is 5 and so on as integer x = 0, 1, 2, etc.

(1 + 2x)^2 = 1 + 4x + 4x^2 = 1 + 4(x + x^2);

x + x^2 = x(x+1), which for x >= 0 is an odd times an even, which is always even
So let x + x^2 = 2k, where k is some positive integer
By substitution, 1 + 4(2k) = 1 + 8k.


Will start working on part 2.

 

[Mugen]

(1 + 2x)^2 = 1 + 4x + 4x^2 = 1 + 4(x + x^2);


lol i got stuck in that... cool stuff Xellos ^_^.\

 

[Dilbert]

Ok so I probably shouldn't be doing math this late.. I can't focus at all. But if anyone could help me with these it would be greatly appreciated. Show me the light!

Proofs
(1) Prove that the square of an odd integer is always of the form 8k + 1, where k is an integer.

(2) Observe that the last two digits of 7^2 are 49, the last two digits of 7^3 are 43, the last two digits of 7^4 are 01, and the last two digits of 7^5 are 07. Prove that the last two digits of 7^201 are 07.

Thanks a lot! I know you can do it GAF!

An idea for part 2: 7^201 = (7^4)^49 * (7^5)

Since the last two digits of 7^4 are 01, I would guess that the last four digits of any power of 7^4 would also be 01. If you can PROVE that guess, then you're in good shape, since the last two digits of (7^4)^n * 7^5 would be the same as 7^5: 07.

Sorry that I can't be more help than that -- number theory is NOT a strong point, and the game is on, after all.

What class is this for?

And, oh yeah, Xellos r0x.

 

[Xellos]

Part 2:

Only last two digits:
7^0 = 01
7^1 = 07
7^2 = 49
7^3 = 343
7^4 = 2401
7^5 = 16807
7^6 = 117649
7^7 = 823543
7^8 = 5764801; etc.

Since 7^0 ends in 01, 7^1 ends in 07, 7^2 ends in 49, 7^3 in 43, and 7^4 begins the pattern again at 01. 7^4 = 2401, so
7^4 =
2401 = 2400 + 01
7^5 =
7 *2401 = 7*(2400 + 01) = (7*2400) + 7*01;
7^6 =
7^2 * 2400 + 7^2 * 01
7^7 =
7^3 * 2400 + 7^3 * 01
7^8 =
7^4 * 2400 + 7^4 * 01 = 2401 * 2401 = (2400 + 01) (2400 + 01)
= ((2400)^2 + 4800) + 01
7^9 =
7 * ((2400^2 + 4800) + 7 * 01

... skipping ahead ...

for n = a multiple of 4 and an integer >= 0

7^n = 2401^n = (2400 + 01)^n, which will always end in 01.



This pattern repeats indeffinitely. So it is possible to determine the last two digits of a power of 7 by dividing the power 7 is being raised to by four, taking the remainder, and then finding the place in the pattern. Divide by four because there are only four elements in the pattern:

r = 0; ends in 01
r = 1; ends in 07
r = 2; ends in 49
r = 3; ends in 43

Example: 7^5; 5/4 = 1 remainder 1, so it will end in 07; 7^5 = 16807 so it works
7^7; 7/4 = 1 remainder 3, so it will end in 43; 7^7 = 823543
7^4; 4/4 = 1 remainder 0, so it will end in 01; 7^4 = 2401

So for 7^201; 201/4 = 50 remainder 1, so it ends in 07.

 

[GG-Duo]

EDIT: (nevermind.)

 

[cvxfreak]

Algebra II/Trig was a NIGHTMARE for my Junior year. I'm glad those days are behind me.

;_;

 

[Diablos]

What's 2+2?

 

[Swordian]

What's 2+2?

2+2={1|}+{1|}={1+2, 1+2|}={3,3|}={3|}=4

 

[thorns]

(2) Observe that the last two digits of 7^2 are 49, the last two digits of 7^3 are 43, the last two digits of 7^4 are 01, and the last two digits of 7^5 are 07. Prove that the last two digits of 7^201 are 07.

I'm really rusty on these things but anyway..

Last two digits of a number is equivalent to finding the remainder when diving that number by 100 (also called modulus),


it is known that ((a mod n)*(b mod n) mod n) = (a*b mod n)

so for example 5 mod 3 = 2, 10 mod 3 = 1, and 50 mod 3 = 2 which is same as (5mod3*10mod3)mod3

so if we factor 7^201, we get
7^201 = 7*7^200 = 7*(7^10)^20 = 7*(7^5*7^4*7)^20

then it's just a matter of substitions..
7^5*7^4*7 mod 100 = 49 mod 100 = 7^2 mod 100

so,
7^201 mod 100 = 7*7^40 mod 100 = 7*(7^4)^10 mod 100

(7^4) mod 100 = 1 mod 100, which means (7^4)^10 mod 100 = 1 mod 100

then

7^201 mod 100 = 7*1 mod 100 = 7 mod 100.

 

[fart]

2+2={1|}+{1|}={1+2, 1+2|}={3,3|}={3|}=4

hahaha, looks like someone's been doing some game theory.

 

[BlackMage]

I before E except after C

 

[Diablos]

2+2={1|}+{1|}={1+2, 1+2|}={3,3|}={3|}=4

Oh jesus christ :lol

 

[Deku Tree]

Chew on this:

Proof:

Given any d> 0, there exists a constant C> 0 such that for every triple of positive integers a,b, c, satisfying a+b=c and gcd(a,b)=1 we have c is less than or equal to C (rad(abc))^{1+d}.

Hint: If you can solve that your a genius.

 

[Diablos]

3489^<43[{39242}[5653]^12049