UNSW maths enrichment question (1 Viewer)

[= Jennifer =]

hey everyone

my brother is in year 8 and he cannot find out these answers maybe with your mathematical minds you could figure this out:

a) 35^2004

b)16^2004

c) 67^2005



thanks

 

[McLake]

Originally posted by = Jennifer =
hey everyone

my brother is in year 8 and he cannot find out these answers maybe with your mathematical minds you could figure this out:

a) 35^2004

b)16^2004

c) 67^2005



thanks

METHOD I:

keep dividing the number by 2, or subtracting 1 if odd.

so 2004 = 1002/2 = 501 - 1 = 500/2 = 250/2 = 125 - 1 = 124/2 = 62/2 = 31 - 1 = 30/2 = 15 - 1 = 14/2 = 7.

so 35^2004 = (((((((((35^7)^2 * 35)^2 * 35)^2)^2 * 35)^2)^2) * 35)^2)^2

METHOD II (WORONG!!!!!!!!!):

A good trick is to square the number a lot of times.

ie: 2^8 = 256 = ((((2)^2)^2)^2

so we find the highest power of 2 that fits in 2004.

2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048 ...

so its 1024 (ie: 2^10)

so 35^2004 = 35^1024*35^980

so we can find 35^1024 by squaring 35 ten times (long multiplication).

now do the same with 980.

so highest power is 512 (2^9)

so 35^2004 = 35^1024*35^512*35^468

and 468 fits in 256

212 in 128

84 in 64

20 in 16

and you can find 35^4

so 35^2004 = 35^1024*35^512*35^256*35^128*35^64*35^16*35^4.

Also not that in finding 35^1024 you already have the other answers.

Similarily for other questions.

Is there a better way?

EDIT: Note that 35^a = 7^a*5^a.

so this question is better done using that trick (far smaller numbers).

I also realised that 5^1024 is rather large, so there may be a better answer ...

 

[gman03]

I think the question is incomplete in some way, like those enrichment question doen't usually ask people to evaluate the number (i mean what's the point)... Does the question ask for the laster digit of the final answer? if so you could use modular arithematics...

(hehe me changing the question)

 

[Xayma]

Originally posted by McLake


Is there a better way?

EDIT: Note that 35^a = 7^a*5^a.

so this question is better done using that trick (far smaller numbers).

I also realised that 5^1024 is rather large, so there may be a better answer ...

Method 2 can sort of be done by a few applications on the calculator, ie by going in scientific notation.

ie.

35²⁰⁰⁴=(35^62.625)³²
=(4.980366205*10⁹⁶)³²
= (4.980366205³²)(10⁹⁶)³²
=(2.052867307*10²²)(10^96*32)
=2.052867307*10^22+96*32
=2.052867307*10³⁰⁹⁴

if you dont understand how I did it (or if Im wrong) just ask.

 

[gman03]

Originally posted by Xayma
= (4.980366205³²)(10⁹⁶)³²
=(2.052867307*10²²)(10⁹⁶⁺³²)

(10⁹⁶)³² = 10^96*32?!?

I'm still not convinced the question is asking for approximation

 

[kimmeh]

whats this got to do with 4 unit? when i re read the question, i wasnt there

 

[Xayma]

Originally posted by gman03
(10⁹⁶)³² = 10^96*32?!?

I'm still not convinced the question is asking for approximation

Bah thats right, Ill go edit it. Thats what I get for doing it in a hurry.

Yeah I dont think it is the full question either, but for yr 8 it cant be that much more difficult.

 

[ngai]

umm, what is the question?
did the question booklet just state three random numbers out of nowhere?
if ur trying to find the last digit, then take congruences mod 10...
in which case a) 5, b) 6...
and c) 7^2 == 9 == -1 mod 10, so 7^2005 == (-1)^1002 * 7 == 7 mod 10, so ends in 7