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

 

[CM_Tutor]

Are you asking for them expressed in scientific notation?

ie. 16²⁰⁰⁴ = 1.138 ... x 10²⁴¹³

 

[Eagles]

0.o
just wondering, Is it one of those questions that asks you to find the last diget (like those stupid & annoying maths comps questions)???

 

[CM_Tutor]

Finding the last digit would be pretty easy for the first two, as (a) must end in 5 and (b) must end in 6.

 

[lucky]

try using a calculator..

 

[gman03]

For part c

You should note that 67²⁰⁰⁵ (mod 10) = 7²⁰⁰⁵ (mod 10)
and 7⁴ mod 10 = 1

so 67²⁰⁰⁵ mod 10
= 7²⁰⁰⁵ mod 10
= 7*(7⁴)⁵⁰¹ mod 10
= 7*(1)⁵⁰¹ mod 10
= 7*1 mod 10
= 7

so the last digit of 67²⁰⁰⁵ is 7

 

[Eagles]

Originally posted by gman03
For part c

You should note that 67²⁰⁰⁵ (mod 10) = 7²⁰⁰⁵ (mod 10)
and 7⁴ mod 10 = 1

so 67²⁰⁰⁵ mod 10
= 7²⁰⁰⁵ mod 10
= 7*(7⁴)⁵⁰¹ mod 10
= 7*(1)⁵⁰¹ mod 10
= 7*1 mod 10
= 7

so the last digit of 67²⁰⁰⁵ is 7

whats mod? do we learn that in high school??

Originally posted by lucky
try using a calculator..

have you tried?

 

[...]

no

mod are not taught in high school

they are taugh in uni

but if u want, some fellas here can give u a rundown, the concept isn't hard

 

[gman03]

sorry, but I used to mod in software....

N (mob D) = R

R is the remainder when N is divided by D...

No drama...

(Know Visual Basic? it is % in VB)