Question

a)If an inter n ends in 3, show why the last digit of n^5 is 3.
b)Prove that, for all integers n, the last digit of n^5 is the same as the last digit of n.
c) Show that n^5n is divisible by 30 for all integers n.


PLEASE HELP ITS VERY URGENT- MUST BE DONE BEFORE TUESDAY :(

Answer 1

btw please giv a detailed response coz im not very smart :P

Answer 2

a) You can form a kind of cycle to find the units digit of numbers like n^5.
Let n be 13.
13^1 --> units digit 3
13^2 --> units digit 3*3 = 9
13^3 --> units digit = units digit of 3*3*3 = units digit of 27 = 7

Answer 3

Keep repeating this till 13^5

Answer 4

Units digit of 13^5 = units digit of 3^5
= units digit of 3*3*3*3*3

Answer 5

Do you know how to do it now?

Answer 6

Are you there @cloe17

Answer 7

oops soz im here now

Answer 8

Did you understand my explanation?

Answer 9

but does that show that every number ending in 3^5 will end in 3?

Answer 10

You're asking for the next question right?
You can try with any number for second question but that doesn't mean proving. I still don't know how to PROVE the second one.

Answer 11

Did you get the first one?

Answer 12

alternate way, if you have done binomial expansion:

if \[n = 10k + 3\]

\[n^5 = (10k + 3)^5 =243+4050 k+27000 k^2+90000 k^3+150000 k^4+100000 k^5\]


\[n^5 = 10(24+405 k+2700 k^2+9000 k^3+15000 k^4+10000 k^5) + 3\]

Answer 13

for the second one you could let n = 10k + r

Answer 14

That's the format for a two digit number.

Answer 15

no its not, i could let k = 15

Answer 16

and get 150 + r

Answer 17

I understand your expansion but how does that help us in proving units digit is n?

Answer 18

well that shows the last digit is three if thats what you mean, because when we divide by 10 we are left with a remainder of 3

Answer 19

Oh yes!! Hope that clears your doubts @cloe17

Answer 20

...

Answer 21

i dont get why n=10k + 3?

Answer 22

this way sorts out a) and b) together:

consider two positive integers, n and m

the last digit of n is p , and the last digit of m is q

so we have
n=10a + p
m = 10b + q

where a and b are integers.

now lets try and find the last digit of nm

nm = (10a+p)(10b+q) = 100ab + 10aq+ 10pb + pq

= 10(10ab + aq+ pb ) + pq

so therefore the last digit of mn is the last digit of pq


eg lets find the last digit of 1412359 x 458789343

well we just need to find the last digit of 9 x 3

9 x 3 = 21

so the last digit of 1412359 x 458789343 is the last digit of 21,

which is 1

Answer 23

would you agree we can write any number ending in 3 as 10k + 3 ?

eg 23 = 10(2) + 3

153 = 10(15) +3

34873843 = 10(3487384) + 3

so we can say that if n ends in 3, then we can write it as 10k +3

Answer 24

sorry i still dont really gwt it, im only in yr 7 so dont worry :D

Answer 25

can you just tell me how to do c?

Answer 26

wait actually i get the 10k + 3 formula now except im not sure why k has to be multiplied bt 10

Answer 27

oh ok sorry, i thought you were older

these are pretty tricky for year 7 im impressed

c) Show that n^5n is divisible by 30 for all integers n

do you mean n^(5n) ? because that doesnt work for n = 1 for example

Answer 28

n^5 - n

Answer 29

ah ok

Answer 30

try to make the answer my level is that okay?

Answer 31

sure

Answer 32

ty

Answer 33

i'm struggling to explain it simpler..

can i try and guide you through the answer i think is the simplest and easiest?

Answer 34

sure

Answer 35

looking at each part of the problem gives us hints for the next part
ok so for a)
we need to think about numbers that end in 3

every number ending in 3 can be split up like this:


would you agree?