Question

can anyone please solve this What is the unit's digit in the product of (7^71*6^59*3^65)????

Answer 1

you need to solve : (7^71*6^59*3^65) mod 10

Answer 2

\( 7^{71} = 7(7^2)^{35} = 7(-1)^{35} = -7 = 3 \mod 10 \)

Answer 3

similarly find the other remainders

Answer 4

3^x can have only three digits in the unit place (leaving the case when x=0)
3,9,1 so for 3^65 you do 65 mod 3 which give 2 so unit place will be 9 for 3^65

similarly 6^x has only 1 value in the unit place so 6^59 will have 6 in the unit place

for 7^x we have 4 values 7,9,3,1 in the unit place so we will find 71 mod 4 which is 3
so 7^71 will have 3 in the unit place

Now we will get 3*6*9 =162
hence 2 will be in the unit place

Answer 5

but the answer to this question is 4

Answer 6

Unit's digit of (3^65)*(6^59)*(7^71) = (Unit's digit of 3^65)*(Unit's digit of 6^59)*(Unit's digit of 7^71)

Any power of 6 ends with 6.
Hence, unit's digit 6^59 = 6

Now for powers of 3,
Unit's digit of 3^(Multiple of 4) = 1
Unit's digit of 3^(Multiple of 4 + 1) = 3
Unit's digit of 3^(Multiple of 4 + 2) = 9
Unit's digit of 3^(Multiple of 4 + 3) = 7

Hence, unit's digit of 3^65 = unit's digit of 3^(4*16 + 1) = 3

Now for powers of 7,
Unit's digit of 7^(Multiple of 4) = 1
Unit's digit of 7^(Multiple of 4 + 1) = 7
Unit's digit of 7^(Multiple of 4 + 2) = 9
Unit's digit of 7^(Multiple of 4 + 3) = 3

Hence, unit's digit of 7^71 = unit's digit of 7^(4*17 + 3) = 3

Hence, unit's digit of (3^65)*(6^59)*(7^71) = unit's digit of 3*6*3 = unit's digit of 54 = 4

Answer 7

ok got it thanks a lot

Answer 8

you are welcome!!

Answer 9

K I made a mistake with 3. Sorry

Answer 10

@beccaboo022a can you help me understand your method please

Answer 11

thanks for helping:)

Answer 12

\(7^{71} = 7(7^2)^{35} \)

Answer 13

@thisSucks

Answer 14

k then?

Answer 15

\(7^2 = 49\) leaves a remainder of -1 when divided by 10

Answer 16

so,

\(7^{71} = 7(7^2)^{35} = 7(49)^{35} = 7(-1)^{35} \mod 10 \)

Answer 17

see if that makes sense

Answer 18

I followed till the point 72=49 leaves a remainder of -1 when divided by 10
But not sure how can we write 7(49)35=7(−1)35 mod 10

Answer 19

we use this property if mods :-
\(a = b \mod n \)

\(\implies \)

\(a^k = b^k \mod n \)

Answer 20

*of

Answer 21

\(49 = -1 \mod 10\)

\(\implies\)

\(49^k = (-1)^k \mod 10\)

Answer 22

I have not learned this property. Also I learned that mod always gives +ve number for positive numbers.

Answer 23

okay, thats just false.

-1 is same as 9 in mod 10

Answer 24

-11 = -1 = 9 in mod 10

Answer 25

mod is defined for all integers

Answer 26

Thanks for the effort but still I am not clear. even when I try it on google calculator -1 mod 10 gives 9

Answer 27

yes, so 9 mod 10 = -1 mod 10

Answer 28

Thanks!
Can you share any online proof for a=b mod n
I don't want to waste your time

Answer 29

i have time,
i can try the proof here. its simple, wont take long time

Answer 30

Sure!

Answer 31

can you state the proof here. I will have a look and if any doubt I will ask you

Answer 32

first a simple definition for mod :-

\(a \equiv b \mod n\)
means, \(n | a-b\)

Answer 33

next few trivial results :-
1) \(a \equiv a \mod n\)
2) if \(a = b \mod n \), then \(b \equiv a \mod n \)
3) if \(a = b \mod n \) and \(c = d \mod n \), then \(a+c = b+d \mod n \) and \(ac = bd \mod n \)

Answer 34

next we can conlcude the proof, if u are okay wid above results :)

Answer 35

K let me comprehend

Answer 36

yahh

Answer 37

to make sense of above results :
for each result, just verify if the definition of mod holds or not.

Answer 38

I feel so stupid...ugh But not sure how 3) point works. It holds true for values though

Answer 39

okay lets prove 3rd result, before moving to our actual proof :)

Answer 40

No No not required I realized it how it works :)

Answer 41

good :) next we prove below using induction :-
to prove : if \(a \equiv b \mod n\), then \(a^k \equiv b^k \mod n \)

Answer 42

clearly, the statement holds when \(k = 1\)

Answer 43

Yeah I think I can prove it with induction. But I never get a good understanding when proofs are done via induction

Answer 44

oh actually induction proofs are easy to *see* than analytical proofs

Answer 45

as for:
Unit's digit of 3^(Multiple of 4) = 1 < same
Unit's digit of 3^(Multiple of 4 + 1) = 3
Unit's digit of 3^(Multiple of 4 + 2) = 9
Unit's digit of 3^(Multiple of 4 + 3) = 7
so, the unit's digit of 3^65 = unit's digit of 3^(4*16 + 1)=3

and when the powers of 7... the unit's:
Unit's digit of 7^(Multiple of 4) = 1 < same
Unit's digit of 7^(Multiple of 4 + 1) = 7 <
Unit's digit of 7^(Multiple of 4 + 2) = 9
Unit's digit of 7^(Multiple of 4 + 3) = 3 <
.....

Answer 46

ever looked at proof for induction method ?

Answer 47

I have proofed it using property 3.

Answer 48

if u did, then u wud love induction proofs

Answer 49

Yes.... property3 is needed for induction here...

Answer 50

I mean there is a proof, which proves that induction method works.

Answer 51

But somehow I don't *see* things with induction but that's OK. But still I say lets divide the step 2 in two parts and say -1 mod 10 is equivalent to 49 mod 10

Answer 52

I would love to see the proof that says induction method works :)

Answer 53

This is a cool method. I liked it Thanks :)
K and help me with one more thing...How can I reward you for all your effort :)
I am in debt now lol

Answer 54

lol im in debt too:)
proof of induction method is simple :-
see if u can get along wid this :

Theorem 1.2 First Principle of Finite Induction. Let S be a set of positive integers
with the following properties:
(a) The integer 1 belongs to S.
(b) Whenever the integer k is in S, the next integer k + 1 must also be in S.

\(\text{Then S is the set of all positive integers.} \)

Answer 55

thats the statement for 'induction method'. we can prove it elegantly wid a small trivial argument.

Answer 56

I know the statement. and It makes sense but how can you say that proving that k+1 is the *sufficient* condition

Answer 57

exactly, the argument involves proving that, 'proving k+1' is sufficient :)

Answer 58

BTW you are really smart. Do you actually remember the definitions also?

Answer 59

Ohh yes I get it now, A bit thoughtful Thanks :)

Answer 60

(a) The integer 1 belongs to S.
so, 1 is in set.
(b) Whenever the integer k is in S, the next integer k + 1 must also be in S.
assuming k is in set, IF WE COULD PROVE that k+1 also is in set --- thats same as proving that the set contains all integers >=1

Answer 61

Yeah I get it now! Thank you so much :)

Answer 62

i have another ques can u ppl pis solve it now

Answer 63

I see np :) thanks for the kind words lol you're super smart xD

Answer 64

yah losan you may close this and ask it in a thread :)

Answer 65

*new thread

Answer 66

Yeah let me try this one with all the extra knowledge I gained today

Answer 67

ok