\(((36472)^{123!})∗((34767)^{76!})\)

Since we have to find last digit, we will find remainder when divided by 10.

For \(((34767)^{76!})\):
34767 fetches 7. Since we have 76! in power thus \(7^2\) gives 49 which gives -1 as remainder. Since 76! is multiple of 4 we get 1 as remainder.

For \(((36472)^{123!})\):
36472 fetches 2. Now we get \({2^123!}\)
We notice every 4th power of 2 fetches 6. Thus, \({2^123!}\) fetches 6.

Hence answer is 6

hello could you explain this in more steps, I did not understand how did we deduce that 76! by 4, by 7's cyclicality right ? but 76! would be too tough to calculate and why did we take \(7^2\) ?

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Hi, Sorry for rushing in the previous comment.

76! = 1*2*3*4*5*6*7*8*.........*76
So, 76! is multiple of 2,4,6,8... and so on. From here I deduced 76! to be multiple of first 2 and then 4.

Now, let me explain each step since I believe some concepts are being missed here:

Concept 1: To find last digit of any term - Find the remainder of term when divided by 10.

Concept 2: When we are finding remainder of any number raised to some power - "For sake of remainder, we can reduce the term to its remainder and raise it to the same power:
Here Remainder(((36472)^123!)/10) = Remainder((2^123!)/10) and Remainder(((34767&76!))/10) = Remainder((7^76!)/10)

Using above two concepts, my question is now reduced to-

(2^(123!)) * (7^(76!))

Now, we have to find Remainder of (2^(123!))/10 -
So here we will apply cyclicity 123! is multiple of 4 as shown above -
We notice every 4th power of 2 fetches 6. Thus, \({2^123!}\) fetches 6

Thus this give me 6.

Now, we have 7^76!- Here we will try to reduce it to 1.

Concept 3 - Negative Remainders - Very Important Concept
Remainder (9/10) = 9 but we can also say (I don't know if it is justified in Official Math, but concept works brilliantly)
Remainder (9/10) = -1 (Caution - Here to find positive remainder, we will have to add the divisor - So if we have -1 as remainder then -1+10 will give me 9 as positive remainder)
**This is essentially a concept that is used often when we divide a negative integer by a positive number - Try on your own or I will share how**

So the easiest way to find remainder is to reduce a number to either 1 or -1 .

Rem(7^(76!))/10 :-
Here we try to reduce 7^76! to some simple number -
We find 7^2 = 49 - We hit gold!
49/10 = -1
Also, 76!/2 is still even so we get (-1)^even = 1

So, 1 is remainder for second term.

Thus, we get 6 as last digit.

I am new to Gmatclub, So, I dont know if there is a comprehensive explainer on concepts for remainder and last digit - But try it on the Gmatclub website and you will find Or else someone can tag the section here! (Btw there are also, a few more concepts that cam handy during exam so do check them)

SOLUTION:

Its a question testing basics from the concepts of Unit Digit

In ((36472)^123!) ,the last two digits of 123! would be 00 as it is a factorial and hence we can say that it is divisible by 4.The unit digit depends on the unit digit of 2^(123!)

Cyclicity of 2 is 4 and hence the unit digit for ((36472)^123!) would be 6 (2^1=2; 2^2 = 4; 2^3 = 8 and 2^4=..6 )

In ((34767)^76!),the last two digits of 76! would be 00 and hence divisible by 4.The unit digit depends on the unit digit of 7^(76!)

Cyclicity of 7 is 4 and hence the unit digit for ((34767)^76!) would be 1 (7!=7 ; 7^2=..9 ; 7^3=..3 ;7^4=..1)

Hence the Unit digit in the expression ((36472)^123!)∗((34767)^76!)

= 6 x 1 = 6 (OPTION E)

Hope this helps
Devmitra Sen(Math)
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