Find the last digit in the expression (36472)^123!*34767^76 ?

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)

I dont understand the unstated fascination with 4 here

Would suggest you to go through Cyclicity of Units Digit and Cyclicity. In this particular question we have the units digits 2 and 7, both have a cyclicity of 4 i.e. the units digit repeats a pattern after the numbers are multiplied with themselves 4 times e.g. 2^1 = 2, 2^2 = 4, 2^3 = 8, 2^4 = 16 | 2^5 = 32, 2^6 = 64, 2^7 = 128, 2^8 = 256 ...and so on, did you notice the pattern 2, 4, 8, 6 ? Similarly it is for 7 => 7, 49, 343, 2401, ---7, ----9, -----3, -----1 and so on.

We know that 123! and 76! are a multiple of 4, hence the last digits of two numbers would be 6 and 1, now 6*1 = 6

The question boil down to 2^123! X 7^ 76!
we need the last digit of 2^123! and 7 ^76!
cyclicity concept:
the cyclicity of 2 is 4 (the last digit repeats after every 4th power).
we divide 123!/4 => Here we notice that any number's(>3) factorial is always divisible by 4.
Here is how:

4!=4X3X2
5!=5X4!
6!=6X5X4!
so.. 123! is divisible by 4. which means the last digit for 2^123! =2^4 = "6"
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same logic for 7 . as 7 also has a cyclicity of 4. and 76! is divisible by 4.
The last digit of 7^76! =7^4 = 1

multiply the last digits: 6X1 =6