What is the units digit of 31467^32 × 97645^23 × (32168^5 + 8652)^479?

Question

A. 0

B. 1
 
C. 2

D. 5

E. 8

chetan2u · Answer

Hi...
Relook at the Q and choices, 0 can never be the units digit here..

For ANSWER let's see only the units digit..
 3^2*2^3*(8^5+5^2)^{479} 
... This means  9*8*(8+5)^{480-1} 
480 is div by 4 so units digit of base will be same as that of 4-1 or 3..
So 72*13^3.....2*3^3=2*7=14..
Units digit is 4

fitzpratik · Answer

Hi  chetan2u

Have made necessary changes!

Please do excuse my clumsiness.

Thank you.

chetan2u · Answer

For ANSWER let's see only the units digit..
 7^{32}*5^{23}*(8^5+2)^{479} 
... This means  7^4*5*(8^1+2)^{480-1} ...... 
after every 4th power, the units digit repeat
 for units digit 0, 1, 5 and 6... it remains the same for every positive integer power
for 2,3,7,8 it repeats after every 4th.. so 2^401 will be same as 2^(4*50+1) or 2^1
4 and 9 repeat after every second.. 
so  7^4*5  will be 5 as ODD *5 will have units digit 5
but  (8^1+2)^{479} = 10^{479}  will always have 0 as units digit..
so overall also it will be 0

A

fitzpratik · Answer

Could not have put it in any better way!!

generis · Answer

chetan2u  , I'm not clear about why you wrote this part:  (8^1+2)^{480-1} . (480/4) is evenly divisible, maybe the (-1) is a negative remainder (and I will never understand negative remainders fully) but that doesn't seem to explain the exponent manipulation, either. . .

Was it to show remainder in case  (8^1+2)  summed to something other than zero? If that is the reason, I still don't follow.

You can't distribute the exponent, so cyclicity of four for addends 8 and 2 can't be the reason.

And cyclicity per se doesn't seem to be the reason: 2, 3, 7, and 8 have cyclicity of four (where 480 is divisible by 4) - but 0, 1, 4, 5, 6, and 9 do not have cyclicity of four. That 8 might have had units digit 2, e.g. Then units digit is 4, and I can't see how  4^{480-1}  helps.

The (-1) part isn't clear, either. I suppose I know that, in the number whose cyclicity of powers is four, I can take the  n^4  term and "move backwards" by one power in the cycle to get the units digit, but 479/4 leaves remainder 3, which is the same thing as "moving one place backward" in the cycle. It seems harder to see, from a divisibility perspective, that (-1) power in cycle = (+3) power in cycle.

I looked only at  8^5 , where 5/4 = remainder 1, units digit of first term is same as  8^1  = 8. Last digit of addend in parentheses is 2. (8 + 2) = 0.

And as you point out, zero to any power (except zero) is 0. So I just wrote (first expression) * 0.

Sorry if this confusion arises because I am missing something obvious. I suspect what you wrote is important for problems different from this one. Please explain?

What is the units digit of 31467^32 × 97645^23 × (32168^5 + 8652)^479?

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