If Z = 1^1 * 2^2 * 3^3*...* 8^8, what is the remainder when Z is divid

Question

If  Z = 1^1 * 2^2 * 3^3*...* 8^8 , what is the remainder when Z is divided by 10?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

CrackverbalGMAT · Answer

The question is simply asking what is the last digit of Z.

We use cyclicity to find the last digit of each number. Cyclicity means after how many cycles the base comes back as itself.

Cyclicity of 0, 1, 5 and 6 is 1.

Cyclicity of 4 and 9 is 2.

Cyclicity of 2, 3, 7 and 8 is 4.

The product of any even number with a 5 will end in 0.

Here the last digit of  2^2 = 4  and the last digit of  5^5 = 5  as the cyclicity of 5 is 1.

4 * 5 = 20 which is divisible by 10. The remainder is therefore 0.

Option A

Arun Kumar

Chatenya · Answer

answer is simple 
we know that 10 has two factors 5 and 2
and both are present in numerator therefore it will be completely divisible by 10

BrushMyQuant · Answer

Given that If  Z = 1^1 * 2^2 * 3^3*...* 8^8  and we need to find the remainder when Z is divided by 10?

Now, to get a 10 we need a 2 and a 5

Since we have both 2^2 and 5^5 in Z
=> Z becomes a multiple of 10

=> Remainder of Z by 10 = 0

So,  Answer will be A .
Hope it helps!

https://www.youtube.com/watch?v=h1HTEhmUb1w

bumpbot · Answer

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If Z = 1^1 * 2^2 * 3^3... 8^8, what is the remainder when Z is divid

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