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Finding the Unit digit

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Finding the Unit digit

by gdk800 » Fri Nov 19, 2010 1:55 pm
Pls help with the following problem.

Unit digit of (264)^103 + (264)^102 = ?


My answer: 5

First Term
4 has a cyclicity of 2. Thus 103/2 would leave a remainder of 1
=>4^1 = 4.

Second Term
4 has a cyclicity of 2. Thus 102/2 would leave a remainder of 0
=>4^0 = 1

Therefore, 4 + 1 = 5 is the unit digit.

Am i doing it correctly??
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by Rahul@gurome » Fri Nov 19, 2010 2:23 pm
gdk800 wrote:Pls help with the following problem.

Unit digit of (264)^103 + (264)^102 = ?


My answer: 5

First Term
4 has a cyclicity of 2. Thus 103/2 would leave a remainder of 1
=>4^1 = 4.

Second Term
4 has a cyclicity of 2. Thus 102/2 would leave a remainder of 0
=>4^0 = 1

Therefore, 4 + 1 = 5 is the unit digit.

Am i doing it correctly??
No. You're not.
For the second term you're doing 4^0, which is wrong. Following your logic 4^2 will have 1 as unit digit! You're blindly applying the cyclical pattern method. Simple cyclical pattern exists for powers of 4. Which is: 4^(even) has 6 as unit digit and 4^(odd) has 4 as unit digit. Thus unit digit of [(264)^103 + (264)^102] is unit digit of (4 + 6) i.e. 0.

Another efficient way to solve this particular problem,
... [(264)^103 + (264)^102]
= [264 + 1]*[(264)^102]
= 265*[(264)^102]
= 5*(even number) => Unit digit is 0
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by rishab1988 » Fri Nov 19, 2010 2:31 pm
Here's how i'd approach this problem.

264^102[ 264+1] - factorize

264^102 (265)

Now since the question is about units digit we care about multiplication in only in units digit.

Units digit in 264 -4 ; take the square of 264 - 4*4=16; 6 in units digit ( coz we care only about multiplication in units digit- 4*4=16, in which units digit is 6).

next take 264^3 - gives 4*4*4= 6(from 4*4) *4 = 4 in units digit;

264^4 - 4 in units digit again.

See the pattern; we get 4 in units place when power is odd and 6 in units place when power is even.

Since 102 is even, 264^102 has 6 in units place.

Now muliply 6*5 (units digit from 265)=30 or 0 in units digit.

What is the OA?
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by gdk800 » Fri Nov 19, 2010 2:51 pm
Thanks Rahul, this means that i can apply this method of even power and odd power applicable to only 9 because it has also got a cyclicity of 2?

How about other numbers such as 2, 3, 5?

Can you provide some more similar questions or probably a link on this topic for practice?
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by goyalsau » Fri Nov 19, 2010 8:07 pm
gdk800 wrote:Pls help with the following problem.

Unit digit of (264)^103 + (264)^102 = ?

I would like to do some change, (264)^103 + (263)^102
What should be the answer now,
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by Rahul@gurome » Fri Nov 19, 2010 8:31 pm
I would like to do some change, (264)^103 + (263)^102
What should be the answer now,

264^103 ends in 4 because 103 is odd.

263^102 will end in 3^102.
Now 102 = 4*25 + 2.
So 3^102 will end in 3^2 which is 9.

4+9 = 13 ends in 3.
So 264^103 + 263^102 ends in 3.
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