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??
Finding the Unit digit
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- Rahul@gurome
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No. You're not.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??
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
Rahul Lakhani
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- rishab1988
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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?
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?
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?
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?
- goyalsau
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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,
Saurabh Goyal
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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.
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.
Rahul Lakhani
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Quant Expert
Gurome, Inc.
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On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)