Problem:
Solution:
Yes I can!
First, note that we only need to find the unit digit, so we can use the binomial theorem to help us to get rid of a lot of noise
$ 264^n \% 10 = (260 + 4)^n \%10 = 4^n \% 10 $.
Next, notice the unit digit pattern in power of 4s.
4
16
64
256
It is just alternating 4 and 6, so the answer unit digit of the first term is 6, and the second term is 4, so the final unit digit is just 0.
By the way, $ 264^{102} + 264^{103} = $
2672061284943160918522454763266223159532013534099073129065764495853122847577901956596137015881471742644503555582762443491090345356529962017230305249153860349231365381334758920717003765678148536467012213716576437712205080690644467304472055027010109440
It takes no time to compute this in python:
>> 264 ** 102 + 264 ** 103
Solution:
Yes I can!
First, note that we only need to find the unit digit, so we can use the binomial theorem to help us to get rid of a lot of noise
$ 264^n \% 10 = (260 + 4)^n \%10 = 4^n \% 10 $.
Next, notice the unit digit pattern in power of 4s.
4
16
64
256
It is just alternating 4 and 6, so the answer unit digit of the first term is 6, and the second term is 4, so the final unit digit is just 0.
By the way, $ 264^{102} + 264^{103} = $
2672061284943160918522454763266223159532013534099073129065764495853122847577901956596137015881471742644503555582762443491090345356529962017230305249153860349231365381334758920717003765678148536467012213716576437712205080690644467304472055027010109440
It takes no time to compute this in python:
>> 264 ** 102 + 264 ** 103
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