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What is the ones digit of 3^23 - 2^18?
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27 Jul 2017, 10:05
27 Jul 2017, 10:05
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Question Stats:
79% (01:12) correct
20% (01:17) wrong based on 88 sessions
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Re: What is the ones digit of 3^23 - 2^18?
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25 Sep 2017, 06:11
25 Sep 2017, 06:11
1
Let's use the fact that unit digits are recurrent in powers so that
\(3^1=3\)
\(3^2=9\)
\(3^3=..7\)
\(3^4=..1\)
\(3^5=..3\)
...
And
\(2^1=2\)
\(2^2=4\)
\(2^3=8\)
\(2^4=..6\)
\(2^5=..2\)
....
Thus, \(3^21=...7\) and \(2^18=...4\). Then, their difference will terminate with a 7-4=3!
\(3^1=3\)
\(3^2=9\)
\(3^3=..7\)
\(3^4=..1\)
\(3^5=..3\)
...
And
\(2^1=2\)
\(2^2=4\)
\(2^3=8\)
\(2^4=..6\)
\(2^5=..2\)
....
Thus, \(3^21=...7\) and \(2^18=...4\). Then, their difference will terminate with a 7-4=3!
Re: What is the ones digit of 3^23 - 2^18?
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25 Sep 2017, 06:30
25 Sep 2017, 06:30
3
Carcass wrote:
Now we know
Ones digit of\(2 ^ {any power}\) = 2,4,8,6 i.e it keeps repeating after every 4 cycle
Ones digit of \(3 ^ {any power}\) = 3,9,7,1 i.e it keeps repeating after every 4 cycle
Now ones digit of \(3^{23}\) = 7 (divide 23/4 , since it repeats after every 4 cycle and the remainder is 3, so we have to consider the third term i.e 7 (3,9,7,1))
Similarly ones digit of \(2^{18}\) = 4 (divide 18/4 , since it repeats after every 4 cycle and the remainder is 2,
so we have to consider the second term i.e 4(2,4,8,6))
Now
the ones digit of \(3^{23} - 2^{18}\) = 7-4 = 3
