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Can anyone please tell me -

  1. What is the least significant digit in $2^{3A}$?
  2. How do we define least significant digit?
Zain Patel
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abipc
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  • I use SO but I am new here.. I do not know which tag should I use.. – abipc Aug 20 '13 at 06:02
  • I suggest you put parentheses in your expression so that it is clear what you mean, It looks as if you mean $(2^3)^A$. If so, you could indicate it without Latex by writing (2^3)^A. If you intend $2^{3A}$ write 2^(3A). – André Nicolas Aug 20 '13 at 06:23
  • You should edit with Latex to make others understand your formula. – eccstartup Aug 20 '13 at 06:56
  • Done.. Please see the edit note.. Let me use Latex next time I ask a Q here.. – abipc Aug 20 '13 at 08:24
  • Now that you have clarified the question, I can write (if you wish) a few lines giving the answer, with reasons. – André Nicolas Aug 20 '13 at 14:16
  • @AndréNicolas .. tnx .. i can grok it now – abipc Aug 20 '13 at 14:18
  • Good. A straightforward way to do it is to see (as I think you did) that we are looking at $8^A$, and see how the last digits of powers of $8$ cycle. Or else we can simply see how the last digits of powers of $2$ cycle. – André Nicolas Aug 20 '13 at 14:27

2 Answers2

4

The least significant digit of a positive integer, when the integer is given in decimal form, is the units digit, the digit furthest to the right. So the least significant digit of $78675$ is $5$. The least significant digit of $3341116$ is $6$. The least significant digit of $17$ is $7$. The least significant digit of $17450$ is $0$.

Another way of putting it is that the least significant digit of the positive integer $n$ is the remainder when you divide $n$ by $10$.

Now that you know what least significant digit means, you can attack the problem. The important thing is that "last digits" cycle. For example, successive powers of $8$ end in $8,4,2,6,8,4,2,6,\dots$.

André Nicolas
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  • Are you sure about that? I would think that the least significant digit is the nonzero digit in the least-power place. E.g., the least significant digit of 127,000 is 7. I am not sure about this, though. – Stephen Herschkorn Aug 20 '13 at 08:30
  • @Stephen This is what i remember from my Physics classes at university.. but i think the context is different there.. I would say Andre is correct.. – abipc Aug 20 '13 at 08:33
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    @StephenHerschkorn, a power of 2 never ends with a zero digit. – lhf Aug 20 '13 at 10:06
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In powers of $2$ you can see a pattern:

$$2^1=2$$ $$2^2=4$$ $$2^3=8$$ $$2^4=16$$ $$2^5=32$$ $$2^6=64$$

So the last digits of powers of 2 are always $2,4,8,6$ in that order.

For example:

$$2^{15}=....8$$ because $15 \mod 4=3$ and the third ending is $8$.

Now you have to calculate $3*10^{100} \mod 4$.

$10^{100}$ means that you have $1....00000$ a $1$ and after that hundred $0$. So $10^{100}$ is divisible by $4$ because it is the multiply of $100$ so $3*10^{100}$ is also divisibly by $4$. That's why $3*10^{100} \mod 4=0$.

$0$ means that your number's last digit is the 4th in the pattern, so $6$.

Edit: In that case if your question wasn't $2^{3A}$ but ${2^3}^A$ than it's also quiet simple. ${2^3}^A=8^A$ and in the power of $8$ you can also see a pattern $(8,4,2,6)$ and you can use the method that I used previously.

Wyctus
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