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How would I go about finding the Units Digit to 7^(2945)?

I know that: 7^0 = 1

7^1 = 7

7^2 = 49

7^3 = 343

7^4 = 2401

...

7^9 = 40353607

JDog
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3 Answers3

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when u are interested only in the unit digit you can multiply step by step like u did in your post but you can forget the other digits.

So you have $$7^2 = 49$$ however you forget what was it exactly and u only remember that $$7^2 = x\cdot 10 + 9 $$ you multiply again $$7^3=7^2 \cdot 7 = 70x+63 = (7x+6) \cdot 10+3$$ yet again you only remembering that it looked like $$7^3=x_3 \cdot 10 +3$$ as you can see you dont have to remember $7^9 = 40353607$, $7^9 =x_9 \cdot 10 +7$ will be enougth

so now we can repeat you what u did but make it simpler. We can note that

  • if the $7^n$ ends with $1$ then $7^{n+1}$ ends with 7
  • if the $7^n$ ends with $7$ then $7^{n+1}$ ends with 9
  • if the $7^n$ ends with $1$ then $7^{n+1}$ ends with 3
  • if the $7^n$ ends with $1$ then $7^{n+1}$ ends with 1

so the sequence of units digits will be $(1,7,9,3,1,7,3,9,\ldots)$

now you want 2945th element of that sequence. The value repeats itself evey 4 steps.$$a_n=a_{n-4}$$ and also $$a_n=a_{n-20}$$ $$a_n=a_{n-100}$$

so u can say that: $$a_{2945}=a_{45}=a_5=a_1=7^1=7$$

Kazz
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1

as you know $7^4$ is a number whose units digit is 1. You can write $7^{2945}$ as $7^{(736\times 4+1)}$ or alternately as $(7^4)^{736}7^1$ as the first factor when simplified will be a number whose units digit is 1 so the answer is a number ending with "7" as its unit digit...

Joffan
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Parminder
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  • Guide to the MathJax formula (and much else) formatting tool: http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Joffan Dec 07 '16 at 02:12
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It is easy to verify that$$7^{4n+k}\equiv 7,9,3,1\pmod{10}\text{ when }k=1,2,3,4 \text{ respectively}$$ Hence $$7^{2945}=7^{4\cdot736+1}=(7^{736})^4\cdot 7\equiv 7\pmod{10}$$

Piquito
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