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- Thread starter Raghav Gupta
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- #1

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- #2

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Can you work out what the last digit is?

- #3

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You have to use congruence modulo theorem.

That is, if ##a,b \in I## and ##m \in I^+##, you can say ##a \equiv b(mod(m))## if a-b is divisible by m.

I had seen this problem somewhere but I'm sure its solved using mod.

That is, if ##a,b \in I## and ##m \in I^+##, you can say ##a \equiv b(mod(m))## if a-b is divisible by m.

I had seen this problem somewhere but I'm sure its solved using mod.

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- #4

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Here is how you can find units digit of ##27^{27}##

##27 \equiv 7mod10##

##27^2 \equiv 47mod10##

##27^2 \equiv -1mod10##

##27^{14} \equiv 1mod10##

##27^{14}*27^{13}=27^{27} \equiv 27^{13}mod10##

Now ##7^{13}=7^{4*3+1}## has a specialty. It's units digit is always 7.

That is ##7^{4k+1}## has units digit 7.

Therefore ##27^{27}## has units digit 7.

##27 \equiv 7mod10##

##27^2 \equiv 47mod10##

##27^2 \equiv -1mod10##

##27^{14} \equiv 1mod10##

##27^{14}*27^{13}=27^{27} \equiv 27^{13}mod10##

Now ##7^{13}=7^{4*3+1}## has a specialty. It's units digit is always 7.

That is ##7^{4k+1}## has units digit 7.

Therefore ##27^{27}## has units digit 7.

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- #5

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Here is how you can find units digit of ##27^{27}##

##27 \equiv 7mod10##

##27^2 \equiv 47mod10##

if 27^2 = -1 (mod 10), then 27^14 is also -1 (mod 10) and not 1.

It can be done even simpler:

27^27 = 3^81

Now use eulers theorem:

[tex] a^{\phi(n)} = 1 (mod n) [/tex]

phi(100) = phi(4) * phi(25) = (4/2) * (25 * 4/5) = 40, so 3^40 = 1 (mod 100)

3^81 = 3^40 * 3^40 * 3 = 1 * 1 * 3 = 3 (mod) 100

- #6

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- #7

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But how ## 27^2\equiv 47mod10## $$ 27^2-47 $$is not divisible by 10.Here is how you can find units digit of ##27^{27}##

##27 \equiv 7mod10##

##27^2 \equiv 47mod10##

##27^2 \equiv -1mod10##

##27^{14} \equiv 1mod10##

##27^{14}*27^{13}=27^{27} \equiv 27^{13}mod10##

Now ##7^{13}=7^{4*3+1}## has a specialty. It's units digit is always 7.

That is ##7^{4k+1}## has units digit 7.

Therefore ##27^{27}## has units digit 7.

- #8

jbriggs444

Science Advisor

Homework Helper

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As to the units digit coming out 3... That is a brain-o. The units digit in ##7^n## cycles through 1, 7, 9, 3 in a cycle of length 4 as n increases sequentially. But 27 is not equal to 4k

So instead of being equal to 7 -- (the cycle element at place 1), it is equal to 3 (the cycle element at place 3).

- #9

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There were certain errors in AdityaDev's post which you corrected.Thanks.But what about second last digit?The topic is to find the sum of last two digits.The last digit could have been figured without modular arithmetic that is the way you posted by the unit cycle method.

As to the units digit coming out 3... That is a brain-o. The units digit in ##7^n## cycles through 1, 7, 9, 3 in a cycle of length 4 as n increases sequentially. But 27 is not equal to 4k+1for some integer k. Instead, 27 is equal to 4k-1for some integer k. (In particular for k = 7).

So instead of being equal to 7 -- (the cycle element at place 1), it is equal to 3 (the cycle element at place 3).

- #10

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Sorry. Typing error.But how ## 27^2\equiv 47mod10## $$ 27^2-47 $$is not divisible by 10.

- #11

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That's okay but what about the second last digit?Sorry. Typing error.

- #12

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Here I got that approach by binomial theorem.Perok why use pencil when pen is in demand.

Can you work out what the last digit is?

- #13

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I don't know. Are you preparing for iit or something?That's okay but what about the second last digit?

- #14

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[tex] 27 ^ {27} = \left( 3^3 \right) ^ {27} = 3 ^ {81}[/tex]

Now finish it. I find easier to work with a power of 3, you may think differently, though.

I provide two solutions below. I wouldn't peek if I were you, though.

Note that [itex]h[/itex] is not constant among the expressions below, the only constraints are those abovementioned.

[tex]27 ^ {27} = (25 + 2) ^ {27} = 25^{27} + 27 \cdot 25 ^ {26} \cdot 2 + \ldots + 2 ^ {27}[/tex]

[tex]25^{27} + 27 \cdot 25 ^ {26} \cdot 2 + \ldots + 2 ^ {27} \equiv 25^{27} + 27 \cdot 25 ^ {26} \cdot 2 + 2 ^ {27} \mod 100[/tex]

[tex]25^{27} + 27 \cdot 25 ^ {26} \cdot 2 + 2 ^ {27} = (25+h) + (27 \cdot 50) + (28 + h) = (25+h) + (50+h) + (28 + h) = 3+h \equiv 3 \mod 100[/tex]

[tex]0 + 3 = 3[/tex]

About [itex]2^{27}[/itex]? Know your powers of 2.

[itex]27 ^ {27} = 3 ^ {81} = 3 ^ {(20 + (20 + (20 + (20 + 1))))} \equiv 3 ^ 1 \mod 100 = 3[/itex]

[itex]0 + 3 = 3[/itex]

- #15

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You are not getting it.I have figured what is the last digit already if you see above posts but not getting the second last digit of the number.I know 3 is the last digit.

[tex] 27 ^ {27} = \left( 3^3 \right) ^ {27} = 3 ^ {81}[/tex]

Now finish it. I find easier to work with a power of 3, you may think differently, though.

I provide two solutions below. I wouldn't peek if I were you, though.

Note that [itex]h[/itex] is not constant among the expressions below, the only constraints are those abovementioned.

[tex]27 ^ {27} = (25 + 2) ^ {27} = 25^{27} + 27 \cdot 25 ^ {26} \cdot 2 + \ldots + 2 ^ {27}[/tex]

[tex]25^{27} + 27 \cdot 25 ^ {26} \cdot 2 + \ldots + 2 ^ {27} \equiv 25^{27} + 27 \cdot 25 ^ {26} \cdot 2 + 2 ^ {27} \mod 100[/tex]

[tex]25^{27} + 27 \cdot 25 ^ {26} \cdot 2 + 2 ^ {27} = (25+h) + (27 \cdot 50) + (28 + h) = (25+h) + (50+h) + (28 + h) = 3+h \equiv 3 \mod 100[/tex]

[tex]0 + 3 = 3[/tex]

About [itex]2^{27}[/itex]? Know your powers of 2.

[itex]27 ^ {27} = 3 ^ {81} = 3 ^ {(20 + (20 + (20 + (20 + 1))))} \equiv 3 ^ 1 \mod 100 = 3[/itex]

[itex]0 + 3 = 3[/itex]

- #16

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Yeah,I am preparing for iit.It seems you also might be preparing?I don't know. Are you preparing for iit or something?

- #17

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And the other digit is 0 (as both my solutions show). IYou are not getting it.I have figured what is the last digit already if you see above posts but not getting the second last digit of the number.I know 3 is the last digit.

- #18

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Sorry for a bit aggressiveness and not understanding you.I was seeing your spoiler of solving it gently and I don't get it.Why we are dividing by 100 and showing remainder like 3.Though we get remainder then it doesn't tells how we got second last digit as 0.Obviously then 0+3=3.And the other digit is 0 (as both my solutions show). Idoget it.

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If the remainder of the division by 100 was 43 I would finish with 4 + 3 = 7.

As the remainder is 3, I end up with 0 + 3 = 3. Get it? It is

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- #21

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Oh, I now get the whole plot. We are doing mod 100 because we have to look at only tens digit and unit digit. So, dividing by 100 would give us remainder of that number last two digits.

## 27^{27}=3^{81}## $$ 3\equiv 3mod100$$ $$ 3^2\equiv 9mod100$$ $$ 3^4\equiv 81mod100$$ $$ 3^5\equiv 243mod100\equiv 43mod100$$ $$ 3^{10}\equiv 1849mod100\equiv 49mod100$$ $$ 3^{20}\equiv 2401mod100\equiv 1mod100$$ Then by your next steps I understood.

But how did you knew [itex]3 ^ {20 + x} \equiv 3 ^ x \mod 100[/itex] so fast. Is it a property or formula known to you earlier for certain numbers?

I was figuring out myself and found out that it is easier to work with ## 3^{81}##. Thanks for imbibing in me mod qualities. See my approach"I want to do it myself and possibly learn something" way:

[tex] 27 ^ {27} = \left( 3^3 \right) ^ {27} = 3 ^ {81}[/tex]

Now finish it. I find easier to work with a power of 3, you may think differently, though.

[itex]27 ^ {27} = 3 ^ {81} = 3 ^ {(20 + (20 + (20 + (20 + 1))))} \equiv 3 ^ 1 \mod 100 = 3[/itex]

[itex]0 + 3 = 3[/itex]

## 27^{27}=3^{81}## $$ 3\equiv 3mod100$$ $$ 3^2\equiv 9mod100$$ $$ 3^4\equiv 81mod100$$ $$ 3^5\equiv 243mod100\equiv 43mod100$$ $$ 3^{10}\equiv 1849mod100\equiv 49mod100$$ $$ 3^{20}\equiv 2401mod100\equiv 1mod100$$ Then by your next steps I understood.

But how did you knew [itex]3 ^ {20 + x} \equiv 3 ^ x \mod 100[/itex] so fast. Is it a property or formula known to you earlier for certain numbers?

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- #22

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Oh, I now get the whole plot. We are doing mod 100 because we have to look at only tens digit and unit digit. So, dividing by 100 would give us remainder of that number last two digits.

I was figuring out myself and found out that it is easier to work with ## 3^{81}##. Thanks for imbibing in me mod qualities. See my approach

## 27^{27}=3^{81}## $$ 3\equiv 3mod100$$ $$ 3^2\equiv 9mod100$$ $$ 3^4\equiv 81mod100$$ $$ 3^5\equiv 243mod100\equiv 43mod100$$ $$ 3^{10}\equiv 1849mod100\equiv 49mod100$$ $$ 3^{20}\equiv 2401mod100\equiv 1mod100$$ Then by your next steps I understood.

But how did you knew [itex]3 ^ {20 + x} \equiv 3 ^ x \mod 100[/itex] so fast. Is it a property or formula known to you earlier for certain numbers?

I do not know if it is a property, I just happen to remember it.

Code:

```
n : 3 ** n % 100
0 : 1
1 : 3
2 : 9
3 : 27
4 : 81
5 : 43
6 : 29
7 : 87
8 : 61
9 : 83
10 : 49
11 : 47
12 : 41
13 : 23
14 : 69
15 : 7
16 : 21
17 : 63
18 : 89
19 : 67
20 : 1
```

For 41, another prime, it is true that [itex]41^{5 + x} \equiv 41^{x} \mod 100[/itex]. Again, this result was obtained by "brute force".

- #23

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- #24

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Glad I could be of assistance.

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