Sum of last two digits of 27^27

  • #1
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Main Question or Discussion Point

It's an amazing question to find sum of last two digits of 27^27 by not using that wolframalpha or calculators.I think binomial theorem would be of help here but not able to apply.can anyone tell me the answer by any method?
 

Answers and Replies

  • #2
PeroK
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If you're allowed paper and pencil, you can just multiply things out and use power reduction.

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.
 
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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.
 
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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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But something's wrong here.According to Wolframalpha the unit digit is coming 3 and second last digit 0.So the sum is 3.Therefore the unit digit is not 7.
 
  • #7
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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.
But how ## 27^2\equiv 47mod10## $$ 27^2-47 $$is not divisible by 10.
 
  • #8
jbriggs444
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Clearly ##27^2## is equal to 49 and hence is congruent to 49 modulo 10. Clearly, this means that it is also congruent to -1 modulo 10. So the "47" is a simple typo.

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+1 for some integer k. Instead, 27 is equal to 4k-1 for 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).
 
  • #9
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Clearly ##27^2## is equal to 49 and hence is congruent to 49 modulo 10. Clearly, this means that it is also congruent to -1 modulo 10. So the "47" is a simple typo.

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+1 for some integer k. Instead, 27 is equal to 4k-1 for 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).
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.
 
  • #10
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But how ## 27^2\equiv 47mod10## $$ 27^2-47 $$is not divisible by 10.
Sorry. Typing error.
 
  • #11
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Sorry. Typing error.
That's okay but what about the second last digit?
 
  • #12
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If you're allowed paper and pencil, you can just multiply things out and use power reduction.

Can you work out what the last digit is?
Here I got that approach by binomial theorem.Perok why use pencil when pen is in demand.
image.jpg
 
  • #13
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33
That's okay but what about the second last digit?
I don't know. Are you preparing for iit or something?
 
  • #14
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"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.

I provide two solutions below. I wouldn't peek if I were you, though.
Let [itex]h[/itex] be any integer such that [itex]100 \mid h[/itex].
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.

I happen to know that [itex]3 ^ {20 + x} \equiv 3 ^ x \mod 100[/itex].
[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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"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.

I provide two solutions below. I wouldn't peek if I were you, though.
Let [itex]h[/itex] be any integer such that [itex]100 \mid h[/itex].
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.

I happen to know that [itex]3 ^ {20 + x} \equiv 3 ^ x \mod 100[/itex].
[itex]27 ^ {27} = 3 ^ {81} = 3 ^ {(20 + (20 + (20 + (20 + 1))))} \equiv 3 ^ 1 \mod 100 = 3[/itex]
[itex]0 + 3 = 3[/itex]
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.
 
  • #16
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I don't know. Are you preparing for iit or something?
Yeah,I am preparing for iit.It seems you also might be preparing?
 
  • #17
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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.
And the other digit is 0 (as both my solutions show). I do get it.
 
  • #18
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76
And the other digit is 0 (as both my solutions show). I do get it.
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.
 
  • #19
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12
Your English is so bad that I can't even understand if you are getting it or not. Add a space after your punctuation.
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 modulo 100 and not modulo 10.
 
  • #20
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As in our high school modular arithmetic is not taught, I am not so acquainted with it. That is one reason for my language looking topsy-turvy. Are you crazy? Why it is looking to you that I am thinking modulo 10 and not modulo 100.
 
  • #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.
"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.

I happen to know that [itex]3 ^ {20 + x} \equiv 3 ^ x \mod 100[/itex].
[itex]27 ^ {27} = 3 ^ {81} = 3 ^ {(20 + (20 + (20 + (20 + 1))))} \equiv 3 ^ 1 \mod 100 = 3[/itex]
[itex]0 + 3 = 3[/itex]
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?
 
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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?
"formula known to you earlier for certain numbers". I know about 3 and 41 by heart.
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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Thank you Mafagafo. My question is solved. In the process you also taught me modular arithmetic and a bit of english like leaving space after punctuations.
 
  • #24
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12
Thank you Mafagafo. My question is solved. In the process you also taught me modular arithmetic and a bit of english like leaving space after punctuations.
Glad I could be of assistance.
 

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