- #1
- 1,011
- 76
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?
AdityaDev said:Here is how you can find units digit of ##27^{27}##
##27 \equiv 7mod10##
##27^2 \equiv 47mod10##
But how ## 27^2\equiv 47mod10## $$ 27^2-47 $$is not divisible by 10.AdityaDev said: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.
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.jbriggs444 said: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).
Sorry. Typing error.Raghav Gupta said:But how ## 27^2\equiv 47mod10## $$ 27^2-47 $$is not divisible by 10.
That's okay but what about the second last digit?AdityaDev said:Sorry. Typing error.
Here I got that approach by binomial theorem.Perok why use pencil when pen is in demand.PeroK said: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?
I don't know. Are you preparing for iit or something?Raghav Gupta said:That's okay but what about the second last digit?
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.mafagafo said:"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]
Yeah,I am preparing for iit.It seems you also might be preparing?AdityaDev said:I don't know. Are you preparing for iit or something?
And the other digit is 0 (as both my solutions show). I do get it.Raghav Gupta said: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.
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.mafagafo said:And the other digit is 0 (as both my solutions show). I do get it.
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 approachmafagafo said:"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]
"formula known to you earlier for certain numbers". I know about 3 and 41 by heart.Raghav Gupta said: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?
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
Glad I could be of assistance.Raghav Gupta said: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.