ParthKohli
  • ParthKohli
A common technique for finding modulo \(10\) of any huge number in the form \(a^b\)?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
ParthKohli
  • ParthKohli
It seems that you have to go through a lot of manipulations... a lot for doing that. Although I was thinking of a technique they use to figure all that out.
ParthKohli
  • ParthKohli
OK, I'm gonna change the topic to something else:\[39^{39} \equiv n\pmod{100}\]
ParthKohli
  • ParthKohli
\(n\) has to be a two-digit number. (I actually have to find the last two digits)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

ParthKohli
  • ParthKohli
Something with a pattern?
shubhamsrg
  • shubhamsrg
39 = -61 mod100 =>39^38 = 61^38 mod100 =>39^39 = 39*61^38 mod100 SInce 61^n always ends in 1, that multiplied by 39, will give last digit as 9, hmm, leme see for the second last.
terenzreignz
  • terenzreignz
Better yet, 39^5 = -1 mod 100 Use that: 39^39 = (39^5)^7 * 39^4 = (-1)^7 * 39^4 mod 100
shubhamsrg
  • shubhamsrg
Seems legit. But on paper, who's gonna go till 39^5 ! :O
ParthKohli
  • ParthKohli
Very nice, can you tell me *how* you came up with that solution? What do you initially think?
terenzreignz
  • terenzreignz
square 39, and then only take the last two digits...
terenzreignz
  • terenzreignz
then multiply 39 to it again.
terenzreignz
  • terenzreignz
and only take the last two digits, again...
ParthKohli
  • ParthKohli
Ahem, that was beautiful. But is there a common way to figure out these solutions?
shubhamsrg
  • shubhamsrg
Ohh, I see.
terenzreignz
  • terenzreignz
The way I think of things, and it's a common trend that the way I do it is highly inefficient (LOL) But when faced with a problem b^p = n (mod k) I try to see if there's an exponent to which b can be raised so that it is 1 (or -1) mod k first.
terenzreignz
  • terenzreignz
That "last two digits" trick only works because you were conveniently working under mod 100... XD
ParthKohli
  • ParthKohli
Nice... can you give me another example?
terenzreignz
  • terenzreignz
How about 6^5000 = n (mod 215)
terenzreignz
  • terenzreignz
(sneakily thinking of simple problems) :>
ParthKohli
  • ParthKohli
Arhmegurd.\[\left(6^{5}\right)^{1000} \equiv (6^2)^{1000} \equiv n \pmod{215}\]
ParthKohli
  • ParthKohli
\[\left(6^5\right)^{400} \equiv (6^2)^{400} \equiv n\pmod{215}\]
ParthKohli
  • ParthKohli
\[\left(6^5\right)^{120} \equiv 6^{240} \equiv n \pmod{215}\]
terenzreignz
  • terenzreignz
I'm not following o.O You may be using a method I'm not familiar with, though... if you are, do continue :)
ParthKohli
  • ParthKohli
\[\left(6^{5}\right)^{48} \equiv 6^{96} \equiv n \pmod{215}\]
ParthKohli
  • ParthKohli
Hmm... I am just using \(6^2 \equiv 6^5 \pmod{215}\)
ParthKohli
  • ParthKohli
What should I do here?
terenzreignz
  • terenzreignz
Hang on...
terenzreignz
  • terenzreignz
I was rather hoping you'd go for \[6^3 = 216 = -1(mod \ 215)\]
ParthKohli
  • ParthKohli
Oh...
terenzreignz
  • terenzreignz
Wait, typo -.-
ParthKohli
  • ParthKohli
Yes, isn't that 1 (mod 215)
terenzreignz
  • terenzreignz
\[\large 6^3 = 216 = 1(\mod \ 215 \ \ \ \ )\] Much better
terenzreignz
  • terenzreignz
And work from there...
terenzreignz
  • terenzreignz
\[\large 6^{5000} = 6^{3(1666)+2}\]
ParthKohli
  • ParthKohli
\[6^{5000}= 6^{4998} \times 36 \equiv \]
terenzreignz
  • terenzreignz
\[1 \times 36 (\mod 215 \ \ \ \ \ )\]
ParthKohli
  • ParthKohli
\[\equiv 36 \pmod{215}\]
ParthKohli
  • ParthKohli
Argh, you type fast. =3
terenzreignz
  • terenzreignz
Awesomeness
ParthKohli
  • ParthKohli
Another n00bish question coming up. Thanks man.
terenzreignz
  • terenzreignz
If you have time, look up Fermat's Little Theorem, might come in handy in a pinch.
ParthKohli
  • ParthKohli
Yesh, that one is cool:\[p^{n - 1} \equiv 1 \pmod{n} \]I'd try to see if that applies in the future.

Looking for something else?

Not the answer you are looking for? Search for more explanations.