Stumped on math problem

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cabri

Diamond Member
Nov 3, 2012
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Can not figure out the logic need to solve

(7^33) Modulo (10)

I can do it brute force; but have been told that is a slippery way to figure this out.
 

TecHNooB

Diamond Member
Sep 10, 2005
7,458
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Write down the 1's place for the first few powers.

7^1, 7^2, 7^3, etc. From the pattern, figure out what the 1's place of 7^33 is supposed to be. That's your answer.
 

dighn

Lifer
Aug 12, 2001
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basically you are looking for the first digit. that repeats in a pattern of 7 9 3 1, so 33 iterations gives you 4 cycles and 1, which is 7

another solution is to halve the exponent every time and also look at only the one's place:

7^33 = (7^2) ^ 16 * 7 = 49 ^ 16 * 7
take one's place only => (9^2) ^ 8 * 7 = 81^8 * 7
take one's place only => 1^8 * 7 = 7
 
Last edited:

_Rick_

Diamond Member
Apr 20, 2012
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You can pull modulo into factors.

So to make dighn's post a bit more understandable:

(7^33)mod 10 = ((7^4)mod 10)^8 * 7 mod 10, where the first term resolves to 1.

But you can also have some fun, and calculate (3^11)mod 10. This teaches you, that 3 and 7 follow the same pattern (well, inversed) for the last digit, when looking at different powers of the numbers.
 

disappoint

Lifer
Dec 7, 2009
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Where did you all learn modulo? No one ever taught me modulo or even mentioned it. Not through high school and not through an accredited BSEE degree. I only know of it after reading a book on it outside of school.
 

DrPizza

Administrator Elite Member Goat Whisperer
Mar 5, 2001
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Where did you all learn modulo? No one ever taught me modulo or even mentioned it. Not through high school and not through an accredited BSEE degree. I only know of it after reading a book on it outside of school.
I think the first time I ran into it was in discrete mathematics - after I had gone through 4 years of engineering & all the math that comes with it.
 

jaedaliu

Platinum Member
Feb 25, 2005
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Where did you all learn modulo? No one ever taught me modulo or even mentioned it. Not through high school and not through an accredited BSEE degree. I only know of it after reading a book on it outside of school.

elementary school math club? Either that or Jr. High.

I can't remember. It's been a while.

I'm pretty sure I saw it in one of my intro-to-programming courses in undergrad.
 

_Rick_

Diamond Member
Apr 20, 2012
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I learned to divide with a remainder in elementary school, but don't recall when I learned that there was an actual interest in it.
Got hit pretty hard with it in fresh man linear algebra, with modulo-classes and later on in a rehash in number theory, I think.
 

HumblePie

Lifer
Oct 30, 2000
14,665
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The remainder operation, Modulo, I learned pretty early. Although never really had a chance or need to apply it to a really large exponential number either. Or if I did back in the day I don't remember doing it much. I still use the operation quite a lot though as a programmer, just again not on very large exponential numbers.
 

Unitary

Member
Dec 7, 2006
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I know I'm a bit late, but I thought I'd share another possible solution.

Again using properties of mod, you first know that 7^2=49 is congruent to -1 (mod 10). In other words 7^2==-1 (mod 10) Take 16 powers of each side, (7^2)^16==(-1)^16 (mod 10), or (7^32)==1 (mod 10)

Multiply each side by 7 to get, 7^33==7 (mod 10)

This method avoids working with powers of 3 and uses powers of -1 instead.
 

JTsyo

Lifer
Nov 18, 2007
11,818
953
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Where did you all learn modulo? No one ever taught me modulo or even mentioned it. Not through high school and not through an accredited BSEE degree. I only know of it after reading a book on it outside of school.

It's used more in programming than basic math. I remember seeing it on my calculator in HS and trying to figure out what it does. I wasn't able to and had to refer to the manual.
 

Daverino

Platinum Member
Mar 15, 2007
2,004
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1*7 = 7 % 10 = 7 (7^1)
7*7 = 49 % 10 = 9 (7^2)
9*7 = 63 %10 = 3 (7^3)
3*7 = 21 % 10 = 1 (7^4)
1*7 = 7 % 10 = 7 (7^5)

Pattern repeats every at 7^4

7^33 = (7^4) ^8 *7
((7^4) ^ 8) % 10 = 1
1 * 7 = 7
 
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