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What is the remainder when 5^137 is divided by 8?

 Dec 19, 2018
 #1
avatar+100800 
+3

1    5,   

2    25,

3    125,           odd

4    625,             even

5    3125,           odd

6    15625           even

7     78125             odd

8    390625           even

 

ok I have a pattern

 

5^137 = 1000n+ 125 = 125*8*n  +   17*8     +  1  =  8(1000n+17) +1       n has a specific integer value

 

Yes I made a small stupid numeric error that i am quite sure hatchet288 was smart enough to correct.

This last line should have been

 

5^137 = 1000n+ 125 = 125*8*n  +   15*8     + 5  =  8(1000n+15) +5       n has a specific integer value

 

so

 

\(5^{137}\div 8 \)        has a remainder of 5

 

It was no big deal, any question asker should have been able to pick up that little error for themsleves.

 

BUT Good effort Tertre, it was a good catch.

 Dec 20, 2018
edited by Melody  Dec 20, 2018
edited by Melody  Dec 20, 2018
edited by Melody  Dec 20, 2018
 #2
avatar+4221 
+2

Melody, you're on the right track, but here is my attempt:

 

You're right about the odd and even switching,

5^3=125

5^4=625

5^5=last three digits of 125

5^6=last three digits are 625

 

So, for odd powers of five, the last three digits are 125 and for even powers five, the last three digits are 625. By the divisibility rule for 8, and we have \(\frac{125}{8}\) since 137 is odd. Thus, the remainder is \(\boxed{5}.\)

.
 Dec 20, 2018
 #3
avatar+100513 
0

Good job, tertre!!!!!

 

 

cool cool cool

CPhill  Dec 20, 2018
 #4
avatar+100513 
+2

Here's how I view this....

 

5^1 / 8 = (8 - 3)^1 / 8   

 

Every term in the expansion is a multiple of 8 except the last  which is

(-3)^1 / 8  = (-3)^1  mod 8  = 5

 

Similarly

 

5^2 /8  =  (8 - 3)^2 / 8

Again we only need consider the last term which is

(3)^2/8  =  (3)^2 mod 8  =   1

 

 

etc....

 

So...we only need to consider  the last term in each subsequent power expansion and note that

 

(3)^n mod 8 =  1      and

 

(-3)^n mod 8 =  5

 

So

 

5^137 / 8    =  (8 - 3)^137 / 8      will  have (-3)^137 / 8  as a last term

And the remainder is  (-3)^137 mod 8  =   5

 

 

 

 

cool cool cool

 Dec 20, 2018
edited by CPhill  Dec 20, 2018
edited by CPhill  Dec 20, 2018
 #5
avatar+100800 
0

Hi Chris, I expect your logic is excellent but I am having trouble following it.

 

You have just done the same (sort of) as Tertre and me haven't you?

I mean you have developed a pattern and used the pattern (odd and even powers) to answer the question.

 

I am just a little confused because you have not shown the pattern but I am assuming that is what etc means.

Melody  Dec 20, 2018
 #6
avatar
+1

Working to mod 8 ,

\(5^{137}\equiv5.5^{136}\equiv5.25^{68}\equiv5.1^{68}\equiv 5.\)

So the remainder will be 5.

 Dec 20, 2018
 #7
avatar+100800 
0

I reallly like that way. It is so short and simple.

Thanks guest.:)

Melody  Dec 20, 2018
 #8
avatar+183 
0

Good solution, but it's \(5*5, \) not \(5.5\) .

azsun  Dec 20, 2018
 #9
avatar+100800 
0

Azsan,    5.5 means 5*5      

I can understand your confusion but the 2 notations are interchangable. 

Guest's answer is excellent.

Melody  Dec 21, 2018

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