Question

Number theory question

Answer 1

Find the number of pairs of \((m,n)\) , \(\ \ 1\leq m\leq 100,\ \ 1\leq n\leq 100\)
such that the number \((2^{m}+2^{n})\) is divisible by \(5\).

Answer 2

Notice that \(2^4 =16\) leaves a remainder \(1\) when divided by \(5\)

Answer 3

yea the fermat little theorem

Answer 4

Therefore the number \(2^{k+4}\) leaves a remainder \(2^k\) when divided by \(5\), yes ?

Answer 5

yyes

Answer 6

yes

Answer 7

Observe that \(2^{4b} = (2^4)^b\) leaves a remainder \(1\) when divided by 5

Answer 8

Therefore the number \(2^{a+4b}\) leaves a remainder \(2^a\) when divided by \(5\), yes ?

Answer 9

yes

Answer 10

2^1 = 2
2^2 = 4
2^3 = 8 = 3
2^4 = 16 = 1
2^5 = 32 = 2
....


as you can see, the remainders repeat after four iterations :

2, 4, 3, 1, 2, ...

Answer 11

Lets take a look at the powers \(m\) that make \(2^m\) leave a remainder \(2\)

Answer 12

\(2^{1},2^{5},2^{9}.....2^{99}\)

Answer 13

they would be of form \(1+4t\)


convince yourself that \(2^{1+4t}\) leaves a remainder \(2\) when divided by \(5\) for all natirual numbers \(t\)

Answer 14

You got it !

Answer 15

\(2^{1},2^{5},2^{9}.....2^{99}\)

These all numbers leave a remainder of \(2\), yes ?

Answer 16

yes of cos

Answer 17

we want \(2^m + 2^n\) to be divisible by \(5\)

since \(m=1,5,9,\ldots\) makes \(2^m\) leave a remainder \(2\),
we may look at \(n\) values that make \(2^n\) leave a remainder \(3\).

That way \(2^m+2^n\) leaves a remainder \(2+3\), which is divisible by \(5\), makes sense ?

Answer 18

yes

Answer 19

2^3 = 8 = 3

therefore \(2^{3+4s}\) leaves a remainder \(3\) for all natural numbers \(s\), yes ?

Answer 20

yes

Answer 21

because we know that \(2^{4s}\) leaves a remainder \(1\) when divided by \(5\)

Answer 22

ok

Answer 23

\(2^{1},2^{5},2^{9}.....2^{99}\)


\(m = 1 + 4t\)

\(1\le m\le 100\)

how many \(m\) values make \(2^m\) leave a remainder of \(2\) ?

Answer 24

i think the last term is not \(2^{99}\) in the series \(2^{1},2^{5},2^{9}.....2^{99}\)

it should be \(2^{97}\)

Answer 25

25 terms

Answer 26

yeah there are 25 pairs of (m, n) such that 2^m leaves a remainder 2,
and 2^n leaves a remainder 3


they add up to 5

Answer 27

similarly for 3 + 2, we get 25 more pairs

Answer 28

similarly for 1 + 4 and 4 + 1 we get 50 more pairs

Answer 29

Overall, is the answer 100 ?

Answer 30

I don't know the answer , I suppose ur logic with answer is all correct.