Last two digits

[huneynumb]

What are the last two digits of

a) 3^1994

b) 7^1994

c) 3^1994 + 7^1994

d) 7^1994 - 3^1994

[RyanJ]

Only advice I can actually offer is too look for a pattern.

 

Let me try an example, nets take [math]2^n[/math]

 

The first few results are:

 

2

4

8

16

32

64

128

256

512

1024

2048

 

But we only want to know the last digit. Then we only need to consider the last digit of the previous term because "larger" digits of the previous term only contribute to "larger" digits of the current term:

 

2

2*2 = 4

4*2 = 8

8*2 = (1)6

6*2 = (1)2

 

As you can see, we're back to 2 again. So the sequence of last digits is 2, 4, 8, 6, 2, 4, 8, 6, 2, 4, 8, 6, etc.

 

Therefore:

 

Last digit of: [math]2^{4n} = 6[/math]

Last digit of: [math]2^{4n+1} = 2[/math]

Last digit of: [math]2^{4n+2} = 4[/math]

Last digit of: [math]2^{4n+3} = 8[/math]

 

Hope that helps... its the best I can do

 

Cheers,

 

Ryan Jones

[matt grime]

One way to solve this is to know the order of 3 mod 100, or a multiple of it such as the number of numbers between 1 and 100 and prime to 100 i.e. phi(100) where phi is Euler's totient function.

[psi20]

That was an interesting problem. I tried RyanJ's way and some interesting things came up.

 

If you try that for 3, the last digit is 1 for exponents in the form 4n (where n is an integer greater than or equal to 0), 3 for (4n +1), 9 for (4n + 2), and 7 for (4n + 3). For 7, they are 1, 7, 9, and 3.

 

It seems like for positive integer powers, the last digit always repeats itself after 4 or less terms upon increasing the power by 1. The reason why they always repeat is because all you're doing is multiplying over and over again. There are only 10 digits (in our counting system). So once you multiply one number by 3 for example, some where along the line you will have to multiply by that number again, getting the same result. So the last digits repeat. Why they repeat after 4 or less terms I haven't looked into.

 

Problem Bb is the easiest so we'll start with that. The second to the last digit also repeats itself. For 7, it repeats like this

01 07 49 43

 

7^0=01 7^1=07 7^2=49 7^3=43

7^4=01 7^5=07 7^6=49 7^7=43

and so on.

For the same reason as stated above, they repeat in this manner. So the last two digits of 7^1994 is 49, because 1994 is in the form (4n + 2).

 

For problem A, doing the same thing is harder, but it works. I start with 3^0 and go across 3^1, 3^2, 3^3, then 3^4 is lined up vertically with 3^0 and so on.

 

01 03 09 27

81 43 29 87

61 83 49 47

41 23 69 07

21 63 89 67

and after this it repeats. So instead of (4n + C) , we're looking for (20n + C)

For 3^1994, the last two digits will be 69.

 

From this, problem C will be 18 after 69 + 49 = 118

Problem D will be 80. It becomes A49 - B69 = C0 where A and B are everything in front of the numbers, and C is the second to the last digit. Seeing that 7^1994 is bigger than 3^1994, you can "borrow" from A and you get 80.

[mike2313]

Ive got the same question, but my teacher said to answer with the use of modulo. I dont understand:confused: Please help

Thanks

[RyanJ]

Ive got the same question' date=' but my teacher said to answer with the use of modulo. I dont understand:confused: Please help

Thanks[/quote']

 

Modulo is when one number is divided by another and the remainder is returned.

 

E.g. [math]3 mod 2 = 1[/math] because 3 is divisible by 2 exactly once leaving 1 as the remainder.

 

-- Ryan Jones