Geometry and Discrete Mathematics (Proofs)
posted by Raj .
Observe that the last two digits of 7^2 are 49, the last two digits of 7^3 are 43, the last two digits of 7^4 are 01, and the last two digits of 7^5 are 07. Prove that the last two digits of 7^201 are 07.
7^2 = ...49
7^3 = ...43
7^4 = ...01
7^5 = ...07
7^201 = ...07
Break 201 into a product of primes. Hint: 201 is divisible by 3.
Yes, but what would I do with 67 and 3?
You should use modular arithmetic. This works as follows. On the set of integers you define a socalled equivalence relation E which in this case is:
n1 E n2 if n1  n2 = 100*k
where k is an integer. So, two integers are equivalent if they differ by a miltiple of 100.
We then use the notation [n] for the set of all integers that are related to n via the equivalence relation.
So, e.g. [3] = {....,97,3,103,...}
There are then 100 such sets:
[0], [1], [2], ..., [99].
Note that [100] and [0] are the same sets.
On the set M = {[0],[1],...[99]} (the socalled quotient set) you can define addition and multiplication:
[n1] + [n2] = [n1 + n2]
[n1]*[n2] = [n1*n2]
Modular artithmetic is just doing calculations on the set M. You can write that [102] = [2] instead of saying that 102 is related to 2. Another notation is:
102 = 2 Mod(100)
Anyway, modular artithmetic (modulo 100 in tis case) allows you to calculate what you want. If we delete the square brackets (or the Mod symbol):
7^4 = 1
Raise both sides to the 50th power
7^200 = 1
and then multiply by 7:
7^201 = 7
In the general case where you have to compute:
X^Y Mod N
and there are no obvious shortcuts, it is simplest to write down the binary representaton of Y:
Y = a_{r} 2^r + a_{r1}2^(r1) + ..
where the a_{r} are either 1 or zero.
This means that:
Y = 2^(r1) + 2^(r2) + ...
for some r1 > r2 > ...
You can then write:
X^Y = X^[2^(r1)]*X^[2^(r2)]
So, all you have to do is to raise X to the appropriate powers of two and then multiply these things together. Raising X to such powers can be done by repeatedly squaring X.
E.g. what is
2^(1133) Mod 41?
Expand 1133 in binary system. You can do that easily by subtracting 1 if you can't divide by 2 and then dividing by 2 repeatedly until the number becomes odd, subtracting 1 again and repeating the process:
1133 = 1 + 1132
1132 = 4*283
283 = 1 + 2*141
141 = 1 + 4*35
35 = 1 + 2*17
17 = 1 + 16
1133 = 1 + 4*(1 + 2*(1+4*(1+2*(1+16))))=
1 + 4 + 8 + 32 + 64 + 1024
Note that the powers of two you get in here are just the "cumulative" product of the factors of two you get when you work your way through reducing 1133 by repeatedly subtracting 1 and dividing by two.
You then calculate the powers:
2^1 = 2
2^4 = 16
2^8 = 16^2 = 10
2^16 = 10^2= 100 = 18
2^32 = 18^2 = 37
2^64 = 37^2 = (4)^2 = 16
2^128 = 16^2 = 2^8 (already alculated)
2^(1024) = 2^64 = 16
So we obtain:
2^1133 = 2*16*10*37*16*16 = 33
Respond to this Question
Similar Questions

math
1. a door code is made up of six unique digits the last digit is 8. 2.oddd and even digits alternate(o is even) 3.the difference of the adjacent is always bigger than 1. 4.the first 2 digits (reads as one number)as well as he two middle … 
Math
A locker combination has three nonzero digits, and digits cannot be repeated. The first two digits are 1 and 2. What is the probability that the third digit is 3? 
Math problems
i am a fourdigit number with the digits in order. my first two digits are the first two odd numbers. my last two digits are the last two singledigit even numbers PLEASE HELP 
Math problems Help Please!!!!!!
i am a fourdigit number with the digits in order. my first two digits are the first two odd numbers. my last two digits are the last two singledigit even numbers 
math
A locker combination has three nonzero digits, and digits cannot be repeated. The first two digits are 1 and 2. What is the probability that the third digit is 3? 
math
Solve the mathematical puzzle. Determine the digits of F from these clues. The digits of F are all the same. The sum of all the digits of F is 12. The answer when any two of the digits are multiplied together is also 9. F is a fourdigit … 
MATH
What is my number if the hundreds digit is less than the tens digit and less than the ones digit and the hundreds digit is even and the tends digit divided by the hundreds digit is 5 and the product of two of the digits is 6 and it … 
Mathematics
A number of two digits is increased by 9. When the digits are reversed, the sum of the digits is 5. Find the number. 
math (5th grade)
There are two 1's, two 2's, and two 3's. There is one digit between the two 1's, there are two digits between two 2's, and there are three digits between the two 3's. Holding number cards, can you duplicate this sequence of numbers? 
math
A locker combination has three nonzero digits, and digits cannot be repeated. The first two digits are 1 and 2. What is the probability that the third digit is 3?