To start this problem, we should get a general idea of the probabilities of each product of the outcomes.
Let's make a table of the products.
1 | 3 | 8 | 8 | |
---|---|---|---|---|
1 | 1 | 3 | 8 | 8 |
3 | 3 | 9 | 24 | 24 |
8 | 8 | 24 | 64 | 64 |
8 | 8 | 24 | 64 | 64 |
The probability of each product then becomes:
\(P(1) = \frac{1}{16}, P(3) = \frac{2}{16}, P(8) = \frac{4}{16}, P(9) = \frac{1}{16}, P(24) = \frac{4}{16}, P(64) = \frac{4}{16}\)
For the decision of side dishes to be fair, we would want each dish to have a \(\frac{1}{4} \) or \(\frac{4}{16}\) probability of being prepared.
Since three products (8, 24, and 64) already have probabilities of \(\frac{4}{16}\), we would want those probabilities being assigned to only one dish each, and the other three products (1, 3, and 9) add up to \(\frac{4}{16}\) for the last dish.
To rephrase, product of 8 goes to one dish, product of 24 goes to another, product of 64 goes to another, and products 1, 3, and 9 go to the last dish.
Answer B satisfies this.
How many integers a from 1 to 1000 are there such that \(a^{100}-1\) is divisible by 1000?
\(\frac{a^{100}-1}{1000}=N\qquad where \qquad N\in (intergers \ge0)\)
My first thought was ... what possibilities are there for the last digit.
The last digit of a^100 must be 1 for any possible 'a' to be considered.
For convenience I will call the last digit k
Only last digit displayed
k^1 | k^2 | k^3 | k^4 | k^5 all have a pattern now | k^100 from pattern |
0 | 0 | 0 | 0 | 0 | |
1 | 1 | 1 | 1 | 1 | |
2 | 4 | 8 | 6 | 2 | 6 |
3 | 9 | 7 | 1 | 3 | 1 |
4 | 6 | 4 | 6 | 4 | 6 |
5 | 5 | 5 | 5 | 5 | 5 |
6 | 6 | 6 | 6 | 6 | 6 |
7 | 9 | 3 | 1 | 7 | 1 |
8 | 4 | 2 | 6 | 8 | 6 |
9 | 1 | 9 | 1 | 9 | 1 |
Since k^100 must end in 1, k must end in 1,3, 7 or 9
But what about the other digits ???
\(\frac{a^{100}-1}{1000}=N\qquad where \qquad N\in (intergers \ge0)\\ \)
I am going to let a=(10x+k) I know that k must be 1,3,7, or 9 but I am not sure about restrictions on x
so using the binomial expansion I have
\((10x+k)^{100}=\displaystyle\sum_{r=0}^{100}\;(10k)^r(k)^{100-r}\)
Since I want to know if this -1 is divisable by 1000 I only care about terms that are not multiples of 1000, that is not many terms.
\((10x+k)^{100}\\=\displaystyle\sum_{r=0}^{100}\;(10k)^r(k)^{100-r}\\ =\binom{100}{0}(10x)^0(k)^{100}\;+\;\binom{100}{1}(10x)^1(k)^{99}\;+\;\binom{100}{2}(10x)^2(k)^{98}\;+\;\dots\\ =k^{100}+100(10x)(k)^{99}\;+\;4950(100x^2)(k)^{98}\;+\;\dots\\ =k^{100}+1000x*(k)^{99}\;+\;495000x^2(k)^{98}\;+\;\dots\\ =k^{100}+\;\dots\\\)
So only k, which is the last digit matters. so long as k^100 ends in 1 then \((10x+k)^{100} -1 \) will be divisable by 1000
So a can be any number at all that ends in 1,3,7,9
There are 1000 numbers between 1 and 1000
10X<1000 so
X<100
There are 100 numbers that end in 1
100 that end in 3
100 that end in 7 and
100 that end in 9
So there are 400 numbers altogether that meet the requirements. Just as our guest already determined.
Find the order of 15 mod 257.
1. 257 is a prime and
2. \(257-1 = 2^8\) is a power of a prime, so
3. the order of 257 must also be a power of two. The order is \(2^k\) for some \(k= 0,1,2,\ldots, 8\)
Test:
\(\begin{array}{|r|r|l|} \hline k & 2^k \\ \hline 0 & 1 & 15^1 \equiv 15 \pmod{257} \\ 1 & 2 & 15^2 \equiv 225 \pmod{257} \\ 2 & 4 & 15^4 \equiv 253 \pmod{257} \\ 3 & 8 & 15^8 \equiv 16 \pmod{257} \\ 4 & 16 & 15^{16} \equiv 256 \pmod{257} \\ 5 & \color{red}32& 15^{32} \equiv {\color{red}1} \pmod{257} \\ \hline \end{array}\)
The order of 15 mod 257 is 32
source see: https://math.stackexchange.com/questions/2271153/calculate-the-order-of-3-modulo-257