Algorithms Q&A - Recent questions and answers in Math Basics

Answered: What is the last digit of number: 2^100

Mon, 14 Oct 2024 02:40:41 +0000

To begin with,

We can calculate all the values of 2^n where 1 <= n <= 10. We get the values, 2, 4, 8, 16, 32, 64, 128, 256, 512, and 1024, respectively.

We can notice a pattern in the last digit for all the numbers, the pattern follows 2, 4, 8, 6 for every 4 "n" values.

Now, we can predict the last digit of 2^100 based on the noticeable pattern from above. Since the last digits of 2^n follow the repeating cycle of 2, 4, 8, 6 every four terms, we can reduce the exponent modulo 4 to get where 100 falls in the cycle.

Here is how we can do that, 100 mod 4 = 0.

Since the remainder is 0, 2^100 corresponds to the last digit of 2^4, 6. Thus, the last digit of 2^100 is 6.

Answered: Probability of rolling at least a 7 with two dice

Mon, 18 Dec 2023 07:48:10 +0000

1. Each die has 6 faces, so there are 6 * 6 = 36 possible outcomes when rolling two dice.

2. If the sum of two dice is at least a 7, the sum can be 7, 8, 9, 10, 11, or 12.

Sum of the dice = 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) - 6 outcomes
Sum of the dice = 8: (2,6), (3,5), (4,4), (5,3), (6,2) - 5 outcomes
Sum of the dice = 9: (3,6), (4,5), (5,4), (6,3) - 4 outcomes
Sum of the dice = 10: (4,6), (5,5), (6,4) - 3 outcomes
Sum of the dice = 11: (5,6), (6,5) - 2 outcomes
Sum of the dice = 12: (6,6) - 1 outcome

3. Add all possible outcomes: 6 + 5 + 4 + 3 + 2 + 1 = 21

4. The probability of rolling at least a 7 when given two dice:

21 / 36 = 7 / 12 ≈ 0.5833

Answered: Prove that there are infinite primes

Sat, 16 Dec 2023 18:14:34 +0000

Start with a counter-assumption: Let's imagine the scenario where the set of prime numbers is limited. Denote these finite primes as P1, P2, P3,...Pn

Create a novel number: Now, form a new number, N= (P1*P2*P3*...Pn )+ 1 Notice that N is greater than any prime in our assumed finite set.

Examine N's prime factors: Every number greater than 1 is either a prime itself or can be decomposed into a unique set of prime factors (Fundamental Theorem of Arithmetic).

Unravel the contradiction:

If N itself is a prime, it's not in our original list, which contradicts our assumption of having listed all primes.

If N is composite, its prime factors cannot be any of P1, P2 ,...Pn because dividing

N by any of these primes leaves a remainder of 1.

Thus, N must be divisible by some prime not in our original list, contradicting our assumption again.

Inference: The initial assumption of a finite number of primes leads us to a contradiction.

Therefore, the only logical conclusion is that the number of primes is infinite.

Answered: Give a Proof by Induction that the sum of squares of first n natural numbers is n. (n+1). (2n + 1) / 6.

Wed, 13 Dec 2023 20:36:54 +0000

Let, S(n) represents the sum of squares of first n natural numbers.

Hypothesis : S(n) = n (n+1) (2n + 1) / 6.

Using mathematical induction:

Base Case

S(1) = 1^2 = 1,   Also, 1 * (1+1) * 3 / 6. Therefore, Base Case is satisfied.

Induction Hypothesis

Let us suppose the hypothesis is true for all values of n < m.

Inductive Step [Here, we should P(n | n < m) => P(m).]

Let us evaluate S(m). By definition, S(m) = S(m -1) + m^2.

By induction hypothesis, S(m-1) = (m-1) * (m - 1 + 1) * (2(m-1) +1) / 6.   This can be used by us, since induction hypothesis allows us to assume hypothesis is true for all values of n < m, and therefore, specifically that it is true for n = m-1.

Therefore, we have

S(m) = (m - 1) * (m) * (2m - 1) / 6 + m^2.   By rearranging terms, we get:

S(m) = m/6 {(m - 1) (2m - 1) + 6m}.

That is,

S(m) = m/6 {2m^2 - 2m - m + 1 + 6m}

S(m) = m/6 {2m^2 + 3m + 1}

S(m) = m/6 (m+1) (2m + 1)

Therefore, the hypothesis is true for m.

Since Induction base is satisfied, and induction step is satisfied, therefore, by PMI, hypothesis is true for all values of n >= 1.

Therefore, we have:

S(n) = n (n+1) (2n + 1) / 6.

Answered: Sum of the Arithmetico-Geometric Progression (AGP)

Thu, 02 May 2019 05:29:45 +0000

k/2*3^(k+1)+1/2