m + n = 2k + 2k = 4k therefore adding even numbers always gives an even number. The above is wrong because.... it doesn't consider the case when adding to odd numbers m and n are different so you cannot use 2k for both it doesn't consider what m and n are for particular values in the integers it doesn't consider the case when m is even and n is odd
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HALP
m + n = 2k + 2k = 4k therefore adding even numbers always gives an even number.
The above is wrong because....
it doesn't consider the case when adding to odd numbers |
||
m and n are different so you cannot use 2k for both |
||
it doesn't consider what m and n are for particular values in the integers |
||
it doesn't consider the case when m is even and n is odd |
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- Correct answer will be upvoted else downvoted. number is called 2050-number if it is 2050, 20500, ..., (2050⋅10k for integer k≥0). Given a number n, you are asked to represent n as the sum of some (not necessarily distinct) 2050-numbers. Compute the minimum number of 2050-numbers required for that. Input The first line contains a single integer T (1≤T≤1000) denoting the number of test cases. The only line of each test case contains a single integer n (1≤n≤1018) denoting the number to be represented. Output For each test case, output the minimum number of 2050-numbers in one line. If n cannot be represented as the sum of 2050-numbers, output −1 instead.The Fibonacci sequence is listed below: The first and second numbers both start at 1. After that, each number in the series is the sum of the two preceding numbers. Here is an example: 1, 1, 2, 3, 5, 8, 13, 21, ... If F(n) is the nth value in the sequence, then this definition can be expressed as F(1) = 1 F(2) = 1 F(3) = 2 F(4) = 3 F(5) = 5 F(6) = 8 F(7) = 13 F(8) = 21 F(n) = F(n - 1) + F(n - 2) for n > 2 Example: Given n with a value of 4F(4) = F(4-1) + F(4-2)F(4) = F(3) + F(2)F(4) = 2 + 1F(4) = 3 The value of F at position n is defined using the value of F at two smaller positions. Using the definition of the Fibonacci sequence, determine the value of F(10) by using the formula and the sequence. Show the terms in the Fibonacci sequence and show your work for the formula.We usually write numbers in decimal form (or base 10), meaning numbers are composed using 10 different “digits" {0,1, ...,9}. Sometimes though it is useful to write numbers hexadecimal or base 16. Now there are 16 distinct digits that can be used to form numbers: {0, 1,...,9, A, B, C, D, E, F}. So for example, a 3 digit hexadecimal number might be 2B8. Assume that digits and letter can be repeated. a. How many 4-digit hexadecimals are there in which the first digit is E or F? b. How many 5-digit hexadecimals start with a letter (A-F) and end with a numeral (0-9)?
- We usually write numbers in decimal form (or base 10), meaning numbers are composed using 10 different “digits" {0, 1,...,9}. Sometimes though it is useful to write numbers hexadecimal or base 16. Now there are 16 distinct digits that can be used to form numbers: {0, 1, ...,9, A, B, C, D, E, F}. So for example, a 3 digit hexadecimal number might be 2B8. a. How many 4-digit hexadecimals are there in which the first digit is E or F? 8192 b. How many 5-digit hexadecimals start with a letter (A-F) and end with a numeral (0-9)? c. How many 3-digit hexadecimals start with a letter (A-F) or end with a numeral (0-9) (or both)?We usually write numbers in decimal form (or base 10), meaning numbers are composed using 10 different "digits" {0,1,...9}. Sometimes though it is useful to write numbers hexadecimal or base 16. Now there are 16 distinct digits that can be used to form numbers: {0,1,....,9,A,B,C,D,E,F}. So for example, a 3 digit hexadecimal number might be 2B8. a. How many 4-digit hexadecimals are there in which the first digit is E or F? b. How many 5-digit hexadecimals start with a letter (A-F) and end with a numeral (0-9)? c. How many 2-digit hexadecimals start with a letter (A-F) or end with a numeral (0-9) (or both)?Print the multiplication of 5 without using * operator. You have to write a program for that.
- Rahul is a maths genius so he came up with a game and as raj is Rahul's best friend so Rahul decided to play the game with raj. Rahul gives raj two numbers LL and RR and asks raj to find the count of numbers in the range from LL to RR (LL and RR inclusive) which are a digit palindromic. A number is a digit palindromic if its first digit is the same as its last digit. As raj is not very good at maths so your task is to help Raj find out how many numbers are a digit palindromic in the range LL to RR. For example if LL = 88 and RR = 2525 .The following numbers are a digit palindromic in the range of LL to RR: 8, 9, 11, and 22. If LL = 12511251 and RR = 12661266. The digit palindromic numbers are 1251 and 1261. Input format The first line contains an integer denoting the number of test cases. Each test case is described by a single line that contains two integers LL and RR. Output format For each test case output, an integer denoting how many a digit palindromic numbers are there in the…Start with a pile of n stones and successively split a pile into two smaller piles until each pile has only one Each time a split happens, multiply the number of stones in each of the two smaller piles. (For example, if a pile has 15 stones and you split it into a pile of 7 and another pile of 8 stones, multiply 7 and 8.) The goal of this problem is to show that no matter how the pile of n stones are split, the sum of the products computed at each split is equal to n(n - 1)/2. Using strong mathematical induction, prove that no matter how the pile of n stones are split, the sum of the products computed at each split is equal to n(n - 1)/2.The greatest common divisor of two positive integers, A and B, is the largest number that can be evenly divided into both of them. Euclid's algorithm can be used to find the greatest common divisor (GCD) of two positive integers. You can use this algorithm in the following manner: 1. Compute the remainder of dividing the larger number by the smaller number. 2. Replace the larger number with the smaller number and the smaller number with the remainder. 3. Repeat this process until the smaller number is zero. The larger number at this point is the GCD of A and B. Write a program that lets the user enter two integers and then prints each step in the process of using the Euclidean algorithm to find their GCD. An example of the program input and output is shown below: Enter the smaller number: 5 Enter the larger number: 15 The greatest common divisor is 5
- We usually write numbers in decimal form (or base 10), meaning numbers are composed using 10 different “digits” {0,1,…,9}.{0,1,…,9}. Sometimes though it is useful to write numbers hexadecimal or base 16. Now there are 16 distinct digits that can be used to form numbers: {0,1,…,9,A,B,C,D,E,F}.{0,1,…,9,A,B,C,D,E,F}.So for example, a 3 digit hexadecimal number might be 2B8. How many 2-digit hexadecimals start with a letter (A-F) or end with a numeral (0-9) (or both)?Let A = {a, b, c} and B = {u, v}. Write a. A × B b. B × APart 2: Binary Arithmetic One of the most common operations we perform on binary numbers (and all numbers) is addition. It can be cumbersome to convert your binary numbers to decimal just to add them and convert them back, so instead we will be learning how to add binary numbers directly. Binary addition works the same way as decimal addition, with the added restriction that each digit can only go up to 1. Let's consider the possibilities for adding the values of any 2 single digits together: 0 + 0 0 + 1 1 + 0 1 + 1 0 1 1 10 (remember that 10 in binary represents the number 2) In that last case, the result is larger than a single digit. When adding larger binary numbers, that means we have to carry the 1 over to the next column. This presents us with another new case: what happens if we have 1 + 1 + carried 1? In that case, the result is 11, which means that column's result is 1, and we carry 1 to the next column. Below is an example of adding two binary numbers that shows all…