Mathematics is a very crucial subject in our life. It gives a power to calculate the daily things. Rajesh is the one of the brightest student in this subject so he has an integer N. His task is to break N into some number of (integer) powers of 2. TO achieve this, he can perform the following operation several times (possibly, zero): Choose an integer X which he already has, and break X into 2 integer parts (Y and Z) such that X = Y + Z. Find the minimum number of operations required by Rajesh to accomplish his task using Python Language. Input 1 3 Output

Mathematics is a very crucial subject in our life. It gives a power to calculate the daily things. Rajesh is the one of the brightest student in this subject so he has an integer N. His task is to break N into some number of (integer) powers of 2. TO achieve this, he can perform the following operation several times (possibly, zero): Choose an integer X which he already has, and break X into 2 integer parts (Y and Z) such that X = Y + Z. Find the minimum number of operations required by Rajesh to accomplish his task using Python Language. Input 1 3 Output

C++ Programming: From Problem Analysis to Program Design

8th Edition

ISBN:9781337102087

Author:D. S. Malik

Publisher:D. S. Malik

Chapter8: Arrays And Strings

Section: Chapter Questions

Problem 21PE

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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…
Given an integer N and a base X, the task is to find the minimum number of operations required to represent N as a sum of the distinct powers of X. In each operation, you can either increment or decrement N. You are allowed to make the given operation any number of times Examples: Input: N = 7, X = 3 Output: 3.
A number in base 2 (binary) is a number such that each of its digits is 0 or 1. To convert frombinary to decimal (base 10), the digits starting from the right are multiplied by powers of 2(starting at 0) and added. For example, the value in decimal of 10011 is calculated as follows:
Maclaurin series are a type of Mathematic series expansion in which all terms are nonnegative real powers of the variable. The Maclaurin series expansion for sin(x) is given by the following formula that is valid for all real values of x such that x is in radians (Note that: radians(x) = x X 1/180): sin(x) = x - 3! 5! Implement a Java program to compute the value of Maclaurin series expansion for sin(x) where x is a nonnegative real value according to the following: a. Write a java method named Factorial that takes as an argument an integer value n and returns the mathematical factorial n! of n as a long value such that: n! = n x (n – 1) x (n – 2) ... × 2 x 1 b. In the main method: i. Ask the user to enter the value of the angle x to be calculate in the Maclaurin series expansion for sin(x) as given above. x Should be entered in degrees, i.e., 0° < x< 360° and then converted into radians using the formula: (radians(x) = x × 1/180). ii. Ask the user to enter the number of terms to be…
Use following series to do the given task; 135 79 11 13 15 1719 21 23 25 27 29 31 ... Given the number N of odd numbers in a certain line, your task is to determine the sum of the last three numbers of that line. For example N = 5 means, the third line, the last three numbers are 13, 15 and 17. The summation of these three numbers is 45. InputThe first line tells the number of test cases. From the next line, the input is a sequence of lines, one odd number N (1 < N <1000000000) per line. OutputFor each input line write the sum of the last three odd numbers written in that line of series with N numbers.
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.
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)?
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
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)?
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.
cancellation error occurs when you aremanipulating a very large number with a very small number. The large numbermay cancel out the smaller number. For example, the result of 100000000.0 +0.000000001 is equal to 100000000.0. To avoid cancellation errors and obtainmore accurate results, carefully select the order of computation. For example, incomputing the following summation, you will obtain more accurate results bycomputing from right to left rather than from left to right: Write a program that compares the results of the summation of the preceding series,computing from left to right and from right to left with n = 50000.