Objective In this challenge, we're going to use loops to help us do some simple math. Task Given an integer, , print its first multiples. Each multiple (where ) should be printed on a new line in the form: N x i = result . Input Format A single integer, . Constraints Output Format Print lines of output; each line (where ) contains the of in the form: N x i = result . Sample Input 2 Sample Output 2 x 1 = 2 2 x 2 = 4 2 x 3 = 6 2 x 4 = 8 2 x 5 = 10 2 x 6 = 12 2 x 7 = 14 2 x 8 = 16 2 x 9 = 18 2 x 10 = 20 Explanation: Here, we just need to use for loops to achieve the result Solution : import java.io.* ; import java.math.* ; import java.security.* ; import java.text.* ; import java.util.* ; import java.util.concurrent.* ; import java.util.regex.* ; public class Solution { public static void main ( String [] args ) t...
Problem:
Given two integers N,d,find the smallest number that number that is a multiple of d that could be formed by permuting the digits of N .You must use all the digits of N.and if the smallest multiple of d has leading zeros,they can be dropped .If no such number exists,output -1.
Input:
A line containing two space separated integers.representing N and d.
Output:
A single line giving the permutation of N that is the smallest multiple of d,without any leading zeros,if any.If not such permutation exists, the output should be -1.
Constraints:
1<=N<=10^12
1<=d<=1000000
Example 1:
Input:
210 2
Output:
12
Example 2:
Input:
531 2
Output:
1
Understanding of Question:
We need to find out the minimum number that are divisible by given divisor from the given number.
Pseudocode:
- Get the input of number and the divisor.
- Extract the digits and store it in a array.
- Sort the digits
- least multiple=-1
- Find the permutations using import itertools
- Convert the array into integer.
- Traverse the array and check if it is divisible by d.
- Print least multiple.
Program:
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