Largest number by which given 3 numbers should be divided such that they leaves same remainder
Given three numbers, our task is to find the largest number by which when given 3 numbers are divided leads to same remainder. It may be assumed that all given numbers are given in increasing order.
Examples:
Input : a = 62, b = 132, c = 237
Output : 35
35 leads to same remainder 27 when divides
62, 132 and 237.
Input : a = 74, b = 272, c = 584
Output : 6
Brute Force Approach:
The brute force approach to solve this problem is to start from the largest number among the three (i.e., c) and decrement it until we find a number that leaves the same remainder when divided by a and b.
We can find the remainder of a, b, and c when divided by any number n using the modulo operator (%). Then, we can check if the remainders of a, b, and c when divided by a candidate number n are equal. If they are equal, then n is a valid solution.
Below is the implementation of above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int sameRemainder( int a, int b, int c)
{
for ( int n = c; n >= 1; n--) {
int remainderA = a % n;
int remainderB = b % n;
int remainderC = c % n;
if (remainderA == remainderB
&& remainderB == remainderC) {
return n;
}
}
return -1;
}
int main()
{
int a = 62, b = 132, c = 237;
cout << sameRemainder(a, b, c) << endl;
return 0;
}
|
Java
import java.util.*;
public class Main {
static int sameRemainder( int a, int b, int c)
{
for ( int n = c; n >= 1 ; n--) {
int remainderA = a % n;
int remainderB = b % n;
int remainderC = c % n;
if (remainderA == remainderB
&& remainderB == remainderC) {
return n;
}
}
return - 1 ;
}
public static void main(String[] args)
{
int a = 62 , b = 132 , c = 237 ;
System.out.println(sameRemainder(a, b, c));
}
}
|
Python3
def same_remainder(a, b, c):
for n in range (c, 0 , - 1 ):
remainder_a = a % n
remainder_b = b % n
remainder_c = c % n
if remainder_a = = remainder_b = = remainder_c:
return n
return - 1
a = 62
b = 132
c = 237
print (same_remainder(a, b, c))
|
C#
using System;
class Program
{
static int SameRemainder( int a, int b, int c)
{
for ( int n = c; n >= 1; n--) {
int remainderA = a % n;
int remainderB = b % n;
int remainderC = c % n;
if (remainderA == remainderB
&& remainderB == remainderC) {
return n;
}
}
return -1;
}
static void Main( string [] args)
{
int a = 62, b = 132, c = 237;
Console.WriteLine(SameRemainder(a, b, c));
}
}
|
Javascript
function sameRemainder(a, b, c) {
for (let n = c; n >= 1; n--) {
let remainderA = a % n;
let remainderB = b % n;
let remainderC = c % n;
if (remainderA == remainderB && remainderB == remainderC) {
return n;
}
}
return -1;
}
let a = 62, b = 132, c = 237;
console.log(sameRemainder(a, b, c));
|
Time Complexity: O(c)
Auxiliary Space: O(1)
Approach:
The idea is based on the fact that if a number leaves same remainder with a, b and c, then it would divide their differences. Let us understand assuming that x is our result. Let a = x*d1 + r where r is the remainder when a is divided by x. Similarly we can write b = x*d2 + r and b = x*d3 + r. So the logic is here we first find differences of all three pairs and after that, we find greatest common divisor of differences to maximize result.
Below is the implementation of above idea.
C++
#include <bits/stdc++.h>
using namespace std;
int gcd( int a, int b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
int sameRemainder( int a, int b, int c)
{
int a1 = (b - a), b1 = (c - b), c1 = (c - a);
return gcd(a1, gcd(b1, c1));
}
int main()
{
int a = 62, b = 132, c = 237;
cout << sameRemainder(a, b, c) << endl;
return 0;
}
|
Java
class GFG {
static int gcd( int a, int b)
{
if (a == 0 )
return b;
return gcd(b % a, a);
}
static int sameRemainder( int a, int b, int c)
{
int a1 = (b - a), b1 = (c - b), c1 = (c - a);
return gcd(a1, gcd(b1, c1));
}
public static void main(String[] args)
{
int a = 62 , b = 132 , c = 237 ;
System.out.println(sameRemainder(a, b, c));
}
}
|
Python3
def gcd(a, b):
if (a = = 0 ):
return b
return gcd(b % a, a)
def sameRemainder(a, b, c):
a1 = (b - a)
b1 = (c - b)
c1 = (c - a)
return gcd(a1, gcd(b1, c1))
a = 62
b = 132
c = 237
print (sameRemainder(a, b, c))
|
C#
using System;
class GFG {
static int gcd( int a, int b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
static int sameRemainder( int a, int b, int c)
{
int a1 = (b - a), b1 = (c - b), c1 = (c - a);
return gcd(a1, gcd(b1, c1));
}
public static void Main()
{
int a = 62, b = 132, c = 237;
Console.WriteLine(sameRemainder(a, b, c));
}
}
|
PHP
<?php
function gcd( $a , $b )
{
if ( $a == 0)
return $b ;
return gcd( $b % $a , $a );
}
function sameRemainder( $a , $b , $c )
{
$a1 = ( $b - $a );
$b1 = ( $c - $b );
$c1 = ( $c - $a );
return gcd( $a1 , gcd( $b1 , $c1 ));
}
$a = 62;
$b = 132;
$c = 237;
echo sameRemainder( $a , $b , $c ) ;
?>
|
Javascript
<script>
function gcd(a, b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
function sameRemainder(a, b, c)
{
var a1 = (b - a), b1 = (c - b), c1 = (c - a);
return gcd(a1, gcd(b1, c1));
}
var a = 62, b = 132, c = 237;
document.write(sameRemainder(a, b, c));
</script>
|
Time Complexity: O( log N )
Auxiliary Space: O(1)
Last Updated :
10 Apr, 2023
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