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Check if a large number is divisibility by 15 in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 217 Views

Here we will see how to check a number is divisible by 15 or not. In this case the number is very large number. So we put the number as string.To check whether a number is divisible by 15, if the number is divisible by 5, and divisible by 3. So to check divisibility by 5, we have to see the last number is 0 or 5. To check divisibility by 3, we will see the sum of digits are divisible by 3 or not.Example#include using namespace std; bool isDiv15(string num){    int n = num.length();    if(num[n - ...

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Check if a large number is divisible by 2, 3 and 5 or not in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 329 Views

Here we will see how to check a number is divisible by 2, 3 and 5 or not. In this case the number is very large number. So we put the number as string.A number will be divisible by 2, 3 and 5 if that number is divisible by LCM of 2, 3 and 5. So the LCM of 2, 3, 5 is 30. We have to check the number is divisible by 30 or not. A number is divisible by 30 when it is divisible by 10 (last digit is 0) and divisible by 3 (sum of all digits ...

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Check if a large number is divisible by 20 in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 264 Views

Here we will see how to check a number is divisible by 20 or not. In this case the number is very large number. So we put the number as string.A number will be divisible by 20, when that is divisible by 10, and after dividing 10, the remaining number is divisible by 2. So the case is simple. If the last digit is 0, then it is divisible by 10, and when it is divisible by 10, then the second last element is divisible by 2, the number is divisible by 20.Example#include using namespace std; bool isDiv20(string num){ ...

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Check if a large number is divisible by 25 or not in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 283 Views

Here we will see how to check a number is divisible by 25 or not. In this case the number is very large number. So we put the number as string.A number will be divisible by 25, when the last two digits are 00, or they are divisible by 25.Example#include using namespace std; bool isDiv25(string num){    int n = num.length();    int last_two_digit_val = (num[n-2] - '0') * 10 + ((num[n-1] - '0'));    if(last_two_digit_val % 25 == 0)       return true;       return false; } int main() {    string num = "451851549333150";    if(isDiv25(num)){       cout

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Check if a large number is divisible by 3 or not in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 535 Views

Here we will see how to check a number is divisible by 3 or not. In this case the number is very large number. So we put the number as string.A number will be divisible by 3, if the sum of digits is divisible by 3.Example#include using namespace std; bool isDiv3(string num){    int n = num.length();    long sum = accumulate(begin(num), end(num), 0) - '0' * n;    if(sum % 3 == 0)       return true;       return false; } int main() {    string num = "3635883959606670431112222";    if(isDiv3(num)){       cout

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Check if a large number is divisible by 5 or not in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 659 Views

Here we will see how to check a number is divisible by 5 or not. In this case the number is very large number. So we put the number as string.To check whether a number is divisible by 5, So to check divisibility by 5, we have to see the last number is 0 or 5.Example#include using namespace std; bool isDiv5(string num){    int n = num.length();    if(num[n - 1] != '5' && num[n - 1] != '0')    return false;    return true; } int main() {    string num = "154484585745184258458158245285265";    if(isDiv5(num)){       cout

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Check if a large number is divisible by 75 or not in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 260 Views

Here we will see how to check a number is divisible by 75 or not. In this case the number is very large number. So we put the number as string.A number will be divisible by 75, when the number is divisible by 3 and also divisible by 25. if the sum of digits is divisible by 3, then the number is divisible by 3, and if last two digits are divisible by 25, then the number is divisible by 25.Example#include using namespace std; bool isDiv75(string num){    int n = num.length();    long sum = accumulate(begin(num), end(num), 0) ...

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Check if a large number is divisible by 8 or not in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 518 Views

Here we will see how to check a number is divisible by 8 or not. In this case the number is very large number. So we put the number as string.A number will be divisible by 8, if the number formed by last three digits are divisible by 8.Example#include using namespace std; bool isDiv8(string num){    int n = num.length();    int last_three_digit_val = (num[n-3] - '0') * 100 + (num[n-2] - '0') * 10 + ((num[n-1] - '0'));    if(last_three_digit_val % 8 == 0)       return true;       return false; } int main() {    string num = "1754586672360";    if(isDiv8(num)){       cout

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Check if a large number is divisible by 9 or not in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 378 Views

Here we will see how to check a number is divisible by 9 or not. In this case the number is very large number. So we put the number as string.A number will be divisible by 9, if the sum of digits is divisible by 9.Example#include using namespace std; bool isDiv3(string num){    int n = num.length();    long sum = accumulate(begin(num), end(num), 0) - '0' * n;    if(sum % 9 == 0)       return true;       return false; } int main() {    string num = "630720";    if(isDiv3(num)){       cout

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Check if a number can be expressed as a^b in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 402 Views

Here we will check whether we can represent a number like ab or not. Suppose a number 125 is present. This can be represented as 53. Another number 91 cannot be represented as power of some integer value.AlgorithmisRepresentPower(num): Begin    if num = 1, then return true    for i := 2, i2

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