CODEWITHSUNDEEP
visual-c++
dave5678
dave5678

Reputation: 19

I can't figure out why it's not printing the first perfect square with the last two digits that are odd

The professor asked us to write a program that printed the first perfect square with the last two digits being odd. Below is my attempt at the problem, I tried different ways of the code to get it it to work but to no avail. I searched around and can't seem to find anyone else with a similar assignment. Any help on this would be greatly appreciated. 

// Project3.cpp : main project file.

#include "stdafx.h"
#include <iostream>
#include <cmath>
using namespace std;

int main() {

    cout << "This program will attempt to find the first perfect square with the last two digits of the square being odd...\n";

    int i = 10;         //1. initialize "i" to 10 since square of 1 to 9 are already known to not have 2 odd last digits.
    int a = i*i;        //2. initialize "a" equal to the square of "i".
    int b = a % 100;    //3. initialize "b" to be the remainder of "a" divided by 100 (modulus operator).
    int c = b / 10;     //4. initialize "c" to be b divided by 10, which gives us the "tens" place.
    int d = a % 10;     //5. initialize "d" to be the remainder of "a" divided by 10, giving us the "ones" place.

    while (((b>10) && (b%2 != 0) && (c%2 != 0) && (d%2 != 0) && ((i*i)%i ==0)) != true){
        if ((b>10) && (b%2 != 0) && (c%2 != 0) && (d%2 != 0) && ((i*i)%i ==0)) {
        cout << "The first perfect square whose last two digits are odd is " << a << ".\n";
        system("PAUSE");
        return 0;
        }   
        else {
        ++i;}
    }    
    system("PAUSE");
    return 0;
}

Thanks, Dave

Upvotes: 1

Views: 2038

Answers (2)

Daniel Fischer
Daniel Fischer

Reputation: 183968

The following program prints out all perfect squares whose last two digits (in base 10) are odd:

int main(void){
    return 0;
}

If the last digit of n2 is odd, then n itself must be odd, n = 2*k + 1 for some k. Then

n^2 = (2*k + 1)^2 = 4*k*(k+1) + 1

is of the form 4*m + 1. But 10 = 2*4 + 2, so if the penultimate digit is odd, that produces a remainder of 2 modulo 4. To get a number of the form 4*m + 1, the last digit thus must have the form 4*s + 3, that means it must be 3 or 7. But no perfect square ends in 3 or 7. Therefore, there is no perfect square whose last two digits (in base 10) are odd.

Maybe your professor asked for the first perfect square whose first two digits are odd?

Upvotes: 0

Joe White
Joe White

Reputation: 97778

Your code is only executing once. You need a loop.

Consider:

  • You run your initialization code.
  • You run an if/else, which either displays output and exits, or increments i.
  • Then you get to the end of your main and your program exits.

Nowhere do you say "now go back and try again with this new value of i". You need a for or a while in there.

Upvotes: 1

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