This is easy to understand math problem which i solved using C++.
Task
Make equations for as many natural numbers (including zero) using only numbers 1, 2, 3, 4, 5 and operators +, -, *, / exactly once. Example: 14 = 5 * 3 - 4 / 2 + 1
I encourage you to try it yourself. :D
Solution (spoiler alert!)
After thinking about it for few days I was able to find only solutions for 19 different numbers. I was told there were more solutions, so programmed C++ program to test it. I got the results and I was correct, there are only solutions for 19 different numbers. Solutions made with my program are listed here:
| Number | One of solutions |
|---|---|
| 0 | 5 / 1 - 4 * 2 + 3 |
| 1 | No solution |
| 2 | 4 / 2 * 3 - 5 + 1 |
| 3 | 5 / 1 - 3 * 2 + 4 |
| 4 | 5 * 1 - 3 + 4 / 2 |
| 5 | 4 / 1 - 5 + 3 * 2 |
| 6 | 5 * 1 - 4 / 2 + 3 |
| 7 | 5 / 1 - 4 + 3 * 2 |
| 8 | 5 / 2 * 4 - 3 + 1 |
| 9 | 5 / 1 * 2 - 4 + 3 |
| 10 | 5 / 1 - 3 + 4 * 2 |
| 11 | 5 / 1 * 2 - 3 + 4 |
| 12 | 5 / 2 * 4 - 1 + 3 |
| 13 | 5 / 1 * 3 - 4 + 2 |
| 14 | 5 * 3 - 4 / 2 + 1 |
| 15 | 5 / 1 - 2 + 4 * 3 |
| 16 | 5 * 3 - 1 + 4 / 2 |
| 17 | 5 / 1 * 3 - 2 + 4 |
| 18 | No solution |
| 19 | 5 / 1 * 4 - 3 + 2 |
| 20 | No solution |
| 21 | 5 / 1 * 4 - 2 + 3 |
Code
I made this code to work not to look pretty or to be compatible with other tasks, so bare with me this code is messy.
#include <iostream>
#include <string>
#include <cstring>
#include <cmath>
using namespace std;
double calculate(string s) {
double br[5];
char op[4];
br[0] = s[0] - '0';
br[1] = s[2] - '0';
br[2] = s[4] - '0';
br[3] = s[6] - '0';
br[4] = s[8] - '0';
op[0] = s[1];
op[1] = s[3];
op[2] = s[5];
op[3] = s[7];
int av[5];
for(int i = 0; i < 5; i++) av[i] = 1;
for(int i = 0; i < 4; i++) {
if(op[i] == '/') {
br[i] = br[i] / br[i+1];
av[i + 1] = 0;
op[i] = 'd';
break;
}
else if(op[i] == '*') {
br[i] = br[i] * br[i+1];
av[i + 1] = 0;
op[i] = 'd';
break;
}
}
for(int i = 0; i < 4; i++) {
if(op[i] == '/') {
int fi, se;
for(int j = i; j >= 0; j--) {
if(av[j]) {
fi = j;
break;
}
}
for(int j = i+1; j < 5; j++) {
if(av[j]) {
se = j;
break;
}
}
br[fi] = br[fi] / br[se];
av[se] = 0;
op[i] = 'd';
break;
}
else if(op[i] == '*') {
int fi, se;
for(int j = i; j >= 0; j--) {
if(av[j]) {
fi = j;
break;
}
}
for(int j = i+1; j < 5; j++) {
if(av[j]) {
se = j;
break;
}
}
br[fi] = br[fi] * br[se];
av[se] = 0;
op[i] = 'd';
break;
}
}
for(int i = 0; i < 4; i++) {
if(op[i] == '+') {
int fi, se;
for(int j = i; j >= 0; j--) {
if(av[j]) {
fi = j;
break;
}
}
for(int j = i+1; j < 5; j++) {
if(av[j]) {
se = j;
break;
}
}
br[fi] = br[fi] + br[se];
av[se] = 0;
op[i] = 'd';
break;
}
else if(op[i] == '-') {
int fi, se;
for(int j = i; j >= 0; j--) {
if(av[j]) {
fi = j;
break;
}
}
for(int j = i+1; j < 5; j++) {
if(av[j]) {
se = j;
break;
}
}
br[fi] = br[fi] - br[se];
av[se] = 0;
op[i] = 'd';
break;
}
}
for(int i = 0; i < 4; i++) {
if(op[i] == '+') {
int fi, se;
for(int j = i; j >= 0; j--) {
if(av[j]) {
fi = j;
break;
}
}
for(int j = i+1; j < 5; j++) {
if(av[j]) {
se = j;
break;
}
}
br[fi] = br[fi] + br[se];
av[se] = 0;
op[i] = 'd';
break;
}
else if(op[i] == '-') {
int fi, se;
for(int j = i; j >= 0; j--) {
if(av[j]) {
fi = j;
break;
}
}
for(int j = i+1; j < 5; j++) {
if(av[j]) {
se = j;
break;
}
}
br[fi] = br[fi] - br[se];
av[se] = 0;
op[i] = 'd';
break;
}
}
return br[0];
}
string f[100];
int has[100];
int main() {
for(int i1 = 1; i1 <= 5; i1++) {
for(int j1 = 1; j1 <= 4; j1++) {
for(int i2 = 1; i2 <= 5; i2++) {
for(int j2 = 1; j2 <= 4; j2++) {
for(int i3 = 1; i3 <= 5; i3++) {
for(int j3 = 1; j3 <= 4; j3++) {
for(int i4 = 1; i4 <= 5; i4++) {
for(int j4 = 1; j4 <= 4; j4++) {
for(int i5 = 1; i5 <= 5; i5++) {
if(i1 == i2 || i1 == i3 || i1 == i4 || i1 == i5
|| i2 == i3 || i2 == i4 || i2 == i5 || i3 == i4 || i3 == i5 || i4 == i5) continue;
if(j1 == j2 || j1 == j3 || j1 == j4 || j2 == j3
|| j2 == j4 || j3 == j4) continue;
string s;
s.push_back(i1 + '0');
if(j1 == 1) s.push_back('+');
else if(j1 == 2) s.push_back('-');
else if(j1 == 3) s.push_back('*');
else if(j1 == 4) s.push_back('/');
s.push_back(i2 + '0');
if(j2 == 1) s.push_back('+');
else if(j2 == 2) s.push_back('-');
else if(j2 == 3) s.push_back('*');
else if(j2 == 4) s.push_back('/');
s.push_back(i3 + '0');
if(j3 == 1) s.push_back('+');
else if(j3 == 2) s.push_back('-');
else if(j3 == 3) s.push_back('*');
else if(j3 == 4) s.push_back('/');
s.push_back(i4 + '0');
if(j4 == 1) s.push_back('+');
else if(j4 == 2) s.push_back('-');
else if(j4 == 3) s.push_back('*');
else if(j4 == 4) s.push_back('/');
s.push_back(i5 + '0');
double rez = calculate(s);
if(rez < 0) continue;
if(rez == trunc(rez)) {
f[(int)rez] = s;
has[(int)rez] = 1;
}
}
}
}
}
}
}
}
}
}
for(int i = 0; i < 50; i++) {
if(has[i]) {
cout << i << " = " << f[i] << endl;
}
else cout << i << " = ?" << endl;
}
system("pause");
return 0;
}
If you love this kind of math, you will really enjoy solving Euler problems on projecteuler. They are really interesting and have a lot to do with prime numbers. Have you ever heard of it?
No I havent, thank you for the recommendation :D
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