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## #1 This is Cool » Solve this: ALFA + BETA + GAMA = DELTA Replace letters with digits » 2017-07-17 11:14:24

ProgrammerCam
Replies: 1

I was just wondering if anyone on this site uses programming languages to solve these types of challenges.
I have worked this one out and found a few more solutions than the https://www.mathsisfun.com/puzzles/alfa-beta-gama-delta.html solution gives.
my solutions are:

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5325 + 0465 + 8575 = 14365
5325 + 0495 + 8575 = 14395
5625 + 0435 + 8575 = 14635
5625 + 0495 + 8575 = 14695
5925 + 0435 + 8575 = 14935
5925 + 0465 + 8575 = 14965
5235 + 0475 + 8565 = 14275
5235 + 0495 + 8565 = 14295
5735 + 0425 + 8565 = 14725
5735 + 0495 + 8565 = 14795
5935 + 0425 + 8565 = 14925
5935 + 0475 + 8565 = 14975
5265 + 0475 + 8535 = 14275
5265 + 0495 + 8535 = 14295
5765 + 0425 + 8535 = 14725
5765 + 0495 + 8535 = 14795
5965 + 0425 + 8535 = 14925
5965 + 0475 + 8535 = 14975
5375 + 0465 + 8525 = 14365
5375 + 0495 + 8525 = 14395
5675 + 0435 + 8525 = 14635
5675 + 0495 + 8525 = 14695
5975 + 0435 + 8525 = 14935
5975 + 0465 + 8525 = 14965
5305 + 2475 + 6595 = 14375
5305 + 2485 + 6595 = 14385
5705 + 2435 + 6595 = 14735
5705 + 2485 + 6595 = 14785
5805 + 2435 + 6595 = 14835
5805 + 2475 + 6595 = 14875
5395 + 2475 + 6505 = 14375
5395 + 2485 + 6505 = 14385
5795 + 2435 + 6505 = 14735
5795 + 2485 + 6505 = 14785
5895 + 2435 + 6505 = 14835
5895 + 2475 + 6505 = 14875
5305 + 6475 + 2595 = 14375
5305 + 6485 + 2595 = 14385
5705 + 6435 + 2595 = 14735
5705 + 6485 + 2595 = 14785
5805 + 6435 + 2595 = 14835
5805 + 6475 + 2595 = 14875
5395 + 6475 + 2505 = 14375
5395 + 6485 + 2505 = 14385
5795 + 6435 + 2505 = 14735
5795 + 6485 + 2505 = 14785
5895 + 6435 + 2505 = 14835
5895 + 6475 + 2505 = 14875
5325 + 8465 + 0575 = 14365
5325 + 8495 + 0575 = 14395
5625 + 8435 + 0575 = 14635
5625 + 8495 + 0575 = 14695
5925 + 8435 + 0575 = 14935
5925 + 8465 + 0575 = 14965
5235 + 8475 + 0565 = 14275
5235 + 8495 + 0565 = 14295
5735 + 8425 + 0565 = 14725
5735 + 8495 + 0565 = 14795
5935 + 8425 + 0565 = 14925
5935 + 8475 + 0565 = 14975
5265 + 8475 + 0535 = 14275
5265 + 8495 + 0535 = 14295
5765 + 8425 + 0535 = 14725
5765 + 8495 + 0535 = 14795
5965 + 8425 + 0535 = 14925
5965 + 8475 + 0535 = 14975
5375 + 8465 + 0525 = 14365
5375 + 8495 + 0525 = 14395
5675 + 8435 + 0525 = 14635
5675 + 8495 + 0525 = 14695
5975 + 8435 + 0525 = 14935
5975 + 8465 + 0525 = 14965

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I used C++ to solve it like this:

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#include <iostream>

using namespace std;

int main(){

//initializing variables at 0 so they can start counting up
int a = 0;
int b = 0;
int d = 0;
int e = 0;
int f = 0;
int g = 0;
int l = 0;
int m = 0;
int t = 0;

while(a != -1){
//this checks to make sure each variable is unique
if((a!=b) && (a!=d) && (a!=e) && (a!=f) && (a!=g) && (a!=l) && (a!=m) && (a!=t)){
if((b!=d) && (b!=e) && (b!=f) && (b!=g) && (b!=l) && (b!=m) && (b!=t)){
if((d!=e) && (d!=f) && (d!=g) && (d!=l) && (d!=m) && (d!=t)){
if((e!=f) && (e!=g) && (e!=l) && (e!=m) && (e!=t)){
if((f!=g) && (f!=l) && (f!=m) && (f!=t)){
if((g!=l) && (g!=m) && (g!=t)){
if((l!=m) && (l!=t)){
if((m!=t)){

//this creates place holder variables to make the math easier and quicker
int ALFA = ((a*1000) + (l*100) + (f*10) + (a*1));
int BETA = ((b*1000) + (e*100) + (t*10) + (a*1));
int GAMA = ((g*1000) + (a*100) + (m*10) + (a*1));
int DELTA = ((d*10000) + (e*1000) + (l*100) + (t*10) + (a*1));

//this checks to see if the initial requirement is true and outputs correct sequences
if(ALFA + BETA + GAMA == DELTA){
cout << a << l << f << a;
cout << " + ";
cout << b << e << t << a;
cout << " + ";
cout << g << a << m << a;
cout << " = ";
cout << d << e << l << t << a;
}
}
}
}
}
}
}
}
}
//This part of the code counts from 0 to the highest number possible with the amount of variables (in this case 9 variables) it then terminates with a -1
if(t < 9){
t++;
}else if((t == 9) && (m < 9)){
m++;
t = 0;
}else if((t == 9) && (m == 9) && (l< 9)){
l++;
m = 0;
t = 0;
}else if((t == 9) && (m == 9) && (l == 9) && (g < 9)){
g++;
l = 0;
m = 0;
t = 0;
}
else if((t == 9) && (m == 9) && (l == 9) && (g == 9) && (f < 9)){
f++;
g = 0;
l = 0;
m = 0;
t = 0;
}else if((t == 9) && (m == 9) && (l == 9) && (g == 9) && (f == 9) && (e < 9)){
e++;
f = 0;
g = 0;
l = 0;
m = 0;
t = 0;
}
else if((t == 9) && (m == 9) && (l == 9) && (g == 9) && (f == 9) && (e == 9) && (d < 9)){
d++;
e = 0;
f = 0;
g = 0;
l = 0;
m = 0;
t = 0;
}else if((t == 9) && (m == 9) && (l == 9) && (g == 9) && (f == 9) && (e == 9) && (d == 9) && (b < 9)){
b++;
d = 0;
e = 0;
f = 0;
g = 0;
l = 0;
m = 0;
t = 0;
}
else if((t == 9) && (m == 9) && (l == 9) && (g == 9) && (f == 9) && (e == 9) && (d == 9) && (b == 9) && (a < 9)){
a++;
b = 0;
d = 0;
e = 0;
f = 0;
g = 0;
l = 0;
m = 0;
t = 0;
}
else if(a == 9){
a = -1;
b = 0;
d = 0;
e = 0;
f = 0;
g = 0;
l = 0;
m = 0;
t = 0;
}
}
}

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Anyone know how to do it better or faster?

## #2 Re: Maths Is Fun - Suggestions and Comments » ABCD x E Puzzle - Solution ? » 2017-07-17 09:10:19

// ABCD * E = DCBA
#include <iostream>

using namespace std;

int main(){
int a = 0;
int b = 0;
int c = 0;
int d = 0;
int e = 0;
for(int i = 0; i < 100000; i++){
if((a!=b) && (a!=c) && (a!=d) && (a!=e) && (b!=c) && (b!=d) && (b!=e) && (c!=d) && (c!=e) && (d!=e)){
if((((a*1000) + (b*100) + (c*10) + (d*1)) * e) == ((d*1000) + (c*100) + (b*10) + (a*1))){
cout << a << b << c << d << " * " << e << " = " << d << c << b << a << endl;
}
}
if(e < 9){
e++;
}else if((e == 9) && (d < 9)){
d++;
e = 0;
}else if((e == 9) && (d == 9) && (c< 9)){
c++;
d = 0;
e = 0;
}else if((e == 9) && (d == 9) && (c == 9) && (b < 9)){
b++;
c = 0;
d = 0;
e = 0;
}
else if((e == 9) && (d == 9) && (c == 9) && (b == 9) && (a < 9)){
a++;
b = 0;
c = 0;
d = 0;
e = 0;
}else if(a == 9){
a = 0;
b = 0;
c = 0;
d = 0;
e = 0;
}
}
cout << "End";
}

----------------------------------------------------------------------------------------------------------------------------------------------------------------------------

This program in c++ will solve it.
to find out the solutions if you ignore duplicates use this:

----------------------------------------------------------------------------------------------------------------------------------------------------------------------------
// ABCD * E = DCBA
#include <iostream>

using namespace std;

int main(){
int a = 0;
int b = 0;
int c = 0;
int d = 0;
int e = 0;
for(int i = 0; i < 100000; i++){
if((((a*1000) + (b*100) + (c*10) + (d*1)) * e) == ((d*1000) + (c*100) + (b*10) + (a*1))){
cout << a << b << c << d << " * " << e << " = " << d << c << b << a << endl;
}
if(e < 9){
e++;
}else if((e == 9) && (d < 9)){
d++;
e = 0;
}else if((e == 9) && (d == 9) && (c< 9)){
c++;
d = 0;
e = 0;
}else if((e == 9) && (d == 9) && (c == 9) && (b < 9)){
b++;
c = 0;
d = 0;
e = 0;
}
else if((e == 9) && (d == 9) && (c == 9) && (b == 9) && (a < 9)){
a++;
b = 0;
c = 0;
d = 0;
e = 0;
}else if(a == 9){
a = 0;
b = 0;
c = 0;
d = 0;
e = 0;
}
}
cout << "End";
}