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May 18, 2010

Substitute each of the capital letters by a different base ten digit from 0 to 9 to satisfy this alphametic equation. None of the numbers can contain any leading zero.

XEROX+ XX + EE + RR = LIGHT + COLOR + L + I + G + H + T

** For an extra challenge, solve this problem without using a computer program

Edited May 18, 2010 by K Sengupta

Glycereine · May 18, 2010

Substitute each of the capital letters by a different base ten digit from 0 to 9 to satisfy this alphametic equation. None of the numbers can contain any leading zero.

XEROX+ XX + EE + RR = LIGHT + COLOR + L + I + G + H + T

** For an extra challenge, solve this problem without using a computer program

I don't have time to do it all now, so here's my start:

First off there are 10 variables all of which must be different base 10 digits, this means each digit is used once
None of the numbers listed can have leading 0's. The only variable that never leads a number is O, so O=0

Second, It is impossible knowing that O=0 for XEROX+XX+EE+RR to result in a 6 digit number, so we know that L+C must equal either X or X+1 if the result of adding XX EE RR rolls over to make the resulting 5 digit number 1 digit higher in the 10,000 column. The only way for this to happen is if X=7, R=8, and E=9. If these cases are true, then XEROX (79807) + XX (77) + EE (99) + RR (88) = 80071. We can fairly quickly see that this possibility is not valid because we now have a zero in two of the resulting digits, whereas we already are adding LIGHT and C0L0R (known zeros in COLOR) to 5 one digit numbers. We cannot get a zero in the 1,000 column of that answer without adding something to I to make 10. This would ONLY work if I was 9 because we are currently adding 0 to it (O=0). 9 is used already in this situation so this is invalid.

So since we know that L+C is not equal to X + 1, L+C must be X. If this is true and L and C are both non-zero then X must be 3 or greater.

So we know that:
O=0
L+C = X
X > 2

May 19, 2010

I had my comp try out every 360000+ possible combinations:

n=check point number
(0,L,C,E,R,I,G,H,X,T);

n=10000
(0,1,3,9,8,4,6,5,7,2)
n=20000
(0,1,5,9,7,6,3,2,8,4)
n=30000
(0,1,7,9,5,8,6,4,3,2)
n=40000
(0,1,9,8,5,3,6,4,7,2)
n=50000
(0,2,3,9,5,7,4,1,8,6)
n=60000
(0,2,5,9,3,8,7,6,4,1)
n=70000
(0,2,7,9,3,4,6,5,8,1)
n=80000
(0,2,9,8,1,6,4,3,7,5)
n=90000
(0,3,2,8,9,7,6,5,4,1)
n=100000
(0,3,5,8,9,2,6,4,7,1)
n=110000
(0,3,7,8,6,5,2,1,9,4)
n=120000
(0,3,9,7,5,8,6,4,2,1)
n=130000
(0,4,2,8,6,3,7,5,9,1)
n=140000
(0,4,5,8,3,7,2,1,9,6)
n=150000
(0,4,7,8,2,9,6,5,3,1)
n=160000
(0,4,9,7,2,3,6,5,8,1)
n=170000
(0,5,2,8,1,7,4,3,9,6)
n=180000
(0,5,4,7,9,8,6,3,2,1)
n=190000
(0,5,7,6,9,2,4,3,8,1)
n=200000
(0,5,9,6,7,4,2,1,8,3)
n=210000
(0,6,2,7,5,9,8,4,3,1)
n=220000
(0,6,4,7,5,2,8,3,9,1)
n=230000
(0,6,7,5,3,8,2,1,9,4)
n=240000
(0,6,9,5,2,8,7,4,3,1)
n=250000
(0,7,2,6,3,4,8,5,9,1)
n=260000
(0,7,4,6,1,8,3,2,9,5)
n=270000
(0,7,6,4,9,8,5,3,2,1)
n=280000
(0,7,9,4,8,2,5,3,6,1)
n=290000
(0,8,2,5,7,6,3,1,9,4)
n=300000
(0,8,4,5,6,9,7,3,2,1)
n=310000
(0,8,6,4,5,2,7,3,9,1)
n=320000
(0,8,9,4,3,6,2,1,7,5)
n=330000
(0,9,2,5,3,8,7,6,4,1)
n=340000
(0,9,4,5,2,3,7,6,8,1)
n=350000
(0,9,6,4,1,7,3,2,8,5)
n=360000
(0,9,8,3,7,6,5,4,2,1)

But nothing came out,here's the heart of the script of you wanna check:


	a[0]=X;

	a[1]=E;

	a[2]=R;

	a[3]=L;

	a[4]=I;

	a[5]=G;

	a[6]=H;

	a[7]=45-X-E-R-L-I-G-H-C; //T is the remaining number

	a[8]=C;

	a[9]=0; //O=0 for reason above poster explained

	con2(a);

//===================================

function con2(a){

	var sum1=0;

	var sum2=0;


	sum1+=10000*a[0]+1000*a[1]+100*a[2]+10*a[9]+a[0];

	sum1+=11*(a[0]+a[1]+a[2]);

	sum2+=10000*a[3]+1000*a[4]+100*a[5]+10*a[6]+a[7];

	sum2+=10000*a[8]+1000*a[9]+100*a[3]+10*a[9]+a[2];

	sum2+=a[3]+a[4]+a[5]+a[6]+a[7];

	if(sum1==sum2){

		trace("FOUND IT");

		for(var i=0;i<10;i++)

			trace(a[i]);

	}

}

May 20, 2010

I had my comp try out every 360000+ possible combinations:

n=check point number
(0,L,C,E,R,I,G,H,X,T);

n=10000
(0,1,3,9,8,4,6,5,7,2)
n=20000
(0,1,5,9,7,6,3,2,8,4)
n=30000
(0,1,7,9,5,8,6,4,3,2)
n=40000
(0,1,9,8,5,3,6,4,7,2)
n=50000
(0,2,3,9,5,7,4,1,8,6)
n=60000
(0,2,5,9,3,8,7,6,4,1)
n=70000
(0,2,7,9,3,4,6,5,8,1)
n=80000
(0,2,9,8,1,6,4,3,7,5)
n=90000
(0,3,2,8,9,7,6,5,4,1)
n=100000
(0,3,5,8,9,2,6,4,7,1)
n=110000
(0,3,7,8,6,5,2,1,9,4)
n=120000
(0,3,9,7,5,8,6,4,2,1)
n=130000
(0,4,2,8,6,3,7,5,9,1)
n=140000
(0,4,5,8,3,7,2,1,9,6)
n=150000
(0,4,7,8,2,9,6,5,3,1)
n=160000
(0,4,9,7,2,3,6,5,8,1)
n=170000
(0,5,2,8,1,7,4,3,9,6)
n=180000
(0,5,4,7,9,8,6,3,2,1)
n=190000
(0,5,7,6,9,2,4,3,8,1)
n=200000
(0,5,9,6,7,4,2,1,8,3)
n=210000
(0,6,2,7,5,9,8,4,3,1)
n=220000
(0,6,4,7,5,2,8,3,9,1)
n=230000
(0,6,7,5,3,8,2,1,9,4)
n=240000
(0,6,9,5,2,8,7,4,3,1)
n=250000
(0,7,2,6,3,4,8,5,9,1)
n=260000
(0,7,4,6,1,8,3,2,9,5)
n=270000
(0,7,6,4,9,8,5,3,2,1)
n=280000
(0,7,9,4,8,2,5,3,6,1)
n=290000
(0,8,2,5,7,6,3,1,9,4)
n=300000
(0,8,4,5,6,9,7,3,2,1)
n=310000
(0,8,6,4,5,2,7,3,9,1)
n=320000
(0,8,9,4,3,6,2,1,7,5)
n=330000
(0,9,2,5,3,8,7,6,4,1)
n=340000
(0,9,4,5,2,3,7,6,8,1)
n=350000
(0,9,6,4,1,7,3,2,8,5)
n=360000
(0,9,8,3,7,6,5,4,2,1)

But nothing came out,here's the heart of the script of you wanna check:
	a[0]=X;

	a[1]=E;

	a[2]=R;

	a[3]=L;

	a[4]=I;

	a[5]=G;

	a[6]=H;

	a[7]=45-X-E-R-L-I-G-H-C; //T is the remaining number

	a[8]=C;

	a[9]=0; //O=0 for reason above poster explained

	con2(a);

//===================================

function con2(a){

	var sum1=0;

	var sum2=0;


	sum1+=10000*a[0]+1000*a[1]+100*a[2]+10*a[9]+a[0];

	sum1+=11*(a[0]+a[1]+a[2]);

	sum2+=10000*a[3]+1000*a[4]+100*a[5]+10*a[6]+a[7];

	sum2+=10000*a[8]+1000*a[9]+100*a[3]+10*a[9]+a[2];

	sum2+=a[3]+a[4]+a[5]+a[6]+a[7];

	if(sum1==sum2){

		trace("FOUND IT");

		for(var i=0;i<10;i++)

			trace(a[i]);

	}

}

In this Wikipedia article on alphametics it is stated that “the leading digit of a multi-digit number must not be zero”

Accordingly, any positive integer having two or more numbers is deemed as a valid number. Therefore, the phrase “none of the numbers can contain any leading zero” means that none of XEROX, XX, EE, RR, LIGHT, COLOR can have any leading zero. However, this does not imply that the individual digit summands (except L) must be nonzero. Consequently, there is no restriction on any one of the digits I, G, H and T to be nonzero.

I am sure that taking due cognizance of the foregoing will lead to a valid assignment of the letters.

May 20, 2010

^ Now that you say that...

X=4

E=3

R=9

O=7

L=2

I=6

G=8

H=5

T=0

C=1

XEROX + XX + EE + RR = LIGHT + COLOR + L + I + G + H + T

43974 + 44 + 33 + 99 = 26850 + 17279 + 2 + 6 + 8 + 5 + 0

44150 = 44150

And if you think solving it with a computer program isn't an "extra challenge" by itself, think again:

function con2(a){

	var sum=0;

	var sum2=0;


	sum+=10000*a[0]+1000*a[1]+100*a[2]+10*a[3]+a[0];

	sum+=11*(a[0]+a[1]+a[2]);

	sum2+=10000*a[4]+1000*a[5]+100*a[6]+10*a[7]+a[8];

	sum2+=10000*a[9]+1000*a[3]+100*a[4]+10*a[3]+a[2];

	sum2+=a[4]+a[5]+a[6]+a[7]+a[8];

	if(sum==sum2){

		trace("FOUND IT");

		for(var i=0;i<10;i++)

			trace(a[i]);

	}

}


var a=new Array();

for(var X=1;X<10;X++)

for(var E=1;E<10;E++){trace("("+X+","+E+")");if(E==X) continue;

for(var R=1;R<10;R++){if(R==X||R==E) continue;

for(var O=0;O<10;O++){if(O==X||O==E||O==R) continue;

for(var L=1;L<10;L++){if(L==X||L==E||L==R||L==O) continue;

for(var I=0;I<10;I++){if(I==X||I==E||I==R||I==O||I==L) continue;

for(var G=0;G<10;G++){if(G==X||G==E||G==R||G==O||G==L||G==I) continue;

for(var H=0;H<10;H++){if(H==X||H==E||H==R||H==O||H==L||H==I||H==G) continue;

for(var C=1;C<10;C++){if(C==X||C==E||C==R||C==O||C==L||C==I||C==G||C==H) continue;

	a[0]=X;

	a[1]=E;

	a[2]=R;

	a[3]=O;

	a[4]=L;

	a[5]=I;

	a[6]=G;

	a[7]=H;

	a[8]=45-X-E-R-O-L-I-G-H-C;

	a[9]=C;

	con2(a);

}

}

}

}

}

}

}

}

stop();

I even had to stand by it and keep canceling the error messages that say that the script is causing my comp to overload every 10 seconds...

:P

Edited May 20, 2010 by Anza Power

May 21, 2010

In this Wikipedia article on alphametics it is stated that “the leading digit of a multi-digit number must not be zero”

Accordingly, any positive integer having two or more numbers is deemed as a valid number. Therefore, the phrase “none of the numbers can contain any leading zero” means that none of XEROX, XX, EE, RR, LIGHT, COLOR can have any leading zero. However, this does not imply that the individual digit summands (except L) must be nonzero. Consequently, there is no restriction on any one of the digits I, G, H and T to be nonzero.

I am sure that taking due cognizance of the foregoing will lead to a valid assignment of the letters.

...wow I was just going to get on to ask if I could use a computer program to simply take the determinant as I had after a couple hours of work derived a 9x9 matrix of equations in mod100000 that would theoretically yield the answer but as I derived those equations under the assumption that o had to be 0 because it was the only letter not used at the beginning of any number,all of that was for naught. I think i'll just try writing a script then. lol way too much work by hand.

May 25, 2010

not sure

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