Friday puzzle -- World Cup Score

Hint

Alex Y

You don't need to (and in fact can't) determine the digits assigned to individual letters in order to answer the question.

In fact, there are 216 distinct ways of assigning the letters, which will all lead to the same result for A+N+G+L+E+D.

Re: Hint

Larry Barrett

AxNxGxLxExD = 2520

Alex Y

Very good.

No one else had indicated that they are working on this, so go ahead and give the answer and how you went about solving it.

Re: 

Larry Barrett

My first observation was that ANGLED was an anagram of ENGLAND, -N, so I thought this might be a trick question and might be easy if I could discover what N was. Alas, this was a dead end.

Next, from the statement of the problem, observe that

(ExNxGxLxAxNxD)/(GxExRxMxAxNxY)=1/4 reduces to (NxLxD)/(RxMxY)=1/4

or (omitting the multiplication symbol), 4(NLD)=(RMY).

Similarly, the second equation reduces to (N+L+D)-(R+M+Y)=-3

So there are three sets of letters, N,L,D, R,M,Y, and A,G,E for which the digits 1 thru 9 must be assigned. Your hint that there are 216 variations, all solving the problem constraints, comes about from observing that three digits can be assigned to three letters in 3!=6 ways, and there are 3 sets of letters; (6x6x6)=216.

So the problem comes down to finding combinations of digits that satisfy 4(NLD)=(RMY), subject to the constraint that (N+L+D)-(R+M+Y)=-3.

I first tried assigning 1 to the first set, and 4 to the second set, so the equation looked like 4(1xy)=(4jk), reasoning that all I needed to find now was a number, such that it was the product of two different sets of digits, neither of which included 1 or 4. The first number I found was 18, which can be expressed as 2x9 or 3x6. I thought this might be the solution, but it turned out that this did not satisfy the second equation.

So kept looking.

Finally found this solution:

4(1x4x9)=(8x3x6). Check that (1+4+9)-(3+6+8)=-3.

Therefore,

N,L,D can be assigned the digits 1,4,9 (in any order); similarly

R,M,Y can be assigned the digits 3,6,8 (in any order), and the remaining letters A,G,E can be assigned the remaining digits 2,5,7.

So A+N+G+L+E+D conveniently = (A+G+E)+(N+L+D)=(2+5+7)+(1+4+9)=28.

Is there a more straight-forward solution?

Re: 

Alex Y

Excellent!

Is there a more straight-forward solution?

Next, from the statement of the problem, observe that

(ExNxGxLxAxNxD)/(GxExRxMxAxNxY)=1/4 reduces to (NxLxD)/(RxMxY)=1/4

or (omitting the multiplication symbol), 4(NLD)=(RMY).

Similarly, the second equation reduces to (N+L+D)-(R+M+Y)=-3

So there are three sets of letters, N,L,D, R,M,Y, and A,G,E for which the digits 1 thru 9 must be assigned. Your hint that there are 216 variations, all solving the problem constraints, comes about from observing that three digits can be assigned to three letters in 3!=6 ways, and there are 3 sets of letters; (6x6x6)=216.

Whew! I aways breathe a sigh of relief when I don't mess up a hint like that!

So the problem comes down to finding combinations of digits that satisfy 4(NLD)=(RMY), subject to the constraint that (N+L+D)-(R+M+Y)=-3.

At this point, you can make few other observations.

1) From the multiplication equation, we can eliminate the larger primes, 5 and 7, since if those multiplicands are on one side, there is not another digit you can put on the other side to make the equation come out. So 5 and 7 are in {A,G,E}

2)This leaves 1,2,3,4,6,8,&9 as candidates for the equations. The additive equation has to have an odd number of odd digits to come out to an odd answer (-3). So the other member of {A,G,E} is an even number.

3) Going back to the multiplication equation, we see from powers of 3 that one side has to have 3 and 6, while the other has 9.

4) Looking at powers of 2, one side has to have two more than the other. One side already has one (in the six), so our possibilities are 6x4 with 2 and 1 on the other side or 6x8 with 4 and 1 on the other side, and only one of those works in the additive equation.