# A bad and boring riddle, and an easy alphametic

First comes the bad and boring riddle (the answer is to be written in all capital letters):

Magnesium plus Gallium yields a joke.

Next comes the easy alphametic:

Solve the answer to the bad and boring riddle as an alphametic.

(Every letter stands for a digit in base-10 representation, different letters stand for different digits, and leading digits are always non-zero.) Which digit does each letter represent? Please present the full analysis how these digits can be determined.

• Are you sure it have only one solution? Mar 4, 2016 at 12:22

The solution to the bad and boring riddle is:

Mg + Ga = Gag, respectively in all capitals: $~~MG+GA=GAG$

The resulting easy alphametic has only a single solution in the decimal system.

(1) Since $10M+G\le99$ and $10G+A\le99$, their sum satisfies $100G+10A+G\le198$. This implies $G=1$.

(2) The unit digits in the addition satisfy $G+A=G$ or $G+A=10+G$. This implies $A=0$.

(3) Finally, $10M+1=101-10=91$ implies $M=9$.

The unique solution to the easy alphametic is:

$~~MG+GA=GAG$ $~~~~~\Longleftrightarrow~~~~~$ $91+10=101$

I tried, I really did, but I can't make myself do alphametics. I did think of another solution to the riddle part which I am willing to share with anyone interested. It is:

Mg + Ga yields GaMg or Gamgee, a name from the Lord Of the Rings books. Gaffer and Sam are, as far as I recall, the only named members of the Gamgee clan. Gaffer is clearly too long so I would propose G A + M G = S A M as the equation.

I am using the mobile app and can't figure out how to make this a wiki post.

It is

MG+GA=LOL
Here Mg is Magnesium and Ga is Gallium and I interpreted joke as LOL.

First of all, it has many solutions. It is

72+29=101, 63+68=101, 54+47=101, 45+56=101

How I got it?

Now $L=1$ because well if two 'n digit numbers' are added then if they yield a $n+1$ digit number then the $(n+1)^{th}$ digit is always $1$. That being said, we now have MG+GA=1O1. All alphabets have different values, so M, G, A, O cannot be equal to $1$.
Now as G and A cannot be 1 so they can have possible values such that their addition always gives 1 at the one's place So they can have possible values of (2 and 9), (3 and 8), (4 and 7) etc. And they will surely leave a carry of 1 at tens place. So,
CASE 1 $\Rightarrow$ if G and A are 2 and 9 respectively. In this case we have M2+29=1O1. Here M have to be greater than or equals to 7. It can't be 9 because A=9. It can't be 8 because then O will be 1 which should not be. So M=7. Therefore O=zero. And therefore 72+29=101
CASE 2 $\Rightarrow$ if G and A are 3 and 8 respectively. In this case we have M3+38=1O1. Here M have to be greater than or equals to 6. If M=9 then O is 3 which should not be. M cannot be 8 because 8 is reserved by A in this case. If M=7 then O=1, which cannot be. So if M=6 then O=zero.

I hope you have got the point by now.