Little Math Problem

While doomscrolling a friend shared this problem for elementary schoolers:¹

Substitute the letters with digits to make this operation correct
$GAG + GAG = FADA$
What’s the maximum value for $FADA$ ? (Keep in mind that for every digit we always have the same letter and that for two different digits we always have different letters. In addition no number starts with zero)

How would you resolve it?

GAG + GAG = FADA

Becomes:

$\begin{aligned} G A G & + \\ G A G & = \\ F A D A \end{aligned}$

First thing first: some terminology to go all on the same plane:

Digit position are counted from right to left: i.e: 345 has 3 as the first digit, 4 as the second and 5 as the third and last.
Square brackets show the possible values that an expression can assume i.e: $T=[34,78]$ means that T can either be 34 or 78
In the long addition representation a small number on top means a carry over in that column i.e:
$\begin{aligned} 1 & 1 \\ 1 & 76 & 4 + \\ 52 & 9 = \\ 2 & 29 & 3 \end{aligned}$

Let’s get our hands dirty then.

The first thing we notice is that our numbers go from 3 digits to 4 digits: then the last digit in GAG needs to be higher than 4 so that we can have a carry in our sum. Our carry then can only be 1.

$\begin{array}{c} F = [1] \\ G = [5, 6, 7, 8, 9] \\ A, D = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9] \end{array}$

$\begin{aligned} 1 \\ G & A & G + \\ G & A & G = \\ F & A & D & A \end{aligned}$

Why does it happen?

Ok, let’s think with smaller numbers, from 1 to 9. $\begin{aligned} 1 & + & 1 & = & 2 \\ 2 & + & 2 & = & 4 \\ 3 & + & 3 & = & 6 \\ 4 & + & 4 & = & 8 \\ 5 & + & 5 & = & 10 \\ 6 & + & 6 & = & 12 \\ 7 & + & 7 & = & 14 \\ 8 & + & 8 & = & 16 \\ 9 & + & 9 & = & 18 \end{aligned}$ You can see that until 4 our sum is always one digit, then it gets 2 digits long but the second digit is never different than 1 in the following sums.

Can you see now how this applies to our case?

Not really

Let’s keep in mind the rule from the previuos box and try this with 3 digits numbers:

$\begin{aligned} 100 & + & 100 & = & 200 \\ 200 & + & 200 & = & 400 \\ 300 & + & 300 & = & 600 \\ 400 & + & 400 & = & 800 \\ 499 & + & 499 & = & 998 \\ 500 & + & 500 & = & 1000 \\ 600 & + & 600 & = & 1200 \\ 700 & + & 700 & = & 1400 \\ 800 & + & 800 & = & 1600 \\ 900 & + & 900 & = & 1800 \\ 999 & + & 999 & = & 1998 \end{aligned}$

To summarize we can see that only the last digit of the addends is important when considering the number of digits of the final sum.

Having F equal to 1 and only 1 means that no other letter can be 1 at the same time so we can trim our possible values a little bit

$\begin{matrix} F = [1] \\ G = [5,6,7,8,9] \\ A, D = [0,2,3,4,5,6,7,8,9] \end{matrix}$

In the first and third position we can see that $G+G=1A$ , that means that A can only be an even number.

$\begin{matrix} F = [1] \\ G = [5,6,7,8,9] \\ A = [0,2,4,6,8] \\ D = [0,2,3,4,5,6,7,8,9] \end{matrix}$

Having the first and third position equal in both the addends and the final sum means that we have no carry in the second position, that can only happen if $A+A+1<10$ .

$\begin{matrix} F = [1] \\ G = [5,6,7,8,9] \\ A = [0,2,4] \\ D = [0,2,3,4,5,6,7,8,9] \end{matrix}$

What....

$\begin{aligned} 1 \\ G & A & G + \\ G & A & G = \\ 1 & A & D & A \end{aligned}$ In our long addition we can assume that a column is “indipendent” as long as we don’t have a carry from the column on the right.

Let’s try with some random numbers:

$\begin{aligned} 4 & 0 & 4 + \\ 4 & 0 & 4 = \\ 8 & 0 & 8 \end{aligned} \begin{aligned} 3 & 1 & 3 + \\ 3 & 1 & 3 = \\ 6 & 2 & 6 \end{aligned}$

We can see the same number in the first and third digits of the addends and the same number in the same positions of the final result. But what happens if in the second column we create a carry?

$\begin{aligned} 1 \\ 4 & 6 & 4 + \\ 4 & 6 & 4 = \\ 9 & 2 & 8 \end{aligned} \begin{aligned} 1 \\ 3 & 8 & 3 + \\ 3 & 8 & 3 = \\ 7 & 6 & 6 \end{aligned}$

Our third columns are no longer independent because we have a carry from the second column so the first and third digits of our sums are different.

To have G in the addends and A in the sum in the first and third columns we have to be sure that the second column creates no carry and we do that by imposing the $A+A+1<10$ rule. (That $+1$ is from the carry by the way)

That also implies that in the second column $D=A+A+1<10$ to preserve $G+G=1A$ in the third and fourth column.

F= [1]

G = [5,6,7,8,9]

A = [0,2,4]

D = [5,9]

This reduces the possible values for G and A too:

With: A = 0

D= 0 + 0+1=1 = F

We have a value collision between D and F that go against the rules of our problem.

$\begin{matrix} F = [1] \\ G = [5,6,7,8,9] \\ A = [2,4] \\ D = [5,9] \end{matrix}$

In the first and third column we can now see $G+G=[12,14]$ so we can further reduce the possible digits for G too.

$\begin{matrix} F = [1] \\ G = [6,7] \\ A = [2,4] \\ D = [5,9] \end{matrix}$

Our space of possible values is small enough that we can just go with trial and error.

First try:

$\begin{array}{r} F = [1] \\ G = [6] \end{array}$

$\begin{array}{r} \begin{aligned} 1 \\ 6 & A & 6 & + \\ 6 & A & 6 & = \\ 1 & A & D & A \end{aligned} \\ \begin{aligned} 1 \\ 6 & 2 & 6 & + \\ 6 & 2 & 6 & = \\ 1 & 2 & D & 2 \end{aligned} \\ \begin{aligned} 1 \\ 6 & 2 & 6 & + \\ 6 & 2 & 6 & = \\ 1 & 2 & 5 & 2 \end{aligned} \end{array}$

Second try:

$\begin{array}{r} F = [1] \\ G = [7] \end{array}$

$\begin{array}{r} \begin{aligned} 1 \\ 7 & A & 7 & + \\ 7 & A & 7 & = \\ 1 & A & D & A \end{aligned} \\ \begin{aligned} 1 \\ 7 & 4 & 7 & + \\ 7 & 4 & 7 & = \\ 1 & 4 & D & 4 \end{aligned} \\ \begin{aligned} 1 \\ 7 & 4 & 7 & + \\ 7 & 4 & 7 & = \\ 1 & 4 & 9 & 4 \end{aligned} \end{array}$

The maximum value for FADA is 1494 then. We could have skipped one calculation by just going

7A7 * 2 > 6A6 *2

But heh, we just confirmed that our model was right by having two sets of sums both correct.

The high difficulty was what led to originally share it btw ↩︎

@ExAnironDec 03, 2022

@ExAniron
Dec 03, 2022