(source)
From Ben Gundry via Eric Emmet, find and replace with a twist:
Riddler Nation has been enlisted by the Pentagon to perform crucial (and arithmetical) intelligence gathering. Our mission: decode two equations. In each of them, every different letter stands for a different digit. But there is a minor problem in both equations.
In the first equation, letters accidentally were smudged on their clandestine journey to a safe room within Riddler Headquarters and are now unreadable. (These are represented with dashes below.) But we know that all 10 digits, 0 through 9, appear in the equation.
E X M R E E K + E H K R E K K ----------------- - K - H - X - EWhat digits belong to what letters, and what are the dashes?
OK, so first off, I feel like it is easier on my brain if I use consecutive letters of the alphabet rather than the random X, M, R, etc.
So, using A, B, C, D, E, and F, and adding thousand separators where necessary,
we get
A , B C D , A A E
+ A , F E D , A E E
---------------------
- E , - F - , B - A
Next thing to notice is that we have the digits A through F, which means the letters
correspond to six different digits. From the instructions, we know âall 10 digits,
0 through 9, appear in the equation.â Because there are four âdashesâ in the final sum,
that means that all those dashes have to be unique, so our A through F plus the
dashes add up to using all 10 digits, 0 through 9.
Since the dashes are all unique, letâs just assign them their own variable letters:
A , B C D , A A E
+ A , F E D , A E E
---------------------
G E , H F I , B J A
OK, so the first thing I notice is that we are adding two seven digit numbers and getting an eight digit sum. That means that the first digit of our sum is a 1, or rather:
G has to be 1.Letâs fill that in
A , B C D , A A E
+ A , F E D , A E E
---------------------
1 E , H F I , B J A
Sticking with that âtwo seven digit numbers summing to an eight digit numberâ idea,
the first digit of each term in our sum is the same (an A), so that tells me A has
to be a 5, 6, 7, 8, or 9.
Think about it, say A was a 4.
Letâs make the largest possible seven digit number that starts with 4: we get 4,999,999.
4,999,999 + 4,999,999 = 9,999,998, which is obviously not an eight digit number.
If we have a seven digit number that starts with 3, 2, or 1, weâll get a similar result.
So, we know A is a 5 or greater.
Next thing I notice, is the last digit of the terms is the same: they are both Es.
Plus, they add up to a number whoâs last digit is A! This is important because, E + E
is the same just writing 2E, which means E + E is an even number.
We donât know if E + E is a one digit or two digit number, but its last digit has to
be even. We know its last digit is A, so A has to be even.
Thus our list for potential digits of A just got way shorter.
If A is a 5 or greater, and it is event, then A is either a 6 or an 8.
OK, letâs guess that A is 8. So E + E has to sum to a one- or two-digit number that ends in 8.
That means our choices are E + E = 8 or E + E = 18, which corresponds to E is either
4 or 9, respectively.
However, look at our main equation. We know the second digit of the final sum is E!
So, if E is a 4 or a 9,
then some number 8,xxx,xxx plus some other number 8,xxx,xxx has to sum to an eight digit
number where the second digit is a 4 or 9.
Think about that, there is no such number! 8,xxx,xxx + 8,xxx,xxx has to equal 16,xxx,xxx or 17,xxx,xxx. AKA, the second digit has to a 6 or a 7!
This can be shown through a similar approach to when we figured out that A had to be a
5 or greater: take the smallest two seven-digit numbers that begin with 8: 8,000,000.
8,000,000 + 8,000,000 = 16,000,000, yielding an eight-digit number where the second digit
is a 6. Next, take the two largest seven-digit numbers that begin with 8: 8,999,999.
8,999,999 + 8,999,999 = 17,999,998, yielding an eight-digit number where the second digit
is a 7. So by induction, summing two seven-digit numbers that begins with an 8 has to have
a sum which has a second digit of either 6 or 7. But the only valid choices for
E are 4 or 9!
That means we have a contradiction, so A canât be an 8.
OK, so by process of eliminitation:
A has to be 6Letâs fill that into our equation:
6 , B C D , 6 6 E
+ 6 , F E D , 6 E E
---------------------
1 E , H F I , B J 6
Looks good so far, now letâs look at E. Using similar logic from before, E + E has to end in
a 6, so that means we solve E + E = 6 or E + E = 16, meaning E is 3 or 8.
Borrowing again from logic above, we know the second digit of our sum (that is, E)
has to be either a 2 or a 3.
That means, looking at the intersection of our two possible lists, we have determined that
E has to be 3Filling that into our equation:
6 , B C D , 6 6 3
+ 6 , F 3 D , 6 3 3
---------------------
1 3 , H F I , B J 6
We have enough digits to fully calculate what B and J are! Doing the math, we find that:
J has to be 9B has to be 2Once again, filling that into the equation gives us:
6 , 2 C D , 6 6 3
+ 6 , F 3 D , 6 3 3
---------------------
1 3 , H F I , 2 9 6
Note that there is an imaginary 1 floating in the thousandâs place that got carried over from the 663 + 633 = 1,296. This isnât super useful at the moment, but weâll use it later.
Ignoring that for now, what is the next obvious step? Well, remember, our full equation has to use
0 through 9, with each number assigned to a unique variable A through J. Weâve already used
a fair amount of digits, letâs see what is left.
Weâve used 1, 2, 3, 6, and 9 so far. That leaves:
Unused digits:
Nothing super obvious is jumping out, but there is another thing we can determine.
Letâs look at the second digit of the second term: F.
We know that 6,xxx,xxx + 6,xxx,xxx has to sum to 13,xxx,xxx. That means that the second digits of each of our terms has to sum to 10 or greater, so we can âcarryâ the 1 into the second digit of our sum!
Put a little more concretely, if we had
6 , X - - , - - -
+ 6 , Y - - , - - -
---------------------
1 3 , - - - , - - -
Then X + Y is greater than or equal to 10.
We already know the second digit of the first term: it is 2. That means, the second digit of the second term is either 8 or 9. Looking at list of unused digits above, only 8 is available, so weâve solved another digit! Namly:
F has to be 8Filling that in:
6 , 2 C D , 6 6 3
+ 6 , 8 3 D , 6 3 3
---------------------
1 3 , H 8 I , 2 9 6
Letâs look at the third digit of our sum: H. We can see from the 2 and the 8 that
it has to be a 0 or a 1 (in the case C + 3 results in a tenâs place digit carrying over).
However, weâve already used 1, so that means:
H has to be 0Getting close! So far we have:
6 , 2 C D , 6 6 3
+ 6 , 8 3 D , 6 3 3
---------------------
1 3 , 0 8 I , 2 9 6
We have three digits left to figure out. Slimming down our unused list from before means we now have:
Unused digits:
Letâs look at D. Since we only have three digits remaining, letâs just plug each of them
(4, 5, and 7) into D and see if any give us a contradiction. Remember the 1 we had to
carry before into the thousandâs place?
1 <- carried over
| \
6 , 2 C D , 6 6 3
+ 6 , 8 3 D , 6 3 3
---------------------
1 3 , 0 8 I , 2 9 6
Letâs make sure we donât forget that when doing the arithmetic.
Starting down the line, what if D equaled 4?
1 + D + D
= 1 + 4 + 4
= 9
That would correspond with I being equal to 9, but 9 isnât available,
so D canât equal 4.
Next, what if D equaled 5?
1 + D + D
= 1 + 5 + 5
= 11
That would correspond with I being equal to 1, but again 1 isnât available,
so D canât equal 5.
By process of elimintation, that means that
D has to be 7Filling that in, we get:
6 , 2 C 7 , 6 6 3
+ 6 , 8 3 7 , 6 3 3
---------------------
1 3 , 0 8 I , 2 9 6
We can now fill in I, which shows that:
I has to be 5Almost done, we have:
6 , 2 C 7 , 6 6 3
+ 6 , 8 3 7 , 6 3 3
---------------------
1 3 , 0 8 5 , 2 9 6
And last but not least, we have only have C remaining, which means:
C has to be 4Finally, filling that in gives us:
6 , 2 4 7 , 6 6 3
+ 6 , 8 3 7 , 6 3 3
---------------------
1 3 , 0 8 5 , 2 9 6
Double checking that equation, the math works out, so we have our answer!