Solving $DEF+FEF=GHH$, $KLM+KLM=NKL$, $ABC+ABC+ABC=BBB$
She visits third class and is $8$ years old (you can imagine how ashamed I felt when I said so to her). I helped her with lots of maths stuff today already but this is very unknowable for me. Sorry it's in German but I have translated it :)
It's saying "Each letter represents a digit. Determine them". First question, what is "them"? The letters I guess?
How shall I determine them when they are unknown? Or is it simply $A=1, B=2, C=3, D=4, E=5, F=6, G=7, H=8, K=11, L=12, M=13, N=14$?
Alright...
With this we gave a) the first try:
It doesn't seem to make sense to set $A=1, B=2, ...$
Or we did something wrong.. Any ideas how this could be solved? :s
1.This is a homework for a third grade girl, so we need to think as a kid but not as an advanced expert, in this case we will be able to explain to this kid step by step how we solve the problem and he will understand the solution in an easy way.
2.I do not think that the purpose of the teacher of a third grade was to make all letters equal to 0 because it will not make any sense to the children and they will not learn anything from this case, so I think that this is not a valid solution at all.
3.FYI: the solutions must be from right to left because this is how children learn to do calculations.
4.The three operations seems to be independents because they are no common letters between them.
5.I have started to solve the third operation (c):
The trick is to firstly we give the value 8 to C then we can find B because 8+8+8=24 which means we write 4 and we put 2 on the top of the second column. Therefore B = 4 then we replace B by 4 in the second column and in the result. In the second column we get 2+4+4+4=14 we write 4 and we put 1 on the top of the last column, in the last column we have the result 4 and we already have 1 on the top so A will be 1 (1+1+1+1=4)
SO A=1 ,B=4 ,C=8
6.For the second operation (b):
We give the value 1 to M then L= M+M = 1+1 = 2 so K = L+L = 2+2 = 4 in the end N = K+K = 4+4 = 8 therefore M=1, L=2, K=4, N=8 .Please notice if we give to M a value of 2,4,6,7,9 so the result will be on more then 3 digits (e.g. M=2 the result will be 1684) Therefore the solutions for M are 1,3,5,8.
Each letter stands for a digit, but not all of them are different. For example, if $A=1, B=4$ and $C=8$, that makes the last equation correct. It's more than standard third grade math, it's a puzzle to try to figure out which letters are which digits.
The problems are meant to be solved by sequential trials beginning in the ones place and noticing the ramifications of carry as the student follows a logical progression and tests the results of each step against the "rules" or requirements of the problem.
One "standard" rule that would be known to the students but is not obvious to a random adult is that within a problem, each letter represents a different digit. It is unfortunate that this is not explicitly stated, but the truth is that nearly any randomly selected homework problem from grade school will depend on a local understanding of what constraints may be assumed and what approaches are to be used.
For problem C we have ABC + ABC + ABC = BBB
The student is expected to notice that
all the digits of the answer must be the same, and that
the middle digit of abc must match all the digits of bbb.
Third grade thinking for Problem C proceeds like this:
Try C = 1
1 + 1 + 1 = 3, so b would equal 3. For the tens place, that means 3 + 3 + 3 would also need to have 3 in the digits place, but 3 + 3 + 3 = 9 so C cannot be 1.
Try C = 2
2 + 2 + 2 = 6, but 6 + 6 + 6 is 18, which can't be right because the digits place has to be 6, because all three digits of the answer have to be the same.
Try C = 3
3 + 3 + 3 = 9, but 9 + 9 + 9 = 27 when the ones place would have to be 9 to fit, so that's wrong too.
Try C = 4
4 + 4 + 4 = 12, so now B would have to be 2 and we need to carry when we test the tens digit: 2 + 2 + 2 (+ 1) = 7, which doesn't match 2, so wrong again.
Try C = 5
5 + 5 + 5 = 15, so B would be 5 and there's another carry: 5 + 5 + 5 (+ 1) = 16, which is wrong because we need the ones place to be 5.
Try C = 6
6 + 6 + 6 = 18, and 8 + 8 + 8 (+ 1) = 25, but 5 is not 8 so this is not the answer.
Try C = 7
7 + 7 + 7 = 21, so now we have to carry 2, and if B = 1 then 1 + 1 + 1 (+ 2) = 5, but we needed it to be 1, so this is wrong.
Try C = 8
8 + 8 + 8 = 24, so carry 2 again and try B = 4. Then 4 + 4 + 4 (+ 2) = 14. Hey! It matched! So then for the hundreds place we have to carry 1, so we need A to be a number that adds up to 4, and luckily 1 + 1 + 1 (+ 1) = 4, so YAY! This is correct!
Just to be sure, also try C = 9.
9 + 9 + 9 = 27, so carry 2 and B would be 7. Then 7 + 7 + 7 (+2) = 23, and this is wrong because 3 is not equal to 7.
At a higher grade level, students would be expected to notice that
3(abc) = bbb
thus
(bbb) / 3 = abc
and the middle digit of abc must be the same as all the digits of bbb.
Then the sequential trials step through all the possible values of bbb, and the thinking goes like this:
111 / 3 = 37 => too few digits
222 / 3 = 74 => too few digits
333 / 3 = 111 => wrong because 1 != 3
444 / 3 = 148 => correct!
555 / 3 => 185 => wrong because 8 != 5
666 / 3 = 222 => wrong because 2 != 6
777 / 3 = 259 => wrong because 5 != 7
888 / 3 = 296 => wrong because 9 !=> 8
999 / 3 = 333 => wrong because 3 != 9
= = = = =
Similarly, for problem B we have KLM + KLM = NKL.
EDIT: Thanks to those who pointed out that I had made an incorrect snap assumption that the final sum would increase monotonically with M; that of course is not true. I have now expanded the series. A third grader obviously would not make such a mistake. :)
The student should notice that when adding M + M,
the ones place digit becomes L. Then that same L
gets plugged into the tens place for the next column of addition,
so the answer to L + L has a tens place digit that becomes K.
The student also must notice that the first two digits of KLM
must match the last two digits of NKL.
The student also must notice that although the ones place and the tens place
can have answers larger than 9 (resulting in a carry), the hundreds place
must sum to less than ten, or else N would represent TWO digits,
rather than one. Another way of saying this would be that NKL can be
up to 999 but no larger.
Here's how the thinking proceeds:
Try M = 1
1 + 1 = 2, so L would be 2. Then 2 + 2 = 4, so K would be 4. Then 4 + 4 = 8, so N is 8. 421 + 421 = 842. The first two digits of 421 DO match the last two digits of 842, so this is correct.
Try M = 2
2 + 2 = 4, so L is 4. Then 4 + 4 = 8, so K is 8. Then 8 + 8 = 16, so N would be 16. 842 + 842 = 1684. The first two digits of 842 do match the last two digits of 1684, but N can't be higher than 9 or else it would not be a digit, so this is wrong.
Try M = 3
3 + 3 = 6, so L would be 6. Then 6 + 6 = 12, so K would be 2. Then 2 + 2 (+1) = 5, so N = 5. Now we have 263 + 263 = 526. The first two digits of 263 match the last two digits of 526, so this is correct.
Try M = 4
4 + 4 = 8, so L = 8. Then 8 + 8 = 16, so K = 6. Then 6 + 6 (+ 1) = 13, but N can't be greater than 9, so this has to be wrong.
Try M = 5
5 + 5 = 10 so L = 0. Then 0 + 0 (+ 1) = 1, so K = 1. Then N = 1 + 1 = 2. We have 105 + 105 = 210, and the first two numbers of KLM do match the last two numbers of NKL, so this is correct.
Try M = 6
6 + 6 = 12, so L = 2. Then 2 + 2 (+ 1) = 5, so K = 5. Then 5 + 5 = 10, but of course N can't be bigger than 9 so this is wrong.
Try M = 7
7 + 7 = 14, so L = 4. Then 4 + 4 (+ 1) = 9, so K = 9. Now that I'm getting the hang of this I can see that if K is 5 or more then N will be greater than the maximum digit of 9, so I can say this is wrong even without calculating that N = 9 + 9 = 18.
Try M = 7
7 + 7 = 14, so L = 4. Then 4 + 4 (+ 1) = 9, so K = 9. That means N will be 18 -- too big -- so this is wrong (even though the first two digits of KLM do match the last two digits of NKL).
Try M = 8
8 + 8 = 16, so L = 6. Then 6 + 6 (+ 1) = 13, so K = 3. Then 3 + 3 (+ 1) = 7, so N is 7. We have 368 + 368 = 736. The first two digits of KLM do match the last two digits of NKL, so this is correct.
Try M = 9
9 + 9 = 18, so L = 8. Then 8 + 8 (+ 1) = 17, so K = 7, and 7 + 7 (+ 1) = 15, which is more than a single digit, so this is wrong.
Try M = 10. Oops, M can't be larger than 9 because it wouldn't be a digit, so we are done.
Isn't it interesting that even though some of the answers were too big for N to squeeze in with the digits, it was still possible to make this pattern work for every value of M from zero to nine?
= = =
Problem A is DEF + FEF = GHH
This one bothers adults because it does not yield a single unique solution, but that won't bother most third grade students. Full credit would be given for any correct answer. This example demonstrates that the student can notice the odd-even implications of carry and the fact that carry from adding in the hundreds place will give too many digits.
The student figures out that F + F gives an answer where
the ones place digit is H. The observation that H in the digits place
will always be an even number, while H in the tens place can only be
an even number if there is no carry from the digits place (thus forcing
F to be less than 5) may be made at the outset or it may be discovered
along the way.
"F must be small enough that when you add F + F there's no carry,
because when you add E + E in the tens place it will also be
an even number, and if you then had to add a carried one,
you'd get an odd number, and then that couldn't match
the H in the ones place, which is even."
If this observation is not made, the student typically begins at the bottom
and works her way upwards. if it *is* made at the outset, the student
typically begins at 4 and works her way downward.
Third grade thought would be something like this:
F has to be less than 5, because 5 + 5 = 10 and that gives a carry digit, and I already know that since the tens place is E + E, I can't also have a carry digit and end up with an even number.
Try F = 4
4 + 4 = 8, so H is 8.
Now E cannot also be 4, since each letter in this problem represents a different digit. That means E has to be 9 because 9 + 9 = 18.
Now I have to carry one over to the hundreds place, and I have to put 4 in for F in the hundreds place.
That means before I even put D in I already have 4 (+1) = 5,
so I have room to let D be anything less than 5 (because if D were 5 or larger, the hundreds place wouldn't be a single digit).
But D cannot be 4 because that's already taken for F, so D can be 1, 2, or 3.
But if D were 3 then we'd have G = 3 + 4 (+1) = 8, and H is already 8, so D can be 1 or 2, making G = 6 or 7.
So the answers (FHEDG) for F = 4 are 48916 and 48927
= = =
Now try F = 3
3 + 3 = 6, so H would be 6 and E + E would have to be 16, so E must be 8.
8 + 8 is 16 so there's a carry to the hundreds place.
I already know F is 3, and 3 (+1) = 4, so D can be anything from 0 to 5 without causing an overflow.
But D cannot be 3 because it's already used for F, so D can be 1, 2, 4, 5.
But if D were 2, G would be 6, and 6 is already used by H. And if D were 4, G would be 8, which is already used by E.
So D can be 1 or 5, which makes G = 5 or 9.
So the answers (FHEDG) for F = 3 are 36815 and 36859
= = =
Try F = 2
2 + 2 = 4 so H = 4. E can't be 2 because F is 2, so E must be 7 because 7 + 7 > = 14. That means there's a carry to the hundreds place.
Putting in 2 for F, we have 2 (+1) = 3 before we add D, so D can be anything up to 6 without giving too many digits.
But D cannot be 2 or 4 because they're already used, so D can be 1, 3, or 5.
But if D = 1 then G = 4, which is taken, so D can be 3 or 5.
So the answers (FHEDG) for F = 2 are 24736 and 24758
= = =
Try F = 1
1 + 1 = 2, so H must be 2. That means E must be 6 because 6 + 6 = 12. [Hey, E was 9, 8, 7, 6 when F was 4, 3, 2, 1!]
Since we have a carry, when we put 1 in for F in the hundreds place we get 1 (+1) = 2, which means D could be any number up to 7 without causing an overflow.
But D cannot be 1 or 2 or 6, because those are already taken, so D can be 3, 4, 5, or 7.
But if D is 4 then G is 6, which is taken by E, so D can be 3, 5, or 7
So the answers (FHEDG) for F = 1 are 12635, 12657, 12679
= = =
Try F = 0
0 + 0 = 0, and that's no good because F and H can't be the same. In fact, if F = 0 then everything has to be zero, which shows that zero isn't at all like other numbers.
So yes, it does make sense, and it can be solved by a third-grader who has been taught this approach to add and carry, and yes there are many other ways to approach this kind of problem.
In this context it's all about add and carry, and about noticing the impact of carry on the next column.
I give the results of Problem A as FHEDG because that's the order in which the child solves the problem.
I am new at this site and do not have the rep to comment, so I'll point out here that Phil's interesting perl solution doesn't fully apply the "standard rule" that for a given problem, each letter represents a different digit. Phil has applied that rule for D, E, F, and H, but has relaxed it for G, which seems a bit arbitrary. If multiple letters can represent the same number, we could also have things like FHEDG = 36336, right?
A good place to start would be with $ABC + ABC + ABC = BBB$. Since $BBB$ can only take one of $6$ values (it must be at least $123+123+123 = 369$ so it's one of $444,555,\ldots,999$), $ABC$ is uniquely determined as one-third of $BBB$. The fact that $B$ appears in both numbers narrows down the field even further, as shown in Billel Hacaine's answer: $A=1, B=4, C=8$.
From $DEF + FEF = GHH$ we have from the one's digit that $H$ is even, hence there is no carry and thus $2F = H$. Since the last digit of $2E$ is also $H$ but $F \ne E$, it follows that $E = F+5$. There us no choice of $F$ for which $\{F,F+5,2F\}$ is disjoint from $\{1,4,8\}$, though.
There are different ways to solve this.
Given the age and school year, I think a logical trial and error approach is what they expect.
Assume that any number is possible for any character, then try them to see what follows. Knowing basic addition, the possibilities are far fewer than it may seem at first.
I'm starting with the ABC problem, since it looks to me like the easiest one (many repetitions, only 3 letters).
Since addition starts with the smallest value (C) that's where we begin testing numbers.
If C=0 then since C+C+C=B we get B=0, giving
A00
+A00
+A00
----
000
The only possibility here is A=0, and this is a valid solution but too simple so I'd say we can ignore it and look for another.
If C=1 then B=3 and we get
A31
+A31
+A31
----
333
The second column isn't correct since 3+3+3=9 so it's safe to say C is not 1.
Continue like this.
C=2 => B=6 same problem
C=3 => B=9 same problem
C=4 => B=2 (now we have a 1 carry) which gives for column 2: 1+2+2+2=2 , also wrong
You'll find that the only possibility is C=8 which gives you
12
A48
+A48
+A48
----
444
now you solve for A which is 1 since 1+A+A+A=4
For the KLM problem try 1 for M giving
421
+421
----
842
Or M=3 giving
1
263
+263
----
526
Or M=5 giving
1
105
+105
----
210
There are a few possibilities here (I didnt try all of them). If M = 2 or 4 on the other hand, you end up with a carry in the first column giving you a 4 digit sum, which isn't the case, so these are not options.
For the DEF problem, assuming that different letters represent different numbers:
One solution is
1
361
+161
----
522