Different integers A, B, C, D consist of the same digits. A+B=C. B+C=D. What are the smallest numbers which satisfy this?
Can it also be solved if we add C+D=E?
What if instead we add A+D=E?
Source: Original.
Kirk Bresniker's 4-digit solutions to part 1:
0198 0891 1089 1980: 198+891=1089. 1089 891+1089=1980
8019 0891 8910 9801: 8019+891=8910. 8910 891+8910=9801
Update 30 Mar 2005: Luke Pebody found a solution for part 2:
Instead of using n/7, use n/17:
A=1764705882352941
B=1176470588235294
C=2941176470588235
D=4117647058823529
E=7058823529411764
This is the smallest E I have found.
Alan's summary of a 6-digit solution for parts 1 and 3:
1)
A = 142857
B = 285714
C = 428571 = A+B
D = 714285 = B+C
3)
A = 285714
B = 142857
C = 428571 = A+B
D = 571428 = B+C
E = 857142 = A+D
Joseph had some interesting analysis as well:
1. Different integers A, B, C, D consist of the same digits. A+B=C.
B+C=D. What are the smallest numbers which satisfy this?
Since they consist of the same digits, they are congruent modulo 9.
Given that, the only way for A+B=C is if the numbers are divisible by 9. So
we need some digits that add up to a multiple of 9.
Write the numbers as A, B, A+B, and A+2B. Since they use the same digits,
they are the same length. A+B must have a larger leading digit than either A
or B, and A+2B must have a larger one still. So we need at least 3 different
starting digits, 4 unless A and B start with the same digit.
Suppose you have 3 digits. Then A and B start with the same digit, and the
three digits must add to 9 or 18. There are only a few possible sets of
digits, all of which are easily ruled out:
1, 2, 6: impossible for 1xx + 2xx = 6xx (D is too large) 1, 3, 5: A and B
must be 135 and 153, which add to 288 1, 4+, X: C is too large 2, 3, 4: C is
too small 2, 4+, X: D is too small 3, 4 or 5, X: C is too small 3, 6, 9: A
and B must be 369 and 396, which add to 765 3, 7+, X: D is too small
4+, X, X: D is too small
So we must have at least 4 digits. A and B might start with different
digits, or they might start with the same digit and one digit is not the
starting digit of any number, or there could be a repeated digit.
If there is a zero, we have one of the above sets of 3 starting digits (all
of which still fail in a similar manner).
Otherwise, if the digits add to 9, they can be 1233, 1224, 1125, or 1134.
In 1125, D is too large compared to C.
In all the other cases, A+B cannot carry in any position, so wherever the
largest digit appears it gives too large a digit in the sum (C).
If the digits add to 18, we have more possibilities, many of which do not
seem to have an easy way of ruling them out.
The digits cannot add to 27. The maximum for the leading digits of A and B
is 7 and 1 (with two other digits 8, 9 yields sum of digits only
25) or 3 and 3 if there is an unused starting digit or duplicate digit, but
that forces digits 3369 which end up with too large a sum for D (any
ordering of A and B leads to D > 10000).
Unfortunately I do not see any easy way to proceed from this point besides a
computer search.
2. Can it also be solved if we add C+D=E?
This gives us a fifth number equal to 2A+3B. This limits B's leading digit
to 2 and A's to 3 (the latter only when B begins with 1 and the second
digits are rather small). It still seems possible, but much harder.
3. What if instead we add A+D=E?
Since D = A+2B, this means E = 2A + 2B and thus also equals 2C.
So C's leading digit is limited to a maximum of 4. This also seems possible.
I started a program running a while ago and it has not yet found any answers
to even part 1.