This puzzle is a variation of a 3 digit addition puzzle I saw recently.
A , B , and C are positive integers between 0 and 9. A,B and C are different integers
Six digit numbers formed using A,B and C are given in the following equation
ABCABB+BBBCAB-BCBABC = CCCCCC
What are the values of A,B and C?
Beastly Gerbil has pointed out (by using a solver) that this has 6 solutions. Thanks. Can you get it without a computer? Please explain how.
From
a*10^5+b*10^4+c*10^3+a*10^2+b*10^1+b*10^0
+ b*10^5+b*10^4+b*10^3+c*10^2+a*10^1+b*10^0
- b*10^5+c*10^4+b*10^3+a*10^2+b*10^1+c*10^0
= c*10^5+c*10^4+c*10^3+c*10^2+c*10^1+c*10^0
We get
a*100010+20002*b+1100*c-10001*c = 111111*c
and we get that:
6*c=5*a+b
Now you just try a=0,a=1 up to a=9, or you can (for integers only, i lost real solutions, but that does not matter):
b=a+6*t, c=a+t, and viable solutions where t is integer are:
b=a+6, c=a+1
and
b=a-6, c=a-1
So 8 solutions are:
(6,0,5)(7,1,6)(8,2,7)(9,3,8)
(0,6,1)(1,7,2)(2,8,3)(3,9,4)
Looking at the last column,
2B = 2C (mod 10), so they differ by 5.
Looking at the first column,
there must be a carry, so C = A + 1
Then from the penultimate column,
there is a carry from the last column, so B = C + 5 = A + 6.
And then the remaining columns just repeat the same constraints.
Solutions :
(A, B, C) is one of (7,1,6), (8,2,7), (9,3,8), (1,7,2), (2,8,3), (3,9,4)
Explanation:
We can simplify the equation by reducing the letters that appear on the same column with + and - and we get
ABC00B + B0CAB - C000C = CCCCCC
B+B and C+C end up with the same digit (from the units column).
This means that B and C differ by 5.
Now we can blindly stab at it.
Let's assume B = 1.
We immediately get C = 6 and now have
A16001 + 106A1 - 60006 = 666666
The value that fits for A is 7.
So we have (A,B,C) = (7,1,6)
Let's move on to B = 2. We get C = 7.
A27002 + 207A2 - 70007 = 777777
We get A = 8.
So we have (A,B,C) = (8,2,7)
For B = 3 we get C = 8.
Following the same logic we get A = 9.
So we have (A,B,C) = (9,3,8)
For B = 4 we get C = 9 and we have
A49004 + 409A4 - 90009 = 999999
This gets us nowhere.
For B = 5 we get C = 0 so this is wrong.
For B = 6 we get C = 1.
A61006 + 601A6 - 10001 = 111111
A should be 0. Which does not work.
For B = 7 we get C = 2 and A = 1
For B = 8 we get C = 3 and A = 2
For B = 9 we get C = 4 and A = 3
After cancelling identical terms (and replacing the corresponding positions by zero=0), we get:
AB000B
+ 0B00AB
- 0C000C
---------
CC00CC
This leaves $2B=2C+10$ and $C=A+1$. The first equation $B=C+5$ implies $0\le C\le4$, and the second equation implies $C\ge1$. This yields the following solutions:
(If you do not want leading zeroes, then the first solution disappears.)
I found 4 of the 6 solutions to this:
A=0, B=6, C=1 A=1, B=7, C=2 A=2, B=8, C=3 A=3, B=9, C=4
This is how I got them:
Initially, we have 0 ≤ A, B, C ≤ 9. From the rightmost column, we get:
1) 2B - C = C + 10 B = C + 5 And the +10 part means we have a carry for the next columns, which leads us to: 2) A + 1 = C Using equation 1, we now have more information on the ranges for B and C 0 ≤ C ≤ 4 5 ≤ B ≤ 9 Equation 2 adds more information: 1 ≤ C ≤ 4 0 ≤ A ≤ 3 6 ≤ B ≤ 9 Now, using this information, I got the 4 sets of values by first writing out the A values, then calculation C, then B A C B
0 0+1 0+1+5 1 1+1 1+1+5 2 2+1 2+1+5 3 3+1 3+1+5
Mostly trial and error. But got my answer in second attempt.
A=3, B=9, C=4
Logic behind my approach
possible combination for the last digit (which is 2B-C = C +/- 10x) where B=0 & C=5, which was eliminated in previous answer, then I just tried for another such possibility which seemed to be B=9, C=4. Had this failed my next attempt would have been B=4, C=9
Another one
A=9, B=3, C=8
C=0,B=5,A=9.I took numbers 0to9 and looked at middle numbers And tried out middle letter as 4 or 5 .and figured out that B is either 4 or 5 .I then looked at last column and added B aND B and figured out C =0 .then I looked at column before last .added 1 from the 10 of last column to B and subtracted the other B in column before last and figured out A to give a total of 10.
