4
$\begingroup$

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

enter image description here

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.

$\endgroup$
  • $\begingroup$ I had a feeling this would have multiple solutions so I ran it through a solver (don't worry, wasn't going to post an answer). I was right - it has 6 solutions. $\endgroup$ – Beastly Gerbil Aug 28 '17 at 20:31
  • $\begingroup$ I guess I am hoping the answerers will not use the computer. Thanks for your comment though $\endgroup$ – DEEM Aug 28 '17 at 20:36
  • $\begingroup$ @Beastly Gerbil even without computers it's very easy to show that there are 17 solutions so I guess you are not counting those where A=B=C or those with leading zeroes? $\endgroup$ – user39583 Aug 28 '17 at 20:50
  • $\begingroup$ @user39583 yeah - the general rule with alphametics are no leading zeroes $\endgroup$ – Beastly Gerbil Aug 28 '17 at 21:24
2
$\begingroup$

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)

$\endgroup$
  • $\begingroup$ I like your logic Jan Ivan $\endgroup$ – DEEM Aug 29 '17 at 18:57
3
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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:

  • $C=1$, $A=0$, $B=6$
  • $C=2$, $A=1$, $B=7$
  • $C=3$, $A=2$, $B=8$
  • $C=4$, $A=3$, $B=9$

(If you do not want leading zeroes, then the first solution disappears.)

$\endgroup$
1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Sorry it seams to work except for the long column ACA $\endgroup$ – Peryhan Molokhia Aug 29 '17 at 4:47
  • $\begingroup$ 1.I looked at the range of integers 0 to 9 and averaged them out .middle number should be 4 or 5. 2)I figured B is between A&CA then B is either 4 or 5. 3)I tried out last column with B =5 and figured out C must be equal to 0 which is part of the number 10. 4) it seems to work EXCEPT FOR column 3 from last!! $\endgroup$ – Peryhan Molokhia Aug 29 '17 at 5:10
  • $\begingroup$ C cannot be 0. It's an alphametic rule $\endgroup$ – Marius Aug 29 '17 at 6:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.