Permutating Digits Cryptogram

Logic Level 2

$\begin{array}{ccccc} & & & A & B&C\\ + & & & C & A &B \\ \hline & & & B & C &A\\ \hline \end{array}$

If $A,B$ and $C$ are not necessarily distinct non-zero digits that satisfy the cryptogram above, is there any integer solution(s) to $(A,B,C)$ ?

Yes No

4 solutions

展豪張
Mar 30, 2016

$\overline{ABC}+\overline{CAB}=\overline{BCA}$
Taking modulus $9$ we have $9 \vert A+B+C$
$\overline{ABC}+\overline{CAB}+\overline{BCA}=2\overline{BCA}$
$111(A+B+C)=2\overline{BCA}$
So $111 \vert \overline{BCA}$
Combined, $333 \vert \overline{BCA}$
Clearly $\overline{ABC}=\overline{CAB}=\overline{BCA}=333,666,999$ are not solutions.
So, no integer solution.
(Actually checking the $111$ is strong enough, but when I do it I thought of the modulus $9$ first, so I type it here too.)

The question says, "If A, B and C are not necessarily distinct non-zero digits...". Then we have an integral solution (0,0,0), don't we?

Aniruddha Bhattacharjee - 5 years, 2 months ago

"If A, B and C are not necessarily distinct non-zero digits..."

展豪張 - 5 years, 2 months ago

That means A=B=C=0 is possible!

Aniruddha Bhattacharjee - 5 years, 2 months ago

@Aniruddha Bhattacharjee – "If A, B and C are not necessarily distinct non-zero digits..." means A, B, C may or may not be distinct non-zero digits, isn't it so?

Aniruddha Bhattacharjee - 5 years, 2 months ago

@Aniruddha Bhattacharjee – They can be distinct or same, but they must be non-zero.

展豪張 - 5 years, 2 months ago

@展豪張 – Oh! Okay! I didn't catch the flow! Thanks! :-)

Aniruddha Bhattacharjee - 5 years, 2 months ago

@展豪張 – Wonderful solution, by the way! thumbs up

Aniruddha Bhattacharjee - 5 years, 2 months ago

@Aniruddha Bhattacharjee – Thank you! ;)

展豪張 - 5 years, 2 months ago

Tran Hieu
Mar 26, 2016

Suppose a solution exist.

We have, at the hundred digit, A+C(+1) = B --> B>C,A

Therefore we have the following equation

Unit digit: C+B=10+A

Ten digit: B+A+1 = 10 + C

Hundred digit: A+C+1 = B (1)

Sum all equations we have A+B+C = 18, replace (1) into that we have 2(A+C)=17, impossible

I got the question right after like 5 minutes, however, I don't really understand the 'sum all equations A+B+C=18' bit. Could you please elaborate? Im not very bright.

Rico Lee - 5 years, 2 months ago

From

Unit digit: C+B=10+A

Ten digit: B+A+1 = 10 + C

Hundred digit: A+C+1 = B

We sum all left sides and right sides and have: C+B+B+A+1+A+C+1 = 10+A+10+C+B

A+B+C+(A+B+C+2) = 18 + (A+B+C+2)

Tran Hieu - 5 years, 2 months ago

Anonymous1 Assassin
Jan 21, 2021

I used code to solve this: from sympy import * from math import* from itertools import permutations

for n in range(1000):
  for m in range(1000):
    if n**2+n**2+n==m**2:
     print(n,m)

for a,b,c in permutations('0123456789',3):
  if int(a+b+c)+int(c+a+b)==int(b+c+a):
    print(a,b,c)```

Mdzulfiqarali Sabuj
May 12, 2017

ABC+CAB=BCA
or C+10A+100B+B+10A+100C=A+10C+100B or 91C+19A+B=0 , A,B,C >0 can't solve this but if A=0, B=0, C=0 then this equation is satisfied . it is to say that the equation has no solution .

Permutating Digits Cryptogram

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