Relevant wiki: Cryptogram - Problem Solving

Let this 5-digit number be denoted by $\overline{ABCDE}$ as shown in the question, where $A,B,C,D$ and $E$ are not necessarily distinct single digits.

Because $\overline{ABCDE} \times 9 = \overline{ABCDE} \times (10-1) = \overline{ABCDE0} - \overline{ABCDE}$ , then the cryptogram is equivalent to the following:

$\begin{array} { c c c c c c c } & A & B & C & D& E & 0\\ - && A & B & C & D& E \\ \hline && E & D & C & B & A \\ \hline \end{array} \Leftrightarrow \begin{array} { c c c c c c c } && \color{#3D99F6} A & B & C & D& \color{#D61F06} E \\ +&& \color{#3D99F6} E & D & C & B & \color{#D61F06} A\\ \hline & A & B & C & D& E & 0 \\ \hline \end{array}$

Since the cryptogram shows the sum of 2 5-digit integers to produce a 6-digit integer, then the carry over in the leftmost column (as highlighted in blue) is exactly 1. In other words, $A = 1$ .

Then, looking at the rightmost column (as highlighted in red) tells us that $E+A \equiv 0 \pmod {10}$ , because we already know that $A = 1$ , then $E = 9$ only. Thus the cryptogram simplifies to

$\begin{array} { c c c c c c c } && 1 & B & C & D& 9 \\ +&& 9 & D & C & B & 1 \\ \hline & 1 & B & C & D& 9 & 0\\ \hline \end{array}$

Since $\overline{ABCDE} \times 9$ is still a 5-digit integer, then $\overline{ABCDE} < \dfrac{10^5}9 = 11111 \dfrac19$ . Because we already know that $A = 1$ , then $B$ is forced to be either 0 or 1 only.

Suppose $B = 0$ , then looking at the second rightmost column gives $B+D+ 1 \equiv 9 \pmod{10} \Rightarrow D = 8$ . Thus the cryptogram simplifies to

$\begin{array} { c c c c c c c } && 1 & 0 & C & D& 9 \\ +&& 9 & D & C & 0 & 1 \\ \hline & 1 & 0 & C & D& 9 & 0\\ \hline \end{array} \Rightarrow \begin{array} { c c c c c c c } && 1 & 0 & C & 8& 9 \\ +&& 9 & 8 & C & 0 & 1 \\ \hline & 1 & 0 & C & 8& 9 & 0\\ \hline \end{array}$

To finish it off, by looking at the third rightmost column, we have $2C \equiv 8 \pmod {10}\Rightarrow C = 4, 9$ . Trial and error shows that $C = 9$ only.

Thus, $(A,B,C,D,E) = (1,0,9,8,9)$ .

Now suppose $B = 1$ , then looking at the second rightmost column gives $B+D+1 \equiv 9 \pmod {10} \Rightarrow 7$ . Thus the cryptogram simplifies to

$\begin{array} { c c c c c c c } && 1 &1 & C & 7& 9 \\ +&& 9 & 7 & C & 1 & 1 \\ \hline & 1 & 1 & C & 7& 9 & 0\\ \hline \end{array}$

However, looking at the third rightmost column tells us that $2C \equiv 7 \pmod{10} \Rightarrow 2C = 7,17$ . But this is impossible because $2C$ has to be an even number. So $B= 1$ cannot be a solution.

Hence, it is true that there is only one 5-digit integer that satisfies this condition, namely $\boxed{10989}$ .

Relevant wiki: Cryptogram - Problem Solving

Cryptarithms

Let number be ABCDE

ABCDE*9=EDCBA

A must be 1 for ABCDE to be 5 digit and so that there will be no carry over (EDCBA will remain 5 digit) Consequently E must be 9 B can be either 0 or 1 since multiplying by 9 should not give a carry over for the thousandths place, (E already 9)

Case 1: B=0 Solving for D, 9*D+8 (carry over)=0 D can only be 8

Solving for C, 9*C+8 (carry over)=C C can only be 9

Rechecking, 10989*9=98901

A similar process can be done for Case 2: B=1, but we will get no solutions.

Therefore, answer is 10989

• Smallest 5 digit number is $10000$ which when multiplied by $9$ is $90000$

• largest 5 digit number that remains a $5$ digit when multiplied by $9$ is $11111$

• obviously $A=1$ and $E=9$ . and since $E=9$ one of $B,C,D$ has to be zero as $10000<ABCDE<11111$

• now, $ABCDE \implies 10000A+1000B+100C+10D+E$ ( $A,B,C,D,E$ all single digit integer) and $EDCBA=10000E+1000D+100C+10B+A$

• $9 \times ABCDE = 90000A+9000B+900C+90D+9E = 10000E + 1000D + 100C + 10B + A$ with $A = 1$ and $E = 9$ .

Putting values

$\implies 80 + 8990 B + 800 C - 910D = 0$ . with all $B,C,D$ positive, single digit natural number with at least one of them $0$ .

• obviously $D$ cannot be zero. (RHS of the above equation is zero and LHS would have all positive quantities)

• Assuming $B = 0 \implies 80 + 800C = 910D$ which gives $C=9$ and $D=8$ as integral solution.

if we assume $C=0$ , no integral solution can be found. so the number is $10989$ .

1. ABCDE x 9 = EDCBA. It's easy to see Ax9 = E. A can't be 0, and Ax9 is 1 digit, so A=1.
2. Ex9 = ?A = ?1, where ? is carry. E=9.
3. Bx9 + ? has no carry, since we know Ax9 = E = 9, so B must be 0.
4. D*9 + 8 = ?0. So, D = 8.
5. 10C89x9 = 98C01, so 9(10089+100C) = 98001 + 100C. Solving, C=9. The answer is ABCDE = 10989

// I wrote a C++ program for this:

include <iostream>

using namespace std;

int main() {

for(int x = 10000; x<=99999; x++){

        int r = x;
       int y = 0;
        while(r != 0){
            y = y*10 + r%10;
            r = r/10;
        }
    if(x*9 == y){
        cout<<x<<endl;
    }

}
return 0;

}

//Answer is 10989

Dammit ! The idiot that wrote this problem did not specify that numbers could be repeated. But this is a quote from the instructions page :

""""For simplicity and consistency, we will assume the following: 1--Each symbol represents a distinct single non-negative digit. """"

Not specifying this caused unnecessary waste of time !!!

After equating ABCDE*9=EDCBA, for both numbers to have same number of digits we get A as 1 and E as 9. Express each digit as their place values, we get 8990B + 800C +80= 910D since 0<=B,D,C<=9 it is possible for B= 0 This implies, 800C + 80= 910D thus we see a clear solution for C=9 and D=8