Fun Little Extension

Relevant wiki: Cryptogram - Problem Solving

Let's approach this like a cryptogram puzzle.

$\text{ABC} \\ \text{ABC +} \\ \overline{\text{CBA}}$

Since CBA has 3 digits, A must be $\leq$ 4.
A must also be even, because $2 \times \text{ABC}$ is even.
Combine these two facts, and A has to be either 2 or 4.

Now look at C. The units digit of C + C must be either 2 or 4. So C has to be either 1, 2, 6, or 7.
C also has to be equal to A + A (if there's no carry over) or A + A + 1 (if there is a carry over). That means C has to be either 4, 5, 8, or 9.

Since the two lists above with the allowed values for C have no overlap, it can be concluded the answer to the question is no. $\square$

Suppose such a number existed and we tried generating it. We know that $a_1 < 5$ and that obviously $2a_1$ is an even digit. For the reversed integer to correspond to the original integer, the first digit of the reversed integer must be even. Since a digit greater than or equal to $5$ would carry-over at most a $1$ and $2a_1 + 1$ is odd, it follows that $a_2 < 5$ . Now generalizing and using the same logic, we can deduce that $a_1, a_2, a_3, a_4, \cdots , a_{n-1}, a_n < 5$ and thus there will be no carry over when the original integer is multiplied by $2$ . Now that we have established this, we can pair each digit in the normal integer to a digit in the reversed integer as such:

$a_1 \rightarrow a_n$

$a_2 \rightarrow a_{n - 1}$

$a_3 \rightarrow a_{n - 2}$

$\vdots$

If we now consider just $a_1$ and $a_n$ , because of the position of the first digits of both the original and the reversed number, it follows that $2a_1 = a_n$ . For the same reason, because of the position of the last digits of both the original and the reversed number, it follows that $2a_n = a_1$ . However we previously stated that $2a_1 = a_n$ and as a result $2a_n \neq a_1$ unless $a_1, a_n = 0$ which cant be possible because you cannot begin writing an integer with the digit $0$ . As a result, there is no number such that if you reverse the digits of said number, the resulting integer is equal to the original number multiplied by two and the answer is $\boxed{\text{No}}$ .

I seem to have over complicated it but solved it using deductions shown below. I started with what the question has hinted for us to start:

ABC + ABC = CBA

Therefore, C+C = XA where X = 1 or 0

This gives B+B+X = YB where Y = 1 or 0

This gives A+A+Y = C

If you calculate all the B+B possibilities, you will find there is no solution where B+B = B if X=0, however if X=1 and you plug in all the possibilities for B where B+B+1 = YB, the only solution that works is when B is 9, so therefore B can only be 9

That means C has to be between 6 -9 to give us a value of X=1 (as 5+5 will give 10 and A cannot be 0).

So we know C+C = 1A and C >=6, when plugging in the numbers leads us to find that A must be 2,4,6 or 8

We also know that as B has to be 9, so we know Y =1, therefore A+A+1 = C. We can use this equation to narrow down our C and A value possibilities

If A = 2, C = 5; C cannot be 5

If A = 4, C = 9

If A = 6, C = 13; C cannot be 13

If A = 8, C = 17; C cannot be 17

We can now try the values for, A = 4, B = 9, C = 9 These values do not work, so it is not possible.

I have a question more than a solution:

Any number can be written in terms of its place, i.e. 23 can be written as (2*10 + 3) So any number AB can be written as 10A + B

For 2 digit numbers, the question is: "In what situation will 2(10A+B) = 10B + A". When we drill the above eqn down to a fraction, we get A/B = 8/19.

Since we know that A and B are definitely single digit integers, we will never have a situation when the above ratio can occur (because the denominator is a double digit number). Hence the problem statement is false (since it says "any" number, if it is false for 2 digits, the statement is false period).

Is there a flaw in this reasoning?

Let's try to approach this as a numeric puzzle. To do so, we'll assume the initial number X has its digits $a_{0}, a_{1}, a_{2}, ... , a_{n}$ and hence it can be represented as $X = 10^{0}\times a_{0} + 10^{1}\times a_{1} + 10^{2}\times a_{2} + ... + 10^{n}\times a_{n} = \sum_{i=0}^n(10^{i}\times a_{i})$ Except of the first digit (an), all the digits are either one of the numbers 0, 1, 2, ... , 9. The first digit, (an) is either one of the numbers 1, 2, ... , 9. For example, the number 3451 has its digits $a_{0}=1, a_{1}=5, a_{2}=4, a_{3}=3$ and can be represented as $3451 = 10^{0}\times 1 + 10^{1}\times 5 + 10^{2}\times 4 + 10^{3}\times 3$ After we swap the digits of the original number X, we'll have the number Y which is represented as $Y = 10^{0}\times a_{n} + 10^{1}\times a_{n-1} + 10^{2}\times a_{n-2} + ... + 10^{n}\times a_{0} = \sum_{i=0}^n(10^{n-i}\times a_{i})$ Given the problem definition, we can write down $2\times X = Y$ $2\times \sum_{i=0}^n(10^{i}\times a_{i}) = \sum_{i=0}^n(10^{n-i}\times a_{i})$ Transform further this equation: $2\times \sum_{i=0}^n(10^{i}\times a_{i}) - \sum_{i=0}^n(10^{n-i}\times a_{i}) = 0$ $\sum_{i=0}^n((2\times 10^{i} - 10^{n-i})\times a_{i}) = 0$ Next examine the term being summed, specifically its part in the parentheses: $2\times 10^{i} - 10^{n-i}$ For all i non equal to n this term is even because both numbers before and after the minus are even. When i is equal to n this term is transformed as: $2\times 10^{i} - 10^{n-n} = 2\times 10^{i} - 1$ which is apparently an odd number and has its last digit either 1 (when i=0) or 9 (when i>0). For simplicity, we can omit the case when i=0 (and n=0 too) because it means our original number X has only 1 digit and its inverse number is X as well and the problem definition gives us the equation $2\times X = X$ which obviously has no positive integer solutions. If we now go back to the equation $\sum_{i=0}^n((2\times 10^{i} - 10^{n-i})\times a_{i}) = 0$ we see that for all i non equal to n we have the products of some even number, as shown above, and the corresponding digit of the original number X. This means all such products are even. The case when i=n gives us the last term of the summation: $(2\times 10^{i} - 1)\times a_{n}$ Here we have the product of two numbers, the first one having its last digit 9, and the second one which can be either one of the digits 1, 2, ... , 9. Such product will give us an odd number. Hence the equation $\sum_{i=0}^n((2\times 10^{i} - 10^{n-i})\times a_{i}) = 0$ represents a summation of n even numbers (when i loops through 0 to n-1) and one odd number (when i equals n). Such summation never results in 0, and hence the equation has no solutions, which consequently means no positive integer numbers can meet the problem definition.

199 A+10 B=98 C, A is even and 2 A<9 so A is 2 or 4. If A is 2, C must be 1 or 6, but If c is 1, 2 A can't be 1, so c is 6, but because A is 2 or 4 ,2 A doesn't end în 6. If A is 4, then C is 8 or 9, but 2*C doesn't end în 4.In conclusion no A,B,C exists.

No!! It's never possible. Because if we look at the no. at position B , it remains unchanged. And in order to be it twice the no. before, it will always be greater than the previous B.
AB1C + AB1C = CBA . B1+B1= B . ie B > B1

ABC + ABC = CBA. Given that B is in the same position twice when B is multiplied by 2 the last digit of this two digit number must be B. This shows B > 5 because anything less doesn’t produce a two digit number and therefore couldn’t have this property. None of the numbers 6 - 9 have this property which means that the only possibility for this being true is C produces a two digit number. If C does produce a two digit number C > 5. Using guess and check we find that a 9 X 2 + 1 = 19 which is the only number between 6-9 with this property meaning B has to be 9. It’s then a simple case of checking all the numbers 6 - 9 for C to prove that this property does not exist for any 3 digit number. There’s probably a more elegant solution than that. If there is, add a comment below! I’d love to here it!

Take a number AB (such that AB+AB = BA) => 20A +2B = 10B + 2A => 9A=4B So it is not possible, for A, B both are integers, but if they are not integers they will be unable to make a number

Let the original number ( No ) be AB...YZ and the number with reversed digits ( Nr ) be ZY...BA

If Nr is twice No then A (as the last digit of Nr ) must be an even number.

If A=2 then for the doubling to work, Z (as the last digit of No ) would have to be 1 or 6. But 1Y...B2 is clearly less than 2B...Y1 and 6Y...B2 is more than double 2B...Y6. So Z can't be 1 or 6.

If A=4 then for the doubling to work, Z would have to be 2 or 7. But 4B...YZ doubled would have to be either 8Y...BA or 9Y...BA. So Z can't be 2 or 7.

If A=6 or A=8 then No doubled would have more digits than Nr .

And A not = 0 is assumed by the definition of No = AB...YZ

when looking at digits of numbers its easier to write " $AB$ " as " $A*10+B$ " so you can actually calculate them. I will first prove why this cant work with a two-digit number: $2*(A*10+B) = B*10+A$ equals $20*A+2*B = B*10+A$ now subtract A from both sides $19*A+2*B = B*10$ and subtract 2*B from both sides $19*A = B*8$ finally divide by 8 $2.375*A = B$ and there you have it. whatever the 2nd digit is, the first has to be 2.375 times as much. This is impossible since digits have to be whole numbers.

Looking a bit further at the 3 digit numbers we can quickly see that they cant work either: $2*(A*100+B*10+C) = C*100+B*10+A$ in this case B will just cancel out and we have the same issue again.

I hope I could help since this is the first solution I am writing please comment if anything is unclear or incorrect. Have a nice day!

Let $A_{1}A_{2}.....A_{n}$ when added with $A_{1}A_{2}.....A_{n}$ gives $A_{n}A_{n-1}.....A_{1}$ [WHERE ALL A's ARE SINGLE DIGIT NUMBERS !!!!!] Since the first digit of our result is $A_{n}$ therefore it is greater than $A_{1}$ it also follows that $A_{n}$ = $2A_{1}+1$ or $2A_{1}$ [Since $A_{2}+A_{2}$ can give at most carry over 1 !!!] $\Rightarrow$ $A_{n}+A_{n}$ = 10 + $A_{1}$ [ SEE THE UNIT DIGIT OF THE NUMBERS ADDED AND THE RESULT !!!] applying the two expressions we got ,the numerical value of $A_{1}$ and $A_{n}$ comes out to be non-positive integers !!!! Which contradicts our assumption and therefore the answer is $\boxed{NO}$

Let $a_1a_2a_3...a_n$ be a number (with every $a_i$ being a digit) such that $a_na_{n-1}...a_1$ = $(a_1a_2a_3...a_n)*2$ . We must have it that $a_1$ is the last digit of $a_n*2$ , but also less than $a_n$ . That means that $a_n$ cannot be less than 6, since all digits less than 6 are such that the last digit of their product by two is either greater than themselves or zero.

We are left with the options of 6,7,8, and 9 for the value of $a_n$ . It can't be 6, because a number that starts with 2 can't be such that its product by two starts with 6. For a similar reason, it can't be 7, (a number that starts with 4 can't be such that its product by 2 starts with 7). It can't be 8 (any number that starts with 6 is such that its product by two starts with 1) and it can't be 9 for the same reason. Therefore, we run out of options, making such a number impossible.

Brute force in python:

1
2
3
4
5

for a in range (1, 10):
    for b in range(1, 10):
        for c in range(1, 10):
            if a * 200 + b * 20 + c * 2 == c * 100 + b * 10 + c:
                print(a, b, c)

Since B is symetric, I can write it just as

AC+AC=CA; {A,C∊{0,1,2,3,4,5,6,7,8,9}}.

Now proof:

2AC=CA

2(10A+C)=10C+A

20A+2C=10C+A

20A+2C-2C-A=10C+A-2C-A

19A=8C

A= $\frac{19}{8}$ C

and that has 0 non-trivial solutions (for given ranges of A,C).

If $ABC+ABC=CBA$ , then either:

$C=2A$ or $C=2A+1$ , depending on whether $B>4$ .

As $C>A$ must be true for $ABC+ABC$ to result in a 3-digit number, we can use $A=2C-10$ to solve $C$ by substitution, which gives us either $C=20/3$ or $C=19/3$ respectively. Because neither of these values of $C$ are integers, a solution is not possible.

Doubling an integer is the same as doubling each digit in the integer, carrying a 1 if the result > 9. So for this to work, C = 2A and A = 2C. A = 4A and C = 4C. A = C = 1. The integer is a palindrome, so reversing it does not change its value.

A+A=C. C+C=A. •°•. 2A/2C=C/A

C^2=A^2. HENCE THE ANSWER IS NO

No solutions!

from itertools import permutations

for A, B, C in permutations(range(0, 9), 3):
    string = str(A) + str(B) + str(C)
    value = int(string)

    if str(value + value) ≡ string[::-1]:
        print(string)
        break

Since we're doubling ABC, we have two possibilities:

a. If C is even, then C = 2A.

That would mean that when we add our two numbers together, we're adding C + C = 4A.

Since 2ABC = CBA, 4A must be a number ending in A. So 4A = A, or 4A = A + 10.

4A = A only works if A = 0, which it can't because then we wouldn't be adding three-digit numbers.

So, 4A = A + 10

3A = 10

And that can't happen because A would not be an integer.

b. If C is odd, then C = 2A + 1.

That would mean that when we add our two numbers together, we're adding C + C = 4A + 2.

Since 2ABC = CBA, 4A + 2 must be a number ending in A. So 4A + 2 = A, or 4A + 2 = A + 10.

4A + 2 = A doesn't work because A would be a negative number.

So, 4A + 2 = A + 10

3A = 8

And that can't happen because A would not be an integer.

Let $a=A...Z$ , $z=Z..A$ and let's assume $2a=z$

Clearly comparing the first digit must be $Z>A$ , so $2Z=A+10$

The possible cases for A,Z are all excluded

So this is not possible, please note that the trivial one digit solution $a=z=0$ is not admitted

Let $\overline{a_n\ldots a_0}$ be the starting number and assume that we can double it by reversing its digits, i.e.

$\begin{aligned} & \overline{a_n\ldots a_0}\\ + & \overline{a_n\ldots a_0}\\ \hline & \overline{a_0\ldots a_n} \end{aligned}$

It's obvious that $n>0$ .

We know that $2a_n \leq a_0$ (it's not necessarily equality because there could have been carrying over). This implies that $4a_n \leq 2a_0$ and thus $2a_0\neq a_n$ . It follows that $2a_0 = a_n + 10$ . Returning to $4a_n \leq 2a_0$ gives us $4a_n \leq a_n + 10$ , i.e. $3a_n \leq 10$ . Since $a_n$ must be even, it follows that $a_n = 2$ and also $a_0 = 6$ . But, now it's clear that we had to carry over $2$ from adding $a_{n-1} + a_{n-1}$ . This is actually impossible (see below), so the answer is $\boxed{\text{No}}$ .

Let $c_0$ be a number that we carry over when we add $a_0 + a_0$ and $c_{i}$ what we carry when we add $a_i+a_i+c_{i-1}$ . The claim is that $c_i \leq 1$ . We can prove it by induction. The base is clear since $a_0 + a_0 < 20$ . Assume that $c_i \leq 1$ . Then we have $a_{i+1}+a_{i+1} + c_i \leq 19$ . Thus, $c_{i+1} \leq 1$ .

It is not possible as in ABC + ABC =CBA as clearly in the first line it is shown that 2C=A,hence A>C which means that A is a greater no. hence twice of that would 2A>C hence it is not possible for the resulting no. to be twice the original no.