Are there any numbers a & b where $a*b = ab$?

Question

Are there any positive integers (non-zero) a & b where $a*b = ab$, as in $3 * 4 = 34$, or $102*7 = 1027$? If not, how do you prove there are no such numbers?

Answer 1

So, basically you are trying to find the solutions of $$a\times b = 10a+b$$ Rearranging, we have $$a=\frac {b}{b-10} $$ I hope you can take it from here.

Answer 2

Let $b^+$ be the smallest power of 10 (or whatever your base is) that is not less than $b$. Ex $54^+ = 100$, $888^+ = 1000$.

$$a \times b \le a \times b^+$$ $$a . b = a \times b^+ + b$$

So $a \times b < a.b$.

Answer 3

Let $a$ and $b$ positive integers such as $b$ has $k$ digits $(10^{k-1}\le b<10^k)$

Multiplying the last inequality with $a$ gives $ab<10^ka$

Then your equality $ab=$'$ab$' becomes $ab=10^ka+b$

So $10^ka+b<10^ka$ which means $b<0$. Contradiction.

Answer 4

This is not possible with any combination of two positive integers (give that 1 is considered the first positive integer, not zero) for a number of reasons. This is the reason that was most obvious to me.

In this case, we will be looking at the first number $a$ which would become the first part of the number $ab$. Let us call the $a$ in $a*b$ just $a$, and $a$ in $ab$ will b notated as $a'$. So that the integer values of the digits of $a$ and $a'$ are equal. Thus:

$$10^k*a=a'$$

Where $k$ is equal to the decimal value (number of digits in) $b$. The original equation can be written:

$$a*b=a'+b$$

and using substitution we can rewrite it as:

$$a*b=10^k*a+b$$

In order for $10^k*a$ to be separate from $b$, $10^k*a$ must be at least 1 greater than the digit value of of $b$, so $k>digit(b)$. This does not work however, because $k$ must then be equal to and also larger than the digit value of $b$ at the same time, which is not possible for values $b>0$ in base ten notation.

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Another way to think about this is a much simpler rule of algebra. In order for $b$ to be equal in both $a*b$ and $10^k*a+b$, the first digit of $b$ must be equal to either $1$ or $6$. This is due to the fact that any positive integer only repeats its first digit when multiplied by $1$ or $6$. A number does not ALWAYS have the same decimal value when multiplied number with a $1$ or $6$ first digit; however, this is the only time when it can happen.

$3*11=33$

$27*41=1,107$

$842*8756=6,614,752$

This works for integer that have any first digit value greater than $0$, however in order for $10^k*a+b$ to end in the same value as $b$, $10^k*a$ must be a value ending in $0$ because $k>1$, therefore, the first digit of $10^k*a+b$ and $a*b$ cannot be equal while $10^k*a$ is true for the value of $a'$.