1
$\begingroup$

It appears that the concatenation of two positive integers is strictly greater than their sum.

I can prove this for the case when the order of magnitude of their sum is less than the order of magnitude of the concatenation, however I am struggling with the case when they are equal.

$\endgroup$
1
$\begingroup$

Let $a\ge b\ge 1$ and $10^{n-1}\le a<10^n$ (that is, $a$ has $n$ digits). And let's denote concatenation of $a$ followed by $b$ by $a|b$.

Then the sum $a+b$ is certainly not greater that $2a$ since $b\le a$, and $2a$, in turn is certainly less than $10^n+a$ since $a<10^n$:

$$a+b\le 2a < 10^n+a$$

On the other hand, the concatenation of $b$ followed by $a$ is certainly at least equal to $10^n+a$, which would mean adding $1$ before $a$, since $b\ge1$:

$$b|a=10^nb+a\ge 10^n+a$$

Hence we have $a+b<b|a$.


Now, is it also true that $a|b>a+b$?

Yes, because

$$a|b\ge 10a+1>2a\ge a+b$$

$\endgroup$
1
$\begingroup$

Order of magnitude is equal implies they have same number of digits, say $n$. Concatenation of integers $a,b$ that have n-digits is equal to $10^na+b$ or $10^nb+a$. Both these values are greater than $a+b$.

$\endgroup$
1
$\begingroup$

Hint: If the number $b$ is $k$ digits long, then $a\oplus b = a\left(10^k\right)+b$ ($\oplus$ represents concatenation).

$\endgroup$

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.