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When adding two positive integers, the result is sure to have at most the same number of digits as the largest of the two terms, plus one.

What about multiplication? Can many more digits can the product have than its factors? I tried to look at a few cases but was unable to find any pattern.

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You tried things like $999999\times 99999$, varying the number of $9$s, and didn't notice a pattern? – Jonas Meyer Nov 30 '11 at 23:58
Obviously, there is an upper bound. – HbCwiRoJDp Nov 30 '11 at 23:58
@JonasMeyer For some reason it didn't occur to me to try the obvious corner case. :| – Paul Manta Dec 1 '11 at 0:01
up vote 2 down vote accepted

If the two numbers being multiplied have $m$ and $n$ digits, then their product can have at most $m+n$ digits.

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True, but that's a vast overestimate. – Robert Israel Dec 1 '11 at 0:00
@RobertIsrael, $mn$ was a typo. – lhf Dec 1 '11 at 0:01

HINT $\rm\ A < 10^{\:M},\ \ B < 10^{\:N}\ \Rightarrow\ A\:B < 10^{\: ??}$

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You can rewrite a number whose decimal form is $a_1a_2\cdots a_m$ as $a_1.a_2\cdots a_m\times 10^{m-1}$, so it is of the form $a\times 10^{m-1}$ with $1\leq a<10$. (This is "scientific notation.")

If $1\leq a<10$ and $1\leq b<10$, then

$$(a\times 10^{m-1})\times (b\times 10^{n-1})=a\times b\times 10^{m+n-2}.$$ Multiplying by the power of $10$ just shifts the digits, so what is needed is to determine how many digits $a\times b$ can have to the left of the decimal point, when $1\leq a<10$ and $1\leq b<10$. There are just two possibilities depending on whether or not $ab$ is less than $10$.

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Hint: a positive integer with $d$ digits in base $b$ is less than $b^d$.

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