2
$\begingroup$

How many $0$-s have at the end number $11^{11}22^{22}55^{55}$. This problem came up my contest math test .I couldn't answer this .I didn't have really any idea how to prove or what theorem could I use. Will be thankful if you can help.

$\endgroup$
4
  • 3
    $\begingroup$ Hint: the number of $0's$ at the end is exactly the order to which $10$ divides the number. $\endgroup$
    – lulu
    Aug 21, 2021 at 21:08
  • 2
    $\begingroup$ @lulu So there will be $22$ $0$'s? $\endgroup$
    – unit 1991
    Aug 21, 2021 at 21:10
  • 1
    $\begingroup$ That's right. $\quad$ $\endgroup$
    – lulu
    Aug 21, 2021 at 21:15
  • $\begingroup$ @unit1991 Right, in fact $\,11^{11}22^{22}55^{55}=5^{33}11^{88}\cdot 10^{22}\,$. $\endgroup$
    – dxiv
    Aug 22, 2021 at 2:13

2 Answers 2

4
$\begingroup$

As noted in the comments, the answer is simply whatever power of $10$ is contained in that number. Since

$$11^{11} 22^{22} 55^{55} = 11^{11} \cdot 2^{22} 11^{22} \cdot 5^{55} 11^{55}$$

we combine what powers of $2$ and $5$ that we can into $10^{22}$, giving $22$ as our answer.

$\endgroup$
3
$\begingroup$

$10 = (2)(5)$, so you're only focusing on those prime factors. Ignore $11$ completely.

After full prime factorisation, you'll get $2^{22}\cdot 5^{55} \cdot 11^k$, where $k$ doesn't matter. You need to pair up a single $2$ with a single $5$ to get one factor of $10$ and, therefore, one trailing zero. The lower exponent is the limiting factor. So you need the minimum of the two exponents of interest, i.e. $\mathrm{min} (22,55) = 22$, so there will be $22$ zeroes, a conclusion you yourself have reached in the comments. The remaining "unpaired" $33$ instances of $5$ won't contribute any zeroes by themselves.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged .