# How many $0$-s have at the end number $11^{11}22^{22}55^{55}$. [closed]

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.

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

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.
$$10 = (2)(5)$$, 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.