# Let’s make a huge number with only tiny numbers (pt2)

Using only three 1s and two 2s (i.e. 1,1,1,2,2), can you generate the smallest 3-digit prime number, 101? You can use all mathematical operators you know, such as floor function, factorial, powers/indices*, etc., but concatenation is not allowed (e.g. '11' from 1 and 1 is not permitted).

• Any powers used must be derived from the initial set of three 1s and two 2s - you don't get to use 'squared' (for example) without accounting for the '2' required when writing this as a formula.

$$\large\Bigl(\frac{1}{.1}\Bigr)^2+1^2$$

I have a solution, but like an answer in the earlier question it uses a decimal point

$$101 = (\frac{1}{.1}) ^ {2} + 2 - 1$$

$$101 = 10^2 + 1$$

Using base 3 without a decimal point

$$101_3 = 10_{10} = 1 + 1 + 2 + (1 + 2)!$$

... but $$101_3$$ isn't a 3-digit prime :(

So going to base 4, without a decimal point

$$101_4 = 17_{10} = (1 + 1) ^ {(2 + 2)} + 1 = 2^4 + 1$$

• I acknowledge that your answer and mine are virtually the same and that you posted your answer before me. I promise you that I discovered my answer independently and when I started to type my answer there were no posted answers. Oct 20 at 21:55
• @WillOctagonGibson thank you – we used different ways to dispose of the unneeded $2$. Oct 20 at 22:00

Only using (1,1,2,2)

$$\left\lceil\sqrt{((2+1)!+1)!\times 2}\,\right\rceil = 101$$

Maybe it doesn't count, but it could be easy if

you use base 3:

((1 + 1) / .2) ^ 2 + 1 =
(2 / .2) ^ 2 + 1 =
10 ^ 2 + 1 =
100 + 1 =
101

Trivially, we can change bases:

2 + 2 + 1 + 1 - 1 = 5 == 101 (mod 2)

We can use basic algebra:

$$\frac{1}{.1^2} + 2 - 1 = \frac{1}{.01} + 2 - 1 = 100 + 2 - 1 = 101$$

We can also use multifactorials:

$$\frac{(((2+1)!)!!! + 2)}{.2} + 1$$ = $$\frac{((3!)!!! + 2)}{.2} + 1$$ = $$\frac{(6!!! + 2)}{.2} + 1$$ = $$\frac{48 + 2}{.2} + 1$$ = $$\frac{50}{.2} + 1$$ = $$100 + 1 = 101$$

Or we can use a little bit of everything:

$$[\sum_{n=1}^{((1+2)!)!!!!} (n+2)] - 1 = [\sum_{n=1}^{12} n+2] - 1 = 102 - 1 = 101$$

• I like that last solution a lot! Oct 20 at 23:39