## Happy 2022!

Happy 2022!!

(16682658282481)^3 +(53569945145481340779317795455694887946455915369273425239949177706463491487662277471929020010421900308946990929361260010315384419607399003466066035401249081639938095775395163723860718729197941694571507147777116105613183307154746871771308554680828)^3 +(3799493282830239438492224482691690909566619472854971357390219606335900386951333682126198574578485187205013112380672962324296382818471396758867711250989008236860018335444893297349120779943212574909820372429439350519300231559090380238923149583054941586289679546165925407521647277798487291823580372782934471884818982016141906)^3 = 3^2022

2. #### Lilac:

Let 2^2021=a*10^b is a b+1 digit number and 5^2021=10/a*10^c is a c+1 digit number with 1<a<10 and b,c in Z so that 10^2021=10*10^(b+c) or b+c=2020 gives n+1=2022 total digits on the page! Happy 2022 to you too!

3. #### hasnoname:

Haha… 2022

its basically ceiling(log (x)) + ceiling (log (y))

4. #### Gennardo:

A product of an m-digit number and an n-digit number has m+n digits or m+n-1 digits (It is funny that I never thought of this before solving this problem). Since 2^2021 * 5^2021 = 10^2021 which has 2022 digits the sum of digits of 2^2021 and 5^2021 must be 2022 or 2023.

In general the (m+n)-digit case occurs when there is a carryover in the highest digit e.g in 34… * 61…..
and the (m+n-1)-digit case occurs when there is no carryover as in 16… * 28….. But 10^x can only be factored
without a carryover in the highest digit when its both factors are also powers of ten. Since 2^2021 ist no power of ten it must be the (m+n)-case which means the sum of the digits of the factors are the same as the digits of the product, so 2022 is the solution.