OpenStudy (anonymous):

(192^3)(5^19)

5 years ago
OpenStudy (anonymous):

What is the sum of the digiits of the number (192^3)(5^19) when the number is written in standard form (that is, when everything is multiplied out)?

5 years ago
OpenStudy (turingtest):

@FoolForMath seems like your type of Q here

5 years ago
OpenStudy (anonymous):

Problem source?

5 years ago
OpenStudy (anonymous):

pardon?

5 years ago
OpenStudy (anonymous):

From where does this problem originated?

5 years ago
OpenStudy (anonymous):

it was a problem on a test

5 years ago
OpenStudy (anonymous):

Yes, which test?

5 years ago
OpenStudy (anonymous):

lol...it's a problem solving course

5 years ago
OpenStudy (anonymous):

5 years ago
OpenStudy (anonymous):

huh?

5 years ago
OpenStudy (anonymous):

The answer is 9. Hint: There are many zeros. Can you find how many?

5 years ago
OpenStudy (turingtest):

IMO problem? apparently not... lol I wanna take a "problem solving" course too if it will teach me how to do this kind of thing

5 years ago
OpenStudy (anonymous):

Extra hint for my dear Turing: Break each term into primes and then ;)

5 years ago
OpenStudy (turingtest):

ah$3^3\cdot2^{18}\cdot5^{19}=5\cdot 3^3\cdot10^{18}$

5 years ago
OpenStudy (turingtest):

FFM makes it look so easy thanks!

5 years ago
OpenStudy (anonymous):

That's right Turing! $192^3\times 5^{19}=2^{18} \times 3^3 \times 5^{19} = 27\times 5 \times 10^{18} = 135 \times 10^{18}$

5 years ago
OpenStudy (turingtest):

I need to learn how to approach these kinds of things; I always feel like I don't know what approach to take and a get paralyzed. with people like you around @FoolForMath I think I'm getting a little more comfortable with these, so thanks again =)

5 years ago
OpenStudy (anonymous):

Turing is too kind and modest, I am just another ordinary Fool.

5 years ago