Sunday

April 20, 2014

April 20, 2014

Posted by **anonymous** on Sunday, August 4, 2013 at 12:07am.

N=(2^(1!)!) + (2^(2!)!) + ... + (2^(1000!)!)?

please help..

- math -
**drwls**, Sunday, August 4, 2013 at 12:27amN = 2^1 + 2^2! + 2^(1000!)!)

The last number has no ending digits as small as 2 or 4

006 would be the last three digits of the complete number.

- math -
**Steve**, Sunday, August 4, 2013 at 5:44amSince any number n! where n>=5 ends in zero,

(2^1)! + (2^2)! + ...

= 2 + 24 + ...

ends in 6

**Related Questions**

Math - 120 Are the last two digits divisible by 4 Are the last three digits ...

Math - 120 Are the last two digits divisible by 4 Are the last three digits ...

Math - Three digits are randomly selected without replacement from 1, 9, 9 and 6...

math - A locker combination has three nonzero digits, and digits cannot be ...

Math - A locker combination has three nonzero digits, and digits cannot be ...

Math - A locker combination has three nonzero digits, and digits cannot be ...

discrete math - How many strings of four decimal digits (Note there are 10 ...

math..probabilities - After three tosses of a fair die, which is more likely to ...

math: probability - A locker combination has three nonzero digits, and digits ...

math - Consider all three-digit numbers that are greater than the sum of the ...