what are the three digits of 5 to the 100 power?
108525
sorry, you asked it again. its 625
463098
Think as follows.\n\n1 is 5^1 = 5\n\n2 is 5^2 = 25\n\n3 is 5^3 = 125\n\n4 is 5^4 = 625\n\nSee the pattern?, For the nth row is 5^n.\n\nHence, for what you wanted,\n\n5 is 5^5 = 3125\n\nAnd buddy, I always think before answering... :)\n\nRegards
791761
3483\n\nThe numbers correspond to the buttons on a telephone\n\n663 - ONE\n896 - TWO\n84733 - THREE\n3687 - FOUR\n3483 - FIVE
609644
100! = 100*99*98...3*2*1\n\nSources of zeroes are multiples of five and multiples of 10:\n10, 20, 30, ... (10 to 90 and 100, 11 zeroes)\n5*2, 25*2, ... (5, 15, 25... 95, will make 10 zeroes)\n25, 50, 75 (another 3 zeroes)\nSo I guess it should be total 24 zeroes.
154621
If you mean 100 inside a square root symbol, the answer is 10.  However, if you mean the true square roots (as opposed the result of the square root function) the answer is 10 or -10.
871981
3 to the 5th power\nor 3*3*3*3*3
779208
I believe you mean "abc" to be a 3 digit number! So, the problem is to find digits a, b and c such that\n\n     100a + 10b + c = a + b^c\n\nthat is, b^c + digit = three digit number (not beginning with 1)\n\nThe only possibilities for b^c are\n\n     256, 512; 243, 729; 256; 625; 216; 343; 512; 729\n\nAfter trying these, find that there is only one answer corresponding to 736:\n\n     a = 7, b = 3 and c = 7
53602
You need to round off the number to 4 digits and use the proper exponent to reflect the entire number:\n\n1.496 x 10^11  Your first answer is correct.
431241
I think the confusion is coming from the fact that 2^3^5^6 is usually considered to be ((2^3)^5)^6 as far as implied precedence in a written statement where you can't see the levels of superscript.  \n\nBut what you seem to be looking for is this:\n2^(3^(5^6)) = 2^(3^15625) ~= 2^(1.04618*10^7455)\n\nMy calculator couldn't take it any further than that.  But if 2^10 is 4 digits, 2^(10^2) = 31 digits, 2^(10^3) = 301 (3*10^2) digits, 2^(10^4) = 3010 (3*10^3) digits...  I'm seeing a pattern.\n\nIt looks like we might have around 3*10^7454 digits, or "a 3 followed by 7454 zeros" digits.  That is an underestimate, though, because:\n2^(1.0*10^5) (30101 digits) < 2^(1.04618*10^5) (31493 digits)\n\nI believe that we can also conclude that it has more than a googol's count "1 followed by 100 zeroes" of digits, but less than a googolplex's count.  And since it has more than a googol's count of digits, that makes the number itself greater than a googolplex:\n2^(3^(5^6)) > 10^(10^100)
798343
The first 3 digits cannot be 800, or 900, or all the numbers from 000 to 099.  That leaves a total of 1000 - 102 = 898 possibilities left.   The next 3 digits can be anything from 000 to 999, or 1000 possibilities.  The last 4 digits must be 0000, or just one possibility.  The answer is the product: 898 x 1000 x 1, or 898,000.
358808
true\n (9/25)^(1/3) = (45)^(1/3)/(5)\n(9/25)^(1/3)= .7113786609\n(45)^(1/3)/(5)= .7113786609
1145815
6\n\nYou are putting the single digit numbers in alphebetical order:\neight, five, four, nine, one, seven, six, three, two, zero
244158
Those other folks are underestimating it hugely.  Written as 9^(9^9), it's equal to 9 to the power 387420489 and is roughly equal to 4.281455x10^369,690,399.  That's a 4 followed by almost 370 million more digits.
937714
Thinking about it, finding the powers of 5 in 1000! will give you the answer.  So...\n\nThere are 200 numbers from 1 to 1000 that are divisible by 5.  (So far we know 1000! is divisible by 5^200.)\n\nOf these 200 numbers, 40 are divisible by 25.  (Now we have 5^240 as a divisor.)\n\nOf these 40 numbers,  8 are divisible by 125.  (Now we have 5^248 as a divisor.)\n\nFinally, 625 has a fourth multple of 5.  \n\nSo, 5^249 is the largest power of 5 that divides 1000!.  2^249 can be easily shown to divide 1000!, therefore 10^249 divides 1000!  From this we can conclude 1000! ends in 249 zeroes.\n\n* * * * * *\n\n(Note to Conta: You need to recount the zeroes in that number.  Thanks for the link, though... it confirms my answer.  :) )
131932
it cant but .99 repeating is equal 2 one!!!! o unless u do a number 2 the 0th power \nex. (5^0)+(2+2)
569275
2 (to the power of) 30,402,457 - 1\n\nIt's ten million digits long....!!!  That's crazy.  It was only discovered last year too.\n\nBy the way, you can find out stuff like that so easily by googling
532000
How about 9! 51, 52, 55, 56, 57, 58, 59, 60, 61