There's a mathematical formula called Stirling's Approximation, which states that
ln(x!) ~ x * ln(x) - x
Using this,
ln(300!) ~ 300 * ln(300) - 300
ln(300) = 5.7037, so
ln (300!) ~ 300 * 5.7037 - 300 = 1411.135
Now for the other side:
ln(100^300) = 300 * ln(100) = 300 * 4.605 = 1381.551
ln(300!) > ln(100^300)
so 300! > 100^300
2007-01-18 01:27:41
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answer #1
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answered by Grizzly B 3
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300! is definitely the larger.
But for a more intuitive way of seeing why:
100^300 = 100 x 100 x 100...
300! = 300 x 299 x ...
Now the series are of equal length but the terms in the second series are larger than the terms in the first series for the first 200 terms, or 2/3 of the time.
2007-01-18 01:32:00
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answer #2
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answered by Anonymous
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Using Stirling's formula, you see that 300! is roughly
300^300 / e^300 times sqrt{2 pi x 300} . Since e<3, 300! is the largest by a factor (3/e)^300 times sqrt{2 pi x 300} .
2007-01-18 01:21:12
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answer #3
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answered by gianlino 7
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Well, 300!=300 x 299 x 298 x ... x 2 x 1
which is 300 terms
so, compute 300! / (100^300) =
3.00 x 2.99 x 2.98 x 2.97 x ... x .02 x .01
which Excel tells me equals 3.06 x 10^14.
Thus 300! is = 3.06 x 10^14 times as much as 100^300.
2007-01-18 01:27:39
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answer #4
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answered by NotEasilyFooled 5
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100^300
2007-01-18 01:18:43
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answer #5
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answered by Prats 1
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100^300
2007-01-18 01:15:17
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answer #6
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answered by Terry T 2
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Let's see:
100^300 cand be written as (10^2)^300 = 10^600
Now, 300!= 300 *299*298*....*3*2*1
which is less then 300*300*300*300..... *300 (300 times)
So we agree that:
300! < 300^300
300! < (3*10^2)^300
300! < 3^300 * 10^600
So obviously:
300! < 10^600
300! < 100^300
QED
2007-01-18 01:25:47
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answer #7
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answered by krumenager 3
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i have no proof, but even 100^2 is greater than 300 so 100^300 is greater!
2007-01-18 01:19:30
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answer #8
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answered by desT 2
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Direct computation seems simple enough to me:
300! = 30605 75122 16440 63603 53704 61297 26862 93885 88804 17357 69994 16776 74125 94765 33176 71686 74655 15291 42247 75733 49939 14788 87017 26368 86426 39077 59003 15422 68429 27906 97455 98412 25476 93027 19546 04008 01221 57762 52176 85425 59653 56903 50678 87252 64321 89626 42993 65204 57644 88303 88909 75394 34896 25436 05322 59807 76521 27082 24376 39449 12012 86786 75368 30571 22936 81943 64995 64604 98166 45022 77165 00185 17654 64693 40112 22603 47297 24066 33325 85835 06870 15016 97941 68850 35375 21375 54910 28912 64071 57154 83028 22849 37952 63658 01452 35233 15693 64822 33436 79925 45940 95276 82060 80622 32812 38738 38808 17049 60000 00000 00000 00000 00000 00000 00000 00000 00000 00000 00000 00000 00000 00000 00000
This is a 615 digit number. 100^300 is a 601 digit number (a 1 followed by 600 zeros). Therefore, 300!>100^300
2007-01-18 01:32:13
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answer #9
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answered by Pascal 7
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logx=300*log100=300*2=600
x=10^600
Calculate how many zeros are there in 300!
Then compare the numbers of zeros
2007-01-18 01:18:37
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answer #10
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answered by iyiogrenci 6
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