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I need to find the approximate value of 15! . In answer sheet i've got following answers:

A) 1 307 xxx xxx xxx

B) 1 207 xxx xxx xxx

C) 1 405 xxx xxx xxx

where x are futher digits in these numbers.

Frankly speaking I've no idea how to approximate this. After some research I've found Stirling's approximation but still without calculator i'm not able to get this result. Can you give me any tips ?

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  • $\begingroup$ Answer is A using a calculator, thought I am pretty sure you can't do that. Working on a solution right now. $\endgroup$ – QuIcKmAtHs Jan 20 '18 at 11:37
  • $\begingroup$ 15! is really small, so I guess you could work it out $\endgroup$ – QuIcKmAtHs Jan 20 '18 at 11:45
  • $\begingroup$ Maybe I would, but it is not the answer which I expect. I need to find approx without multiplying each number - which is very time consuming $\endgroup$ – Michael213 Jan 20 '18 at 11:50
  • $\begingroup$ I guessed that... I am currently unable to come up with something substantial to answer you $\endgroup$ – QuIcKmAtHs Jan 20 '18 at 11:50
  • $\begingroup$ Even Stirling's approximation is not going to make the calculations easier ... $\endgroup$ – rtybase Jan 20 '18 at 11:54
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It is a question of grouping the factors into chunks which multiply to form "nice" numbers which are close to numbers having many zeros.

$$ 15! = \underbrace{7 \times 13 \times 11} \times \underbrace{7 \times 9 \times 8 \times 2} \times 1296 \times 1000 \\ = 1001 \times 1008 \times 1296 \times 1000 $$

Now, it is really easy : note that $1001 \times 1008 \geq 1000^2$, but not by much. This gives exactly the first four digits being greater than or equal to $1296$, but not by much, so the answer should be expected to be $1 307...$

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  • $\begingroup$ $1.001 \times 1.008=(1+0,001)(1+0.008)=1+0.001+0.008+0.0000\ldots=1.009+0.0000\ldots$, so add $0.9%$ to $1296$ to get the first digits $1296+13*-9=1307,7$ $\endgroup$ – miracle173 Jan 20 '18 at 16:13
  • $\begingroup$ Exactly, that does it. $\endgroup$ – астон вілла олоф мэллбэрг Jan 21 '18 at 2:48
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I would group the higher components together using difference of squares from 10^2.

$15! = (15×5)×(14×6)×(13×7)×(12×8)×(11×9)×10×4×3×2$

$\approx 10^{11}×0.75×0.84×0.9×24$

$\approx 10^{11}×0.57×24$

$\approx 1.3×10^{12}$

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In addition of what's said above, There is another way to approach nearly to less calculations:

$15!=3*2*(3*2)*7*2^3*3^2*11*(3*2^2)*13*(7*2)*3*10^3$

Occurrence of primes $2$ and $3$ necessarily is dense, this suggests we figure a way to route them all to one side. let's denote it $\delta=2^8*3^6=12^4*9$

$\delta/9=(5*2+2)^4=10^4+2\binom{1}{4}10^3+2^2\binom{2}{4}10^2+2^3\binom{3}{4}10+2^4$=$10^4+2*4*10^3+4*6*10^2+8*4*10+16$ $=10^4+8*10^3+24*10^2+32*10+16$

$15!=7*7*13*(11*9)*10^3(delta/9)$=$49*13*(10^2-1)(10^7+8*10^6+24*10^5+32*10^4+16*10^3)$

You can see until this point, calculations are achieved only by hand.

$15!=(637*100-637)*polynomial=(630*100+700-637)*pol=(63000+63)*pol$

You see now the trick is easy you may just evaluate the polynomial this way:

$1 0 0 0 0$

$0 8 0 0 0$

$0 2 4 0 0$

$0 0 3 2 0$

$0 0 0 1 6$

It sums to $20736$, multiplied to $63063$ then to 10^3 .It equals approximately $10^3(130.7*10^7)$. This is the only point when you need extra-tools to multiply both factors.

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