How many digits are in 99^10?

Question

irina · Accepted Answer

continuing (100)^10*(0.99)^10
0.99^10 = (1 -  0.01)^10 = sum of 10_choose_i_combination * 0.01^i = 1 - 10*0.01 + 45*0.01^2 - ...
if we compare the magnitude of 2 consecutive terms, i.e. the absolute value of the i+1th term / the i the term
this ratio = (10-i)/(i+1)*0.01 <= 0.1
therefore 0.99^10 is bounded by (0.9, 1)
99^10 is bounded by (9e19, e20) that is 20 digits

jw · Answer

think of this.
10^1 = 10, 2 digit.
10^2 = 100, 3 digit.. -> so 10^n has n + 1 digit.

For the 99 ^ 10, we can rewrite as: (100 - 1) ^ 10 => (10^2 - 1)^10
if you expand that, you will see that 10^20 - WHATEVER - WHATEVER terms.
You can see that two WHATEVER terms are relevant enough to shift 1 digit back in. (again think of this: 10^2 - 1 = 99... same logic here.)

So, the conclusion is: 21 digit - 1 (by trailing WHATEVER terms) = 20 digit.

Anonymous · Answer

it's 20 digits anyway... dont know how to solve it

90,438,207,500,880,449,001

Brad · Answer

Find the pattern - 99*99 - something with 4 digits... something with 4 digits * something with 2 digits.= something with 6 digits.
10* 2 = 20 digits

Anonymous · Answer

99^10 is (100^10)*(0.99^10) = (10^20)*(0.99^10). Since (0.99^10) > 0.1, we don't "lose" a digit because the factor is not enough to knock us down by a magnitude of 10. 10^20 has 21 digits so we get 20 digits.

Anonymous · Answer

log_10 99^10=10log_10 99～19.X
20 digits

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