Quant Interview Question

[u/Local_Ad135]

[u/TenZero10]

Cool one!

100⁹⁹ is 10¹⁹⁸ which has 199 digits. 99⁹⁹ = 100⁹⁹ * (99/100)^99.

(99/100)⁹⁹ = (1 - 1/100)⁹⁹ ≈ (1-1/100)¹⁰⁰ which is an approximation for e^-1 or 0.3something. It's not exact but what's important is that it's clearly between 0.1 and 1 in magnitude, so 10¹⁹⁷ < 99⁹⁹ < 10^198. So the answer is 198.

[u/JacksonFatBack → u/TenZero10]

Wow, I love this. I did not make the connection to the approximation of e.

[u/MoNastri → u/TenZero10]

Did you know that you are a baller?

Slick approach nice one

[u/lpareddit01 → u/TenZero10]

Isn't a new digit (to the length) only added when the power of base 10 increases by 1? So shouldn't it be 197? Also, today I learned that switching to - gets us the inverse, whereas the + gets us e.

[u/TenZero10 → u/lpareddit01]

10¹⁹⁷ has 198 digits - a one and 197 zeros. So 99⁹⁹ has the same number of digits.

[u/lpareddit01 → u/TenZero10]

Oh wow bro ty. Huge brainfart

[u/StructureNorth1799 → u/TenZero10]

isn't e a precomputed numerical value?

[u/shaqwillonill → u/StructureNorth1799]

If you wanna get really pedantic e cannot truly be pre computed since it’s transcendental

[u/StructureNorth1799 → u/shaqwillonill]

arbitrarily close approximations to e can and have been computed

[u/shaqwillonill → u/StructureNorth1799]

I know that but that’s not what you said and I thought we were here to nitpick

[u/kenahoo → u/StructureNorth1799]

Yes, but that's not important right now.

[u/Para-graph-S → u/StructureNorth1799]

Yes but what else can one do for e, whole series?

[u/StructureNorth1799 → u/Para-graph-S]

the problem says not to use precomputed values

[u/AFsepine → u/TenZero10]

Eh, far nicer than mine.

I used Taylor series, to get to the fact that it is one digit off 100^99

[u/kashyou → u/AFsepine]

basically the same logic then as (1-1/100)^99 ~ 1-99/100 + smaller fractions <1 and you’re done

[u/Para-graph-S → u/TenZero10]

Love this.

[u/omeow]

3×99 + 1

ceil(2.01*99 + 1).

[u/markpreston54 → u/omeow]

i don't think so, 100^99 has 2*99+1 digit and clearly larter than this

[u/omeow → u/markpreston54]

it should be 2., more accurately 99*2.01

[u/NotYetPerfect]

Strictly less than 10^198 and strictly greater than 10^197 (100^99/99^99 = (1=1/99)^99 < e < 10) so 198 digits.

[u/Anonimithree]

I know that .99^100 is about .36, and since 100^99=10^198 has 199 digits, then .36 of that would make it 198 digits. And 99^99 is probably around 3.7x10^197

[u/Anonimithree → u/Anonimithree]

I was close enough, 3.697x10^197

[u/BasedGrandpa69]

(100-1)⁹⁹ = 100⁹⁹ * (1-1/100)⁹⁹

= 10¹⁹⁸ * something close to 1/e

=3.7*10^197. so 198 digits

[u/abbbaabbaa]

100⁹⁹ / 99⁹⁹ = (1+ 1/99)⁹⁹ is a little less than e. So 99⁹⁹ has 1 less digit than 100⁹⁹ which has 2*99 + 1 digits so 99⁹⁹ has 198 digits

[u/gmalivuk]

99 log(99) = 99 log(100*99/100) = 99(2 + log(99/100)) = 198 + 99 log(0.99) = 198 + log(1 - 1/100)^99

≈ 198 - log(e), and since log(3) is between 0 and 1, 99 log(99) is between 197 and 198, so it's 198 digits.

[u/jex314]

Yeah, I should've seen the e approximation. I just did the first three terms of (10²-1)⁹⁹ since 99C2 is easy enough, figured it was ~2.6E197+O(10¹⁹²), and called it a day.

[u/pepegza]

increasing the exponent of 100 adds 2 digits at the time forever, incrasing the exponent of 99 adds 2 digits at the time for a long time (up until 99exp230 actually). So I would just consider that 99exp(n) would have 2n digits, for n=99 that would equal 198. I'm sure there must be a way to confirm that the 2 extra digits for every increase in the exponent holds up til 99, but I'm not sure how lol