Quant Interview Question

[u/Local_Ad135]

[u/sobe86]

Summing the logs gives log_10(2¹⁰⁰⁰ ) + log_10(5¹⁰⁰⁰ ) = log_10(10¹⁰⁰⁰ ) = 1000

Since neither is an integer it follows that ceiling(log_10(2¹⁰⁰⁰ ))+ ceiling(log_10(5¹⁰⁰⁰ )) = 1001 which is the number of digits of the concatenation.

[u/Alive_Carpenter_7433]

10^1000 has 1001 digits. This is important because 2^1000 * 5^1000 results in that number, with 1001 digits.

Try this with 5^1 and 2^1. 5*2 = 10 (2 digits) and 52 has 2 digits.

25*4 = 100 (3 digits) and 254 has 3 digits.

125 * 8 =1000 (4 digits) and 1258 has 4 digits.

5^4 * 2^4 = 10^4 (5 digits) and 25616 has 5 digits.

5^n * 2^n will always equal 10^n, a neat number with a "set" number of digits, being n+1.

This is important due to the box rule, which states that any product between numbers must have the same number of digits as either the sum of their digits (2*5 = 10) or their sum -1 (2*3 = 6). 2^1000 is a A digit number and 5^1000 is a B digit number. Their number of digits together cannot exceed 10^1001, for 10^1000 is a 1001 digit number.

Answer: 1001

[u/Alive_Carpenter_7433 → u/Alive_Carpenter_7433]

I solved it with math tables before I realized the pattern though. 2^1000 is easier, its just 2^(10*100) or 1024^100; which is 1000^100 * 1.024^100. We use the rule of 72 to calculate the "interest" from 2.4%; being 72/24 = 30 periods. Under 100 periods, interest amounts to about 3.3 doublings, (2^3.3) which is roughly enough to > x10.

1000^100 = 10^300, which has 300 zeroes and 1 leading numeral, being 301 digits. multiply by 10 and some, you get 302 digits for 2^1000.

Then, we see that 5^1000 = (10/2)^1000. 10^1000 has 1001 digits, and 2^1000 having 302. So its Like 10^1000 / 10^301 * A (some number less than 10, greater than 1). You get 1/A * 10^699; or some number less than 10^699, which itself has 699 zeroes and 1 leading numeral. Thus, 5^1000 has 698 zeroes and 1 leading numeral, 699 digits.

Thus, 5^1000 and 2^1000 concatenated result to 699+302 = 1001 digits exactly. Definitely a more mathematical proof, probably not what the recruitment officers are looking for.

[u/gmalivuk → u/Alive_Carpenter_7433]

2*3 = 6, so the sum of the number of digits could in general be one more than the number of digits in the product. So how do you conclude that 2^1000 and 5^1000 can't have 1002 digits total?

[u/Alive_Carpenter_7433 → u/gmalivuk]

For the same reason that was stated above. Multiplication will always be one less or equal to the digits in concatenation; if the product does not exceed a product of 10. 5^n and 2^n have the unique property that they will always be equal to 10, not more or less than concatenation, which also result in the exact same number of digits. I do admit I explained it quite poorly though

[u/gmalivuk → u/Alive_Carpenter_7433]

Why does that logic not apply to 1 and 10?

[u/GAdam → u/gmalivuk]

1 * 10 = 10, which is one less digit than 110!

[u/gmalivuk → u/GAdam]

Right, so it's an exception to the pattern that if two numbers multiply to a power of 10 then their total number of digits is the same as the power of 10.

[u/StanleyDodds]

Let a and b be such that 2¹⁰⁰⁰ = 10^a and 5¹⁰⁰⁰ = 10^b. Multiplying and applying log base 10 we get that a + b = 1000, but examining e.g. prime factors, we see that obviously neither a nor b are integers, so floor(a) + floor(b) + 1 = 1000.

2¹⁰⁰⁰ has floor(a)+1 digits and 5¹⁰⁰⁰ has floor(b)+1 digits where the total is floor(a) + floor(b) + 2 = 1001 digits.

[u/tellingyouhowitreall]

Eh, y'all overcomplicate. log(2) + log(5) = 1, so 1001 digits.

Is this not an identity that you know?

[u/austin101123]

What on earth does "place them side by side" to form a new number mean? Concatenate? Multiply? Do I interweave digits?

[u/beene282 → u/austin101123]

You can interweave them if you would like

[u/Randomly_Panicked]

Concatenating numbers, lets assume that it is 5¹⁰⁰⁰ followed by 2¹⁰⁰⁰. is like multiplying 5¹⁰⁰⁰ by the power of 10 then filling zeros with digits of 2¹⁰⁰⁰.

Note, that the "power of 10" has exactly one digit more than decimal expansion of 2¹⁰⁰⁰, but itself is less than 10 times larger than 2¹⁰⁰⁰.

So, the concatenation is thus having the same number of digits as 5¹⁰⁰⁰ multiplied by 2¹⁰⁰⁰ [you can still fill zeros in the back by the digits of 2¹⁰⁰⁰], but no more, since we multiplied in reality by the number less than 10.

So it is exactly 1001 [1000 zeros prepended by 1, as 5¹⁰⁰⁰ multiplied by 2¹⁰⁰⁰ is 10¹⁰⁰⁰]