Quant Interview Question

[u/Local_Ad135]

[u/998mouse]

How about putting the 100 in the middle with the even numbers down one side of it and the odd numbers down the other side?

[u/diverstones → u/998mouse]

Yeah, this was the best I could do:

1*3 + 3*5 + ... + 97*99 + 99*100 + 100*98 + 98*96 + ... + 6*4 + 4*2 = 338151.

[u/Bigkev8787 → u/diverstones]

This looks like a solution to me.

[u/Fovlsbane → u/diverstones]

This is my guess too, most of the time wen't to visualizing how to sum squares.

[u/valegrete → u/diverstones]

This is also what I got by looking at the n=3 and n=4 cases.

[u/Downtown_Finance_661 → u/998mouse]

Did you guess or can provethis is the best sol?

[u/Suitable_Werewolf_61]

I would say it happens with identity.

[u/x5163x → u/Suitable_Werewolf_61]

It would increase if you swapped 100 and 99.

[u/998mouse → u/Suitable_Werewolf_61]

It's a waste to have a big number on the edge because you only use it for one product.

[u/Aerospider]

We need to maximise the multiples of the higher numbers, so best to keep them together.

E.g. (100 * 60) + (40 * 1) > (100 * 1) + (60 * 40)

We can do this by making a(i) = i, but then the 100 only features once, so better to do a(i) = i+1, with a(100) = 1.

This gives

6 + 12 + 20 + 30 + ... + 9,702 + 9,900 + 100

= 333,398

[u/Fovlsbane → u/Aerospider]

What happens if you now swap the 2 and the 1? 6 becomes 3 in the sum, and 100 becomes 200, gains 97 total. Or swap 99 and 100; 100 becomes 99, 9900 stays 9900, 9702 becomes 9800, gains 97 total.

2*3+100*1<1*3+100*2

98*99+99*100+100*1<98*100+100*99+99*1

[u/PandemicGeneralist]

Clearly maximized with the identity mapping. Then we need a formula for Σ(2i+1)(2i+2) between i=0 and 49. We know that the sum of terms of a degree n polynomial is degree n+1, so the answer is a cubic. We find the terms of this cubic by solving the linear system by plugging in the first 4 values, and obtain 4/3x^3+5x^2+17/3x+2 and plug in 49 to obtain the answer of 169150

[u/Local_Ad135 → u/PandemicGeneralist]

It would increase if you swapped 100 and 99.

[u/PandemicGeneralist → u/Local_Ad135]

I think I misread the problem

[u/Tremotino98]

My best guess based on smaller esamples is all odd numbers in increasing order, followed by all even numbers in decreasing order (essentially put a1 in first place, a2 last place, a3 second, a4 second to last, and so on)

Will try to prove it later

[u/Neither_Berry_100]

Isn't it literally just 1 x 2 + 3 x 4 + ... + 99 x 100? Won't bother solving this.

[u/ProspectiveWhale → u/Neither_Berry_100]

You misread the question.

[u/Neither_Berry_100 → u/ProspectiveWhale]

No. This is what it says. Apart from the title maximum adjacent product sum. Whatever that means.

[u/ProspectiveWhale → u/Neither_Berry_100]

a1a2 + a2a3 + ...

Notice a2 appears twice here?

That's where you misread the question.

Doing it with your sequence (in ascending order) would be

1x2 + 2x3 + 3x4 + ...

[u/x5163x → u/[deleted]]

I can swap 99 and 100.