How large is the first tank? If ....

Question

There are 10 tanks, A, B, ...J.

Tank B has the same capacity as tank A.

Tank C has 1/2 the capacity of tank B,

Tank D has 1/3 the capacity of tank C.

Tank E has 1/4 the capacity of tank D. etc....

Tank J has 1/9 the capacity of tank I (capital i).

Now if the total capacity of all the tanks together is 2 + (26065/36288) L, what is the capacity of tank A alone?

Answer 1

Let us look for a mathematical pattern in the tank capacities. It is convenient to do this from an arbitrary simple initial value for tank A, let us say, 1 L.

Then the successive volumes, in L, are 1/0!, 1/1!, 1/2!, 1/3!, .... 1/9!.

There is no shorthand expression for Σ (1/n!) over limited ranges for n (for n=1 to ∞, the limit is e ), so just add it up. (there are efficient ways of writing the sum, namely as a column (1 + 9 + 72 +...) for the third through tenth tank, all with 9! as denominator.)

Surprise, that total is 2 + (26065/36288), therefore the initial choice of 1L for tank A was correct.

Moral: Not all series which can be concisely expressed have formulas for calculating their values over limited ranges (some do to ∞ with unexpected involvement of π or e , but that would not be SAT-level math). In a way, the discrete series expressions over limited ranges (vs. to ∞ ) bear a similarity to stepwise integration of n^k power functions (which yields long polynomial expressions) vs. calculating the exact definite integral (a much simpler polynomial expression).

-- Cheers, --Mr. d.

Answer 2

Write all the tank sizes in terms of A

A

B = A

C = 1/2B = 1/2A

D = 1/3C = 1/6A

E = 1/4D = 1/24A

F = 1/5E = 1/120A

G = 1/6F = 1/720A

H = 1/7G = 1/5040A

I = 1/8H = 1/40320A

J = 1/9I = 1/362800A

Create equation so that you can solve for A:

Sum of all the A terms = 2 + (26065/32688) L