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Explicitly Stated (Posted on 2004-08-18) Difficulty: 3 of 5
A certain bank doesn't believe in interest and gives none the whole year. However, they do two things as a gift at the end of the year. They put money into your account such that it has 5 times as much as it did before. Then, they put 8 dollars in the account after that.

Jack gets one of these accounts at the start of year 1, and puts in 6 dollars. Assuming there are no other withdrawals or deposits into that account, figure out how much money is in that account at the beginning of year x, even if you don't know how much was in the account any of the previous years.

For example, on the beginning of year 1, he would have 6 dollars. On the beginning of year 2, he would have 38 dollars, and on the beginning of year 3 he would have 198 dollars.

What if you put in A dollars to start at the beginning of the first year, the bank put money into your account at the end of the year such that it was B times as much as before, and then put in C more dollars after that; how much money would you have at the beginning of year x, assuming everything else is normal and there are no withdrawals or deposits, even if you don't know how much was in the account any of the previous years?

No Solution Yet Submitted by Gamer    
Rating: 4.0000 (5 votes)

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Solution re: Another question: answered? | Comment 6 of 11 |
(In reply to Another question: by Gamer)

So Gamer, is M(x) = A*B^(x-1) + C*[1 Ė B^(x-1)]/(1ĖB) the right answer for the generalized question?

I looked at the difference between two consecutive balances, and played around with the A, B, and C values to come up with a formula. I got that the balance increase from year n to year n+1 is [A*B + C Ė 2]*B^(n-1).

How did I get that? Well, first, I took the A=6, B=5, C=8 problem that was given. Gamer had pointed out that the differences were all multiples of 32. So I wanted to see if there was a pattern to the multiples. There was, and it was 32*5^(n-1). I confirmed this by changing 5 to other numbers, and what I got was some number Y time B^(n-1) everytime. So I had one part down. Now how do I find Y?

I basically just played around with numbers and tried to find a pattern. It was hard because I started off with A=1 and C=1, so I thought that Y = A*C+B. But when I started using higher numbers, that didnít work. Then I kept B=2 for a while, and played around with A and C. Then I thought that Y = A+B+C-2. But as soon as I changed B, that didnít work any more either.

Finally I started to get some real information. No matter what A and B were, if I changed C by one, Y always changed by one. So I knew that C was an addend all by itself (no multiplying to it).

I have to go now, so Iíll explain later how I came up with the full Y = A*B + C Ė 2 part.


  Posted by nikki on 2004-08-18 17:06:17
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