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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(2): Another question: answered! | Comment 9 of 11 |
(In reply to re: Another question: answered? by nikki)

So I was wrong before about the balance increase from year to year being [A*B + C Ė 2]*B^(n-1). Itís actually [A*(B-1)+C]*B^(n-1).

Ok so the way that I got C had to be an addend all by itself is still true (no matter what A and B were, when I changed C by one, Y always changed by one).

Then I left A and C alone, and played with B. As I played with B, Y always changed by a constant amount. If I changed A to another fixed number and played with B again, Y still changed by a constant amount, but now it was a different amount. If I changed C to another fixed number and played with B again, Y changed by the same constant amount as before. Then I started paying attention to what that amount was Ė it was always equal to A. So if I changed B by one, Y would change by A. This told me that there was an A*B addend in there somewhere.

Now I left B and C alone, and just played with A. Similar to playing with B, if I changed B, Y always changed by a constant amount. If I changed B to another fixed number and played with A again, Y still changed by a constant amount, but now it was a different amount. If I changed C to another fixed number and played with B again, Y changed by the same constant amount as before. Then I started paying attention to what that amount was. It took me a minute to make sure, but it turns out the constant amount was always equal to B-1. So if I changed A by one, Y would change by (B-1). So this told me there was an A*(B-1) addend in there somewhere.

At first I just put all the addends together to see if I got a good equation. Y = C + A*B + A*(B-1). Turns out that wasnít right. Then I said "duh!" because the A*(B-1) addend already has the A*B portion that I needed. So I simplified it to Y = A*(B-1) + C.

My mistake was because I hadnít played around enough with other numbers. It turned out I was playing with A=2 when I came up with the other formula. Y = A*(B-1)+C = A*B-A+C = A*B+C-A, and when A = 2 THEN Y = A*B+C-2 would be true. So that was my mistake.


  Posted by nikki on 2004-08-19 08:14:24
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