# geometric progression

• Dec 1st 2010, 10:11 AM
killykilly
geometric progression
Hi ! I'm trying to brush up on my algebra for an exam next week. I stumbled on this one.

"A rich man called his seven sons. He had with him a number of pebbles, each pebble representing a gold bar. To his first son, he gave half of the pebbles that he initially had and one pebble more. To his second son, he gave half of the remaining pebbles and one pebble more. He did the same to each of his five other sons and then found out that he had one pebble left. How many pebbles were there initially ?"

Book says the correct answer is 382. I tried solving it using geometric progression but I keep getting with a decimal, which of course is incorrect. Could someone please give me a hint on how to solve this one ? Many thanks !
• Dec 1st 2010, 11:15 AM
Quote:

Originally Posted by killykilly
Hi ! I'm trying to brush up on my algebra for an exam next week. I stumbled on this one.

"A rich man called his seven sons. He had with him a number of pebbles, each pebble representing a gold bar. To his first son, he gave half of the pebbles that he initially had and one pebble more. To his second son, he gave half of the remaining pebbles and one pebble more. He did the same to each of his five other sons and then found out that he had one pebble left. How many pebbles were there initially ?"

Book says the correct answer is 382. I tried solving it using geometric progression but I keep getting with a decimal, which of course is incorrect. Could someone please give me a hint on how to solve this one ? Many thanks !

Here's a logical way to look at this...

If 1 remained, then 2 remained before the extra one was given to the 7th son.
2 is half of 4, so the 7th son got 2+1=3.

Therefore 4 remained after the extra 1 was given to the 6th son,
so half of the "then" remainder was 5, so there were 10 before giving 5+1 to 6th son.

Following that train of logic back...

5th son got 12 as 22 remained before giving him 11+1.

4th son got 24 as 46 remained before giving him 23+1.

3rd son got 48 as 94 remained before giving him 47+1.

2nd son got 96 as 190 remained before giving him 85+1.

1st son got 192 as 382 were available before giving him 191+1.
• Dec 1st 2010, 11:16 AM
emakarov
This problem has to be solved backwards. Suppose the man had x pebbles before he gave some to the last son. Then x - x/2 - 1 = 1. Then repeat this, replacing 1 in the right-hand side with the value of x you found.

In fact, it is easy to show that the answer is $x_7$ where the sequence $x_n$ is determined by a recurrence relation: $x_0 = 1$ and $x_{n+1}=2(x_n+1)$. Here $x_n$ is the number of pebbles the man had before dealing with son number 7 - n + 1 (so he had $x_7$ before dealing with son #1, $x_6$ before son #2, etc. and $x_1$ before son #7).