Puzzle : Draw The Maximum Sum

Assume you are blindfolded and placed in front of a large bowl containing currency in $50, $20, $10, and $5 denominations. You are allowed to reach in and remove bills, one bill at a time. The drawing stops as soon as you have selected four-of-a-kind— four bills of the same denomination. 

What is the maximum sum of money you could accumulate before the drawing ends?

THIS should be the maximum sum! 

Source

The Maximum That You Can Get!

What was the question?

The quick response to the question by any body would be 3 x 50 + 1 x 20 = $170 would be the maximum as next pick of highest currency of $50 will stop drawing of currencies. But that 5th attempt may not be of $50 & could be of $20,$10 or $5. So this case doesn't count those possibilities. This is worst case scenario where 3 out of 4 picks resulted into accumulation of 3 currencies of same denomination of $50.

So ideally, after drawing three $50 s, three $20 s, three $10 s, three $5 s and finally drawing one anything will stop drawing attempts as that will accumulate 4 bills of same denomination (of $50/$20/$10/$5). The best that can be picked is $50 to get the maximum.

That is total maximum of 50 x 3 + 20 x 3 + 10 x 3 + 5 x 3 + 50 = $305 can be accumulated in 12 attempts & stopping after 13th attempt. And this will be the best case.

Optimize Weighing Balance

You can place weights on both side of weighing balance and you need to measure all weights between 1 and 1000. For example if you have weights 1 and 3,now you can measure 1,3 and 4 like earlier case, and also you can measure 2,by placing 3 on one side and 1 on the side which contain the substance to be weighed.

So question again is how many minimum weights and of what denominations you need to measure all weights from 1kg to 1000kg. 

We require only......Click here to know! 

Source 

Optimisation Of Weighing Balance

What was the task given? 

Just for a moment, let's assume we have to weigh 1 to 30 Kg. Now you can weigh all those with

1,          2,          4,         6,          8,         10..........30

1           2, (2+1), 4, (4+1),6, (6+1), 8,(9+1),10..(29+1),30   .......For middle weights.


Now 6 can be weighed as 2 + 4 & 10 can be weighed as 2 + 8. We can eliminate those. That means we require only

1,        2,         4,          8,         16,

So we need weights of powers of 2.

Now if subtraction is allowed then,we require

1,        3,         6,          9,          12,          15,         18,...............30

For all weights ,

1, (3-1),3, (3+1), (6-1), 6, (6+1), (9-1),9,         12,         15,............30   

But 6 can be weighed as 9-3 , 12 as 9+3, 15 as 27-(9+3). So we can eliminate 6,12,15... This leaves only

1,         3,        9,        27,

In short, we need power of 3 only.      

For the given problem we need to weight 1 to 1000 Kg with subtraction allowed. So the maximum power of 3 that is less than 1000 is 7. To conclude, we require only 7 weights as below.

1,3,9,27,81,243,729