dgreening

I am not sure if this satisfies the requirements about not impacting the eco-system.

very elegant solution(s)! My hat is off to both of you!

I think you are both on the right track.

Hi I have been reading this site for sometime, but only recently started actually posting anything. I enjoy puzzles [crossword, KenKen and logic are my current faves], world travel and good novels audio or visual. Looking forward to more great puzzlers!

The OP does not constrain the boy's arrival to be on the minute. Between 7:58 and 7:59, for example, p decreases linearly from 1 to 0.8; it does not drop abruptly to 0.8 at 7:58. Similarly for the next two minute-intervals. The boy does not arrive later than 8:01. Make a graph of p vs boy's arrival time and find its average value. good point! Now that I re-read the problem I see your point. In fact the boy does not necessarily arrive on the minute and the bus does not necessarily arrive on the minute either. I think your approach is better.

I agree, we were working in parallel.

I asked some others and after talking it over, I realized that we cannot mix m [distance] with m3 of volume. The m/ m3 is actually shorthand for m / m3 of fuel. Dividing through would leave the expression 1/ m2 of fuel. Bottom line: I think this was a trick question. I am going to forget about this one.

Good one

I think DeGe's strategy is sound. The term "toggle" may be adding confusion. Prisoner 1 always turns the light OFF [or leaves it off] Every other prisoner is allowed to turn the light ON only once [after they turn it on once, the leave it unchanged]. Other prisoners are not allowed to ever turn the light off. after day 1 prisoner 1 counts the times he turns it off and after 99 he can safely assume that all have been in the room

not sure what to make of "Well, I think that you have to add units to the numerator (as well as the denominator) to make any sense of things. in your example, 25 miles/gal = 118,400,000/m2=118,400,000 parsecs/parsecs-m2. Now that makes physical sense!"

when I first started mulling this over, I realized that miles/ gal or km/ liter are skipping several steps. We really mean that gal of gas is sufficient to propel this vehicle so many miles. I took a typical value for a US car [25 mpg] and started converting it. Values are approx 25 miles ~ 32 km 1 Gal ~ 3.7 Liters ~ 3.7/1000 Meters 3 25 Mile/ gal = 32 km/ 0.0037 meters 3 ~ 32,000/ 0.0037 /meters 2 therefore 25 miles/ gal ~ 118,400,000 /meter 2 I am at a loss for how this related to the parallel piped, though I cannot find anything else that makes sense.

I think the smallest square would be a 3 x 3. If we start with the most compact area to contain 4 unit squares, it would be a 2 x 2 square. The addition of another unit square will force it to add at least one row or column - thus forcing you to a 2x3 configuration. To meet the criteria of the problem, we must expand that to a 3 x 3 There are multiple configurations [many are just reflections of others # # # # # - - - - or # # # # - - # - - or # - # - # - # - # or ...

We also do not want the Dr to come into contact with the blood from any of the 3 patients One approach

I hit the wrong key 20 = - T1 ( 0.917 mi/ min) + T2 ( 0.667 mi/ min) [Eqn 1] 100 = T1 +T2 [Eqn 2] Multiplying all terms in Eqn 1 by {- 1/0.667} yeilds -30 = 1.375 T1 - T2 Adding this to Eqn2 results in 70 = 2.375 T1 Therefore T1 = 70 /2.375 = 29.47 min, so D1 = 29.47 min * ( 0.917 mi/ min) = 27.02 D2 = D1 + 20 = 47.02 Therefore Total distance = 74.04 Miles

X = miles at 55 mph X+20 = miles driven at 40 mph and 55 mi/ hr = 0.917 mi/ min 40 mi/ hr = 0.667 mi/ min D = Total Distance = D1 (at 55 mph) + D2 (at 40 mph) T = Total Time = T1 (at 55 mph) + D2 (at 40 mph) D1 = T1 ( 0.917 mi/ min) D2 = T2 ( 0.667 mi/ min) Therefore D = D1 + D2 = T1 ( 0.917 mi/ min) + T2 ( 0.667 mi/ min) We also know that D1 +20 = D2 These can be rearranged into 2 simultaneous equations in the form 20

The problem statement says "His truck can only hold 1000 apples." ​the update says that the store is 1000 miles away If the driver starts with a full load [1000 apples] and drives 1000 miles, he will have 0 apples left upon arrival. Thus, unless he can ferry apples to some point(s) in between and then move these forward, there is no way to every deliver even 1 apple. If we assume that there is a point halfway [500 miles] then he could do the following: leave the warehouse with 1000 apples drive 500 miles [now has 500 apples] leave the apples at the halfway point [now has 500 apples] drive back to the ware house leave the warehouse with 1000 apples drive 500 miles [now has 500 apples] Picks up 500 the apples at the halfway point [now has 1000 apples in the truck and 0 apples at the halfway point]] Drives the remaining 500 miles to the store and delivers 500 apples Thus for 2 trips[500 and 1000 miles one way (3000 miles total round trip)] 500 apples can be delivered to deliver 3000 apples would require: an interim storage area at 500 miles 6 "short" round trips between the warehouse and the interim storage area [1000 miles R/T each] 6 "long" round trips from the warehouse [stopping at the interim storage area to reload] to the store [2000 miles R/T each] Notes: you can break this up many ways it is not clear what the optimal interim storage area distance [from the ware house] is [and if more than 1 helps]

jhawk has it right