Some three years ago, Problematics ran a desert-crossing puzzle in which vehicles travelling in a desert without fuel stations transferred their fuel to one another. This allowed one of them to travel the maximum possible distance, while the rest could still retain enough fuel to return home.
I remember there were a few wrong answers, for the puzzle was slightly on the tougher side. The following variation, which comes from Henry Ernest Dudeney, is somewhat simpler, but still requires some work when viewed in isolation. All said, it should be a fun solve.
#Puzzle 192.1
A certain model of car is so designed that it can carry only 2 litres in its fuel tank, which allows it to travel 40 km @ 20 km per litre. Agreed, no real cars can have such a low range, but here at Problematics we welcome absurdist fiction. Outside of the tank, however, the car has enough space to accommodate an additional 18 litres in petrol cans, which means another 360 km. The problem is, our motorist is travelling in a desert with no fuel stations on the journey. That means he can travel only 200 km using 10 litres and use the remaining 10 litres to remain home.
The motorist gets eight assistants to increase the length of his journey. Each car carries 2 litres in its tank plus the additional 18 km in cans. The arrangement works like this: after going a certain distance, one assistant transfers his remaining fuel among the other cars, keeping just enough in his tank to return home. After going a further distance, a second assistant follows the same principle, and returns home after retaining exactly the right amount of fuel for the return leg. And so on. This goes on until our original motorist is the last one remaining. He completes a certain distance and then returns home using whatever fuel he has remaining.
{{/usCountry}}The motorist gets eight assistants to increase the length of his journey. Each car carries 2 litres in its tank plus the additional 18 km in cans. The arrangement works like this: after going a certain distance, one assistant transfers his remaining fuel among the other cars, keeping just enough in his tank to return home. After going a further distance, a second assistant follows the same principle, and returns home after retaining exactly the right amount of fuel for the return leg. And so on. This goes on until our original motorist is the last one remaining. He completes a certain distance and then returns home using whatever fuel he has remaining.
{{/usCountry}}What is the maximum distance that our motorist can cover this way?
#Puzzle 192.2
A magician lays out 20 matchsticks on the table and invites a member from the audience to the stage. Turning his back, he gives the following instructions:
(1) Take any number of matchsticks less than 10 from the table and put them into your pocket.
(2) Since you took a single-digit number away from 20, the matches remaining on the table are a two-digit number. Now form that two-digit number on the table using as many of the remaining matches as you need. For example, if there are 12 matches, form 12 with 1 match for the tens place and 2 matches for the units place.
(3) Put these matches representing the two-digit, number, too, into your pocket.
(4) From the remaining matches, take any number you need and hold them in your fist.
The magician now turns around, counts the matches remaining at this stage, and announces: “The number of matches in your fist is ___.”
This one is very simple, of course, and some of the steps are superfluous. But how is the answer worked out?
MAILBOX: LAST WEEK’S SOLVERS
#Puzzle 191.1
Hi Kabir,
The contestants and their other details are given in tabular form, in increasing order of age.
In the mathematical puzzle (#191.2), the values of (a, b, c, x, y, z) are (2, 3, 11, 13, 21, 22). This is a great puzzle, which I solved by hit and trial, but I request you to also share the formula for this.
— Shishir Gupta, Indore
To address Shishir Gupta’s question, the following formula gives you any set of 12 numbers satisfying the given conditions:
(m – 11)ⁿ + (m – 6)ⁿ + (m – 5)ⁿ + (m + 5)ⁿ + (m + 6)ⁿ + (m + 11)ⁿ
= (m – 10)ⁿ + (m – 9)ⁿ + (m – 1)ⁿ + (m + 10)ⁿ + (m + 9)ⁿ + (m + 1)ⁿ
where m is any integer, and n = 1, 2, 3, 4 or 5. The following solution from Dr Sunita Gupta shows us why the formula holds good when m = 12 and n = 2.
#Puzzle 191.2
Hello Kabir,
This is an example of the Prouhet–Tarry–Escott problem, in which two different sets of integers gives the same sums for like powers, Here,
1 + 6 + 7 + 17 + 18 + 23 = 72 = a + b + c + x + y + z
1² + 6² + 7² + 17² + 18² + 23² = 1228 = a² + b² + c² + x² + y² + z²
From the first equation, the average of the six numbers is 72/6 = 12. The method for finding the values of the variables requires us to express the second equation in terms of this average. In other words,
(12 – 11)² + (12 – 6)² + (12 – 5)² + (12 + 5)² + (12 + 6)² + (12 + 11)² = 1228
Say p = 11, q = 6, r = 5. Then the equation becomes
(12 – p)² + (12 – q)² + (12 – r)² + (12 + r)² + (12 + q)² + (12 + p)² = 1228
Or, (12 – p)² + (12 + p)² + (12 – q)² + (12 + q)² + (12 – r)² + (12 + r)² = 1228
Or, [(12 – p)² + (12 + p)²] + [(12 – q)² + (12 + q)²] + [(12 – r)² + (12 + r)²] = 1228
Or, 2(12² + p²) + 2(12² + q²) + 2(12² + r²) = 1228
Simplifying, p² + q² + r² = 182
Our assigned values (p = 11, q = 6, r = 5) will obviously satisfy the above, i.e. 11² + 6² + 5² = 182. If we find another set of integers such that p² + q² + r² = 182, these too will satisfy the initial conditions. The only other such set is (10, 9, 1), i.e. 10² + 9² + 1² = 182. Using these values for p, q, r,
(12 – p)² + (12 – q)² + (12 – r)² + (12 + r)² + (12 + q)² + (12 + p)² = 1228
=> (12 – 10)² + (12 – 9)² + (12 – 1)² + (12 + 10)² + (12 + 9)² + (12 + 1)² = 1228
Or, 2² + 3² + 11² + 22² + 21² + 13² = 1228
Therefore the integers (a, b, c, x, y, z) are (2, 3, 11, 13, 21, 22).
1 + 6 + 7 + 17 + 18 + 23 = 72 = 2 + 3 + 11 + 13 + 21 + 22
1² + 6² + 7² + 17² + 18² + 23² = 1228 = 2² + 3² + 11² + 13² + 21² + 22²
These equations hold good even when the powers are raised to 3, 4 or 5.
— Dr Sunita Gupta, New Delhi
Solved both puzzles: Shishir Gupta (Indore), Dr Sunita Gupta (Delhi), Vinod Mahajan (Delhi), Sabornee Jana (Mumbai), Yadvendra Somra (Sonipat), Professor Anshul Kumar (Delhi)
Solved #Puzzle 191.1: Kanwarjit Singh (Chief Commissioner of Income-tax, retired), Anil Khanna (Ghaziabad)
Solved #Puzzle 191.2: YK Munjal (Delhi), Ajay Ashok (Delhi)