Puzzling as a pastime peaked in the 19th and early 20th centuries, with many of the greatest in the field emerging during this period. We have discussed the Englishman Henry Ernest Dudeney’s puzzles several times in Problematics, and his American counterpart Sam Loyd less frequently. The two were briefly collaborators but their relationship soured when Dudeney alleged that Loyd had presented the Englishman’s puzzles as his own. Loyd was also known to curate puzzles from global sources and pass off some as his own inventions.
That said, some of Loyd’s puzzles were truly original. One such gem includes a battle between two species. I first came across this puzzle several decades ago but had held it back from Problematics so far, because it was difficult to adapt it into a new form. In fact, it is best not to tamper with it in any way. All I have done is change the names of the species. The effort was probably unnecessary, because readers who like solving puzzles will not go to the internet searching for the original version and the solution.
Apart from using different species, I have added some notes to pre-empt any misunderstanding. Otherwise, all the numbers below are exactly as Loyd had specified a century ago.
#Puzzle 190.1
{{/usCountry}}#Puzzle 190.1
{{/usCountry}}Two insect species are at war. The giant wasps are more powerful, while the honeybees rely on expert coordination. A giant wasp will take very little time to finish off one honeybee, or even a pair of them. But when three honeybees get together, their combined strength equals that of one giant wasp. In other words, a fight between 1 giant wasp on one side and 3 honeybees on the other will go on forever, without result.
The real equations begin when 4 honeybees fight against 1 giant wasp. These 4 little ones will take exactly 3 minutes to kill the single giant. If more honeybees join in, the time will decrease proportionately. For example, 5 honeybees will finish off 1 giant wasp in 4/5 x 3 = 12/5 minutes, and 6 honeybees will require 4/6 x 3 = 2 minutes. And so on.
In one epic battle, 4 giant wasps are facing 13 honeybees. We know that 4 giant wasps are equal in strength to 12 honeybees, so that 1 extra member gives the advantage to the honeybees. Eventually, if they manage their resources properly, the honeybees will win.
Managing their resources means, for one, that all 13 honeybees will not assemble against one giant wasp, because they would then be at the mercy of the other two wasps. Nor will they ever go in singles or pairs to attack a giant wasp; they know there must be at least three of them for equality, and more for victory. And if 12 honeybees split into four equal groups and attack the four giant wasps, honeybee #13 will not just sit and watch the endless battles. It would join one of the groups.
After any giant wasp is finished off, the killers will go and join other honeybees fighting, again distributing their resources in the most efficient manner. After one more giant is killed, these honeybees will again split and join other teams. Ans so on.
Another aspect to note: when one side is in the process of killing the other, the resistance of the weaker side reduces with time. Take the basic case of 1 giant wasp vs 4 honeybees. The wasp’s end is due in 3 minutes, so after 1 minute has passed, 1/3 of its resistance is gone. If a fifth honeybee joins the team at this stage, the team will be effectively fighting against a giant wasp at 2/3 of its strength. As we discussed above, 5 honeybees would normally need 12/5 minutes to kill 1 giant wasp. But if the giant wasp is at 2/3 of its strength (you might say it is 1/3 dead and 2/3 alive), the time required will proportionately decrease.
Try working out the entire battle plan. If the honeybees manage their resources in the best possible way, how long will they take to kill all four giant wasps?
#Puzzle 190.2
It’s midnight, and the hour and minute hands sit together on the dial of an old-fashioned. How many more times will the hands meet before midnight tomorrow?
MAILBOX: LAST WEEK’S SOLVERS
#Puzzle 189.1
Before going to the answer, here is a recap of the statements given so that we can refer to them by their numbers:
(1) Black Widow: Scarlet Witch is elder to me
(2) Captain Marvel: I am the eldest among the three of us
(3) Scarlet Witch: The difference between Black Widow's age and mine is 3
(4) Black Widow: Scarlet Witch is not 27
(5) Captain Marvel: All three of us are above 24
(6) Scarlet Witch: If you double Black Widow's age and add 2, you get the sum of Captain Marvel's age and my age
(7) Black Widow: Either Captain Marvel is 25 or Scarlet Witch is 26 or I am 24 — that is to say, exactly one of these three ages is correct
(8) Captain Marvel: Black Widow is the youngest among the three of us
(9) Scarlet Witch: If Black Widow is the youngest among us, then I am not the eldest — but if she is not the youngest, then I am the eldest
Now here is the answer, received by mail as usual:
Since two of the three women are always telling the truth, clearly if two statements by two different women are in agreement, it will mean they are both telling the truth.
Statement (1) by Black Widow says Scarlet Witch is older to her. The same conclusion emerges from Scarlet Witch’s statement (9). Therefore, they are both telling the truth. That means Captain Marvel is the liar.
From statements (3) and (9) by Scarlet Witch, it is clear that she is 3 years older than Black Widow. Statement (6) means that Scarlet Witch and Captain Marvel are collectively 2 years older than twice Black Widow’s age. This means Captain Marvel is a year younger than Black Widow and thus the youngest. In statement (7), the option of Captain Marvel being 25 is wrong as it will make her statement of all being above 24 correct. Black Widow cannot be 24 as it will mean Scarlet Witch is 27, which is wrong given statement (4) by Black Widow. Hence Scarlet Witch is 26, and Black Widow is 23 and Captain Marvel is 22.
— Kanwarjit Singh, Chief Commissioner of Income-tax (retired)
# Puzzle 189.2
Dear Kabir,
The minute hand moves at an angular speed of 360 °/60=6° per minute, and the hour hand at 30°/60 = 0.5° per minute.
Let the hour hand move a° from its 8 o'clock position to the time when both hands make an equal angle from vertical. At 8 o'clock the hour hand was at 120° to the vertical. Therefore, in the time the hour hand moves a°, the minute hand must move (120 – a)° for both to be making the same angle with the vertical.
Equating both the times of movement in minutes, we get
a/0.5 = (120 – a)/6
=> 12a = 120 – a
=> a = 120/13
Time taken by the hour hand to move a° = (120/13)/0.5 = 240/13 minutes.
So, the time is 240/13 minutes past 8 o'clock (or 8:18:27 9/13).
— Yadvendra Somra, Sonipat
Solved both puzzles: Kanwarjit Singh (Chief Commissioner of Income-tax, retired), Yadvendra Somra (Sonipat), Amarpreet (Delhi), Vinod Mahajan (Delhi), Dr Sunita Gupta (Delhi), Sabornee Jana (Mumbai), Ajay Ashok (Delhi), Professor Anshul Kumar (Delhi), Shishir Gupta (Indore)
Solved # Puzzle 189.2: Dr Vivek Jain (Baroda), YK Munjal (Delhi)