nyuuzyou/wikis huggingface dataset:
Henry Ernest Dudeney/Puzzles and Curious Problems/141 - The First Boomerang Puzzle/Historical Note

== Historical Note on Henry Ernest Dudeney: Puzzles and Curious Problems $141$ - The First Boomerang Puzzle ==
In the words of :
:One of the most ancient forms of arithmetical puzzle is that which I call the "Boomerang."
:Everybody has been asked at some time or another to "Think of a number,"
::and after going through some process of private calculation, to state the result,
::when the questioner promptly tells you the number you thought of.
:There are hundreds of varieties of the puzzle.
:The oldest recorded example appears to be that given in the Arithmetica by , who died about the year $120$.
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $141$. -- The First "Boomerang" Puzzle
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $213$. The First "Boomerang" Puzzle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/141 - The First Boomerang Puzzle/Solution

== by : $141$ ==
;The First "Boomerang" Puzzle
== Solution ==
Let $n$ be the number that was thought of.
Let $a$ be the remainder after dividing $n$ by $3$.
Let $b$ be the remainder after dividing $n$ by $5$.
Let $c$ be the remainder after dividing $n$ by $7$.
Then:
:multiply $a$ by $70$
:multiply $b$ by $21$
:multiply $c$ by $15$
Add these up.
If you get a number over $100$, subtract $105$, and continue doing so until you get a number between $1$ and $100$.
This is the number $n$ which was first thought of.
== Proof ==
We have:
The result follows.
== Historical Note ==
== Sources ==
* : Solutions: $141$. -- The First "Boomerang" Puzzle
* : Answers: $213$. The First "Boomerang" Puzzle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/142 - Longfellow's Bees

== by : $142$ ==
;Longfellow's Bees
:If one-fifth of a hive of bees flew to the ladambra flower,
::one-third flew to the slandbara,
::three times the difference of these two numbers flew to an arbour,
::and one bee continued to fly about, attracted on each side by the fragrant ketaki and the malati,
::what was the number of bees?
=== Click here for solution ===
== Historical Note ==
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $142$. -- Longfellow's Bees
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $214$. Longfellow's Bees
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/142 - Longfellow's Bees/Historical Note

== Historical Note on Henry Ernest Dudeney: Puzzles and Curious Problems $142$ - Longfellow's Bees ==
In the words of :
:When was Professor of Modern Languages at he was accustomed to amuse himself by giving more or less simple arithmetical puzzles to the students.
:Here is an example:
and indeed so follows the puzzle posed here.
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $142$. -- Longfellow's Bees
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $214$. Longfellow's Bees
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/142 - Longfellow's Bees/Solution

== by : $142$ ==
;Longfellow's Bees
== Solution ==
There were $15$ bees in the hive.
== Proof ==
Let $n$ be the number of bees.
We have:
== Historical Note ==
== Sources ==
* : Solutions: $142$. -- Longfellow's Bees
* : Answers: $214$. Longfellow's Bees
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/143 - Lilivati, A.D. 1150

== by : $143$ ==
;"Lilivati", A.D. $1150$
:Beautiful maiden, with beaming eyes, tell me which is the number that, multiplied by $3$,
::then increased by three-fourths of the product,
::divided by $7$,
::diminished by one-third of the quotient,
::multiplied by itself,
::diminished by $52$,
::the square root found,
::addition of $8$,
::division by $10$,
::gives the number $2$?
=== Click here for solution ===
== Historical Note ==
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $143$. -- "Lilivati", A.D. $1150$
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $215$. Lilivati, $1150$ 
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Category:Bhaskara II Acharya/Lilavati
Henry Ernest Dudeney/Puzzles and Curious Problems/143 - Lilivati, A.D. 1150/Historical Note

== Historical Note on Henry Ernest Dudeney: Puzzles and Curious Problems $143$ - "Lilivati", A.D. $1150$ ==
 quotes this in its entirety as an example of one of the problems in , by , dating from approximately 
In 's words:
:Here is another little morning problem from Lilivati (A.D. $1150$)...
:... This, like so many of those old things, is absurdly easy when properly attacked.
It will be noted that he refers to it as Lilivati, which is most probably incorrect.
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $143$. -- "Lilivati", A.D. $1150$
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $215$. Lilivati, $1150$ 
Category:Historical Notes
Henry Ernest Dudeney/Puzzles and Curious Problems/143 - Lilivati, A.D. 1150/Solution

== by : $143$ ==
;"Lilivati", A.D. $1150$
== Solution ==
:$28$
== Proof ==
We simply reverse the order of operations, calculating as we go:
Hence the result.
== Historical Note ==
== Sources ==
* : Solutions: $143$. -- "Lilivati", A.D. $1150$
* : Answers: $215$. Lilivati, $1150$ 
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/144 - Biblical Arithmetic

== by : $144$ ==
;Biblical Arithmetic
:If you multiply the number of 's sons by the number of times which the compassed on the seventh day,
::and add to the product the number of measures of barley which gave ,
::divide this by the number of 's sons,
::subtract the number of each kind of clean beasts that went into the ,
::multiply by the number of men that went to seek after he was taken to ,
::subtract from this 's age at the time he stood before ,
::add the number of stones in 's bag when he killed ,
::subtract the number of furlongs that was distant from ,
::divide by the number of anchors cast out when was shipwrecked,
::subtract the number of persons saved in the ,
::and the answer will be the number of pupils in a certain Sunday school class.
:How many people in the class?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $144$. -- Biblical Arithmetic
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $216$. Biblical Arithmetic
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/144 - Biblical Arithmetic/Solution

== by : $144$ ==
;Biblical Arithmetic
== Solution ==
There were $7$ in the Sunday school class.
== Proof ==
Let us gird up our armour and proceed.
From Genesis $35 : 23$:
: had $12$ sons while living in Aram-Naharaim.
From Joshua $6 : 15$:
:On the $7$th day, the army got up at daybreak. They marched slowly around the same as they had done for the past $6$ days, except on this day they went around $7$ times.
From Ruth $3 : 15$:
:[] also said, "Bring me the shawl you are wearing and hold it out." When she did so, he poured into it $6$ measures of barley and placed the bundle on her.
From Esther $9 : 10$:
:... the $10$ sons of son of Hammedatha, the enemy of the Jews.
From Genesis $7 : 2$:
: Take with you $7$ pairs of every kind of clean animal, a male and its mate
(Technically this is $14$, but the answer given in the book reveals that $7$ is what is meant.)
From $\text {II}$ Kings $2 : 17$:
:And they sent $50$ men, who searched for $3$ days but did not find him.
From Genesis $41 : 46$:
: was $30$ years old when he entered the service of king of Egypt.
From $\text I$ Samuel $17 : 14$:
:Then [] took his staff in his hand, chose $5$ smooth stones from the stream, put them in the pouch of his shepherd's bag and, with his sling in his hand, approached the Philistine.
From John $11 : 18$:
:Now was nigh unto , about $15$ furlongs off.
From Acts $27 : 29$:
:Then fearing lest we should have fallen upon rocks, they cast $4$ anchors out of the stern, and wished for the day.
From Genesis $7 : 13$:
:In the selfsame day entered Noah, and Shem, and Ham, and Japheth, the sons of Noah, and Noah's wife, and the $3$ wives of his sons with them, into the ark
That makes $8$, then.
It remains to perform the arithmetic.
Hence:
== Sources ==
* : Solutions: $144$. -- Biblical Arithmetic
* : Answers: $216$. Biblical Arithmetic
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/145 - The Printer's Problem

== by : $145$ ==
;The Printer's Problem
:A printer had an order for $10 \, 000$ bill forms per month,
::but each month the name of the particular month had to be altered:
::that is, he printed $10 \, 000$ "JANUARY", $10 \, 000$ "FEBRUARY", $10 \, 000$ "MARCH", etc.;
::but as the particular types with which these words were to be printed had to be specially obtained, and were expensive,
::he only purchased just enough movable types to enable him, by interchanging them,
::to print in turn the whole of the months of the year.
:How many separate types did he purchase?
:Of course, the words were printed throughout in capital letters, as shown.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $145$. -- The Printer's Problem
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $217$. The Printer's Problem
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/145 - The Printer's Problem/Solution

== by : $145$ ==
;The Printer's Problem
== Solution ==
:$\text A, \text A, \text B, \text C, \text D, \text E, \text E, \text E, \text F, \text G, \text H, \text I, \text J, \text L, \text M, \text N, \text O, \text O, \text P, \text R, \text R, \text S, \text T, \text U, \text U, \text V, \text Y$
== Proof ==
You need at least as many of a particular letter as it appears in any given month.
Hence, for example:
:$2 \times \text A$ as it appears twice in JANUARY
:$3 \times \text A$ as it appears $3$ times in both SEPTEMBER and DECEMBER
:$2 \times \text U$ as it appears twice in AUGUST
and so on.
The details are tedious.
== Sources ==
* : Solutions: $145$. -- The Printer's Problem
* : Answers: $217$. The Printer's Problem
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/146 - The Swarm of Bees

== by : $146$ ==
;The Swarm of Bees
:The square root of half the number of bees in a swarm has flown out upon a jessamine bush;
::eight-ninths of the whole swarm as remained behind;
::one female bee flies about a male that is buzzing within the lotus flower into which he was allured in the night by its sweet odour,
::but is now imprisoned in it.
:Tell me the number of bees.
=== Click here for solution ===
== Historical Note ==
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $146$. -- The Swarm of Bees
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $218$. The Swarm of Bees
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Category:Bhaskara II Acharya/Lilavati
Henry Ernest Dudeney/Puzzles and Curious Problems/146 - The Swarm of Bees/Historical Note

== Historical Note on Henry Ernest Dudeney: Puzzles and Curious Problems $146$ - The Swarm of Bees ==
 reports:
:Here is an example of the elegant way in which , in his great work Lilivati, in $1150$, dressed his little puzzles.
Note his continued misspelling of .
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $146$. -- The Swarm of Bees
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $218$. The Swarm of Bees
Category:Historical Notes
Henry Ernest Dudeney/Puzzles and Curious Problems/146 - The Swarm of Bees/Solution

== by : $146$ ==
;The Swarm of Bees
== Solution ==
There were $72$ bees.
== Proof ==
Let $n$ be the number of bees.
We have that:
The result follows as ( notwithstanding) half a bee is not viable.
== Historical Note ==
== Sources ==
* : Solutions: $146$. -- The Swarm of Bees
* : Answers: $218$. The Swarm of Bees
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/147 - Blindness in Bats

== by : $147$ ==
;Blindness in Bats
:A naturalist was investigating (in a tediously long story) whether bats are in fact actually blind.
:He discovered that blindness varied.
:Two of his bats could see out of the right eye,
::just three of them could see out of the left eye,
::four could not see out of the left eye,
::and five could not see out of the right eye.
:He wanted to know the smallest number of bats that he could have examined in order to get these results.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $147$. -- Blindness in Bats
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $219$. Blindness in Bats
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/147 - Blindness in Bats/Solution

== by : $147$ ==
;Blindness in Bats
== Solution ==
The puzzlingly obvious solution is $7$.
One is left wondering whether this is a trick question, but no.
== Proof ==
There must be at least $7$ bats, because they can be partitioned in $2$ ways::
:with or without sight in the right eye
:with or without sight in the left eye.
Both of these lead to $7$ bats.
That these conditions can be fulfilled by $7$ bats, let us suggest:
:$2$ bats can see out of both eyes
:$1$ bat can see out of the left eye but not the right
:$4$ can not see out of either eye
and the conditions are consistent.
== Sources ==
* : Solutions: $147$. -- Blindness in Bats
* : Answers: $219$. Blindness in Bats
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/148 - A Menagerie

== by : $148$ ==
;A Menagerie
:A travelling menagerie contained two freaks of nature -- a four-footed bird and a six-footed calf.
:An attendant was asked how many birds and beasts there were in the show, and he said:
::"Well, there are $36$ heads and $100$ feet altogether.
::You can work it out for yourself."
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $148$. -- A Menagerie
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $220$. A Menagerie
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/148 - A Menagerie/Solution

== by : $148$ ==
;A Menagerie
== Solution ==
There are $12$ beasts and $24$ birds.
== Proof ==
Let $a$ and $b$ be the number of beasts and birds respectively.
If the anomalies had the usual number of legs, there would be $96$ feet altogether.
Hence we have:
== Sources ==
* : Solutions: $148$. -- The Menagerie
* : Answers: $220$. The Menagerie
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/149 - Sheep Stealing

== by : $149$ ==
;Sheep Stealing
:Some sheep stealers made a raid and carried off one-third of the flock of sheep, and one-third of a sheep.
:Another party stole one-fourth of what remained, and one-fourth of a sheep.
:Then a third party of raiders carried off one-fifth of the remainder and three-fifths of a sheep,
::leaving $409$ behind.
:What was the number of sheep in the flock?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $149$. -- Sheep Stealing
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $221$. Sheep Stealing
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/149 - Sheep Stealing/Solution

== by : $149$ ==
;Sheep Stealing
== Solution ==
The flock originally contained $1025$ sheep.
== Proof ==
Let $n$ be the number of sheep in the flock.
Let $n_1$ and $n_2$ be the numbers remaining after the first and second raids respectively.
We have:
== Sources ==
* : Solutions: $149$. -- Sheep Stealing
* : Answers: $221$. Sheep Stealing
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/14 - Horses and Bullocks

== by : $14$ ==
;Horses and Bullocks
:A dealer bought a number of horses at $\pounds 17, 4 \shillings$ each,''
::and a number of bullocks at $\pounds 13, 5 \shillings$ each.
:He then discovered that the horses had cost him in all $33 \shillings$ more than the bullocks.
:Now, what is the smallest number of each that he must have bought?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Money Puzzles: $14$. -- Horses and Bullocks
* : Arithmetical and Algebraical Problems: Money Puzzles: $20$. Horses and Bullocks
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/14 - Horses and Bullocks/Solution

== by : $14$ ==
;Horses and Bullocks
== Solution ==
:$252$ horses and $327$ bullocks.
== Proof ==
We have that:
:$\pounds 17, 4 \shillings = 344 \shillings$
:$\pounds 13, 5 \shillings = 265 \shillings$
Hence we are to find the solution to the Diophantine equation:
:$344 h - 265 b = 33$
To quote :
:This is easy enough if you know how, but we cannot go into the matter here.
If one decides to go into the matter, one finds that a standard method to solve this equation is to use the Euclidean Algorithm on $344$ and $265$:
Now we reverse the equations:
Therefore:
:$33 = \paren {33 \times 104} \times 344 - \paren {33 \times 135} \times 265$
which gives the solutions:
:$h' = 3432, b' = 4455$
but this solution can be reduced.
We can subtract $265$ from $h'$ and $344$ from $b'$ simultaneously to obtain smaller solutions.
We now get:
:$h = 3432 - 12 \times 265 = 252$
:$b = 4455 - 12 \times 344 = 327$
which is minimal, and we check that
Hence the dealer has bought $252$ horses and $327$ bullocks.
== Sources ==
* : Solutions: $14$. -- Horses and Bullocks
* : Answers: $20$. Horses and Bullocks
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/150 - Sheep Sharing

== by : $150$ ==
;Sheep Sharing
:A correspondent (C. H. P.) puts the following little question:
:An Australian farmer dies and leaves his sheep to his three sons.
:Alfred is to get $20$ per cent more than John,
::and $25$ per cent more than Charles.
:John's share is $3600$ sheep.
:How many sheep does Charles get?
:Perhaps readers may like to give this a few moments' consideration.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $150$. -- Sheep Sharing
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $222$. Sheep Sharing
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/150 - Sheep Sharing/Solution

== by : $150$ ==
;Sheep Sharing
== Solution ==
Charles gets $3456$ sheep.
== Proof ==
Let $A$, $J$ and $C$ be the number of sheep each of Alfred, John and Charles got respectively.
We have:
== Sources ==
* : Solutions: $150$. -- Sheep Sharing
* : Answers: $222$. Sheep Sharing
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/151 - The Arithmetical Cabby

== by : $151$ ==
;The Arithmetical Cabby
:The driver of the taxi-cab was wanting in civility, so Mr. Wilkins asked him for his number.
:"You want my number, do you?" said the driver.
:"Well, work it out for yourself.
:If you divide by number by $2$, $3$, $4$, $5$, or $6$ you will find there is always $1$ over;
::but if you divide it by $11$ there ain't no remainder.
:What's more, there's no other driver with a lower number who can say the same."
:What was the fellow's number?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $151$. -- The Arithmetical Cabby
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $223$. The Arithmetical Cabby
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/151 - The Arithmetical Cabby/Solution

== by : $151$ ==
;The Arithmetical Cabby
:The driver of the taxi-cab was wanting in civility, so Mr. Wilkins asked him for his number.
:"You want my number, do you?" said the driver.
:"Well, work it out for yourself.
:If you divide by number by $2$, $3$, $4$, $5$, or $6$ you will find there is always $1$ over;
::but if you divide it by $11$ there ain't no remainder.
:What's more, there's no other driver with a lower number who can say the same."
:What was the fellow's number?
== Solution ==
The cabbie's number was $121$.
== Proof ==
Let $n$ be the driver's number.
We know that $n - 1$ is divisible by $2$, $3$, $4$, $5$ and $6$.
Hence we know that:
:$n - 1 = k \times \lcm \set {2, 3, 4, 5, 6} = 60 k$
where $k$ is an integer.
We see immediately that:
:$k = 0 \implies n = 1$
which is not divisible by $11$
:$k = 1 \implies n = 61$
which is not divisible by $11$
:$k = 2 \implies n = 121$
which is $11 \times 11$ and so is the number we want.
== Sources ==
* : Solutions: $151$. -- The Arithmetical Cabby
* : Answers: $223$. The Arithmetical Cabby
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/152 - The Length of a Lease

== by : $152$ ==
;The Length of a Lease
:A friend's property had a $99$ years' lease.
:When I asked him how much of this had expired, the reply was as follows:
::Two-thirds of the time past was equal to four-fifths of the time to come,
:so I had to work it out for myself.
:My bad for asking intrusive personal questions.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $152$. -- The Length of a Lease
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $224$. The Length of a Lease
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/152 - The Length of a Lease/Solution

== by : $152$ ==
;The Length of a Lease
== Solution ==
$54$ years have elapsed, leaving $45$ years to go.
== Proof ==
Let $t$ be the time elapsed on this lease.
We have:
and we see that:
:$\dfrac 2 3 54 = 36 = \dfrac 4 5 45$
== Sources ==
* : Solutions: $152$. -- The Length of a Lease
* : Answers: $224$. The Length of a Lease
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/153 - A Military Puzzle

== by : $153$ ==
;A Military Puzzle
:An officer wished to form his men into $12$ rows, with $11$ men in every row,
::so that he could place himself at a point that would be equidistant from every row.
::"But there are only one hundred and twenty of us, sir," said one of the men.
:Was it possible to carry out the order?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $153$. -- A Military Puzzle
* : Geometrical Problems: Triangle, Square & Other Polygon Puzzles: $266$. A Military Puzzle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems