nyuuzyou/wikis huggingface dataset:
Henry Ernest Dudeney/Modern Puzzles/213 - The Six Pennies/Solution

== by : $213$ ==
;The Six Pennies
== Solution ==
First lay out the pennies as follows, noting that they have been numbered in order to identify them.
:
Move penny $6$ to touch $4$ and $5$:
:
Move penny $5$ to touch $2$ and $3$:
:
Finally, move penny $3$ to touch $5$ and $6$ as indicated by the dotted circle in the diagram above.
== Also see ==
* by : $329$ - The Four Pennies
== Sources ==
* : Solutions: $213$. -- The Six Pennies
* : Answers: $383$. The Six Pennies
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/214 - Folding Postage Stamps

== by : $214$ ==
;Folding Postage Stamps
:Take a $4 \times 2$ sheet of $8$ postage stamps, labelled $1$ to $8$, as shown in the diagram.
:It is an interesting exercise to count how many ways they may be folded up so they will all lie under the one stamp, as shown.
:
:There are in fact $40$ ways to do this so that No. $1$ is always on the top.
:Numbers $5$, $2$, $7$ and $4$ will always be face down.
:You can always arrange for any stamp except No. $6$ to lie next to $1$,
::although there are only two ways each in which $7$ and $8$ can be made to lie in that position.
:They can be folded in the order $1$, $5$, $6$, $4$, $8$, $7$, $3$, $2$ and also $1$, $3$, $7$, $5$, $6$, $8$, $4$, $2$, with $1$ at the top face upwards,
::but it is a puzzle to work out how.
:Can you fold them like that without tearing any of the perforations?
=== Click here for solution ===
== Sources ==
* : Unclassified Problems: $214$. -- Folding Postage Stamps
* : Geometrical Problems: Paper Folding Puzzles: $367$. Folding Postage Stamps
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/214 - Folding Postage Stamps/Solution

== by : $214$ ==
;Folding Postage Stamps
== Solution ==
;To get the order $1$, $5$, $6$, $4$, $8$, $7$, $3$, $2$
Fold $7$ over $6$.
Lay $4$ flat on $8$.
Tuck them both in betwen $7$ and $6$ so these four are in order $7 \, 8 \, 4 \, 6$.
Bring $5$ and $1$ under $6$, and it is done.
;To get the order $1$, $3$, $7$, $5$, $6$, $8$, $4$, $2$
Fold so that $5 \, 6 \, 7 \, 8$ only are visible with faces uppermost.
Fold $5$ on $6$.
Between $1$ and $5$ you have to tuck $7$ and $8$, so $7$ lies on top of $5$ and $8$ ebnds under $6$.
The stamps will now be in the required order.
== Sources ==
* : Solutions: $214$. -- Folding Postage Stamps
* : Answers: $367$. Folding Postage Stamps
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/215 - An Ingenious Match Puzzle

== by : $215$ ==
;An Ingenious Match Puzzle
:Place $6$ matches as shown, and then shift just one match without touching the others so that the new arrangement shall represent a fraction equal to $1$.
:
:The match forming the horizontal fraction bar must not be the one moved.
=== Click here for solution ===
== Sources ==
* : Unclassified Problems: $215$. -- An Ingenious Match Puzzle
* : Match Puzzles: $498$. An Ingenious Match Puzzle
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/215 - An Ingenious Match Puzzle/Solution

== by : $215$ ==
;An Ingenious Match Puzzle
== Solution ==
The dotted line is where the match was moved from, and the red line is where it was moved to.
:
That is, $\dfrac 1 {\sqrt 1}$ which is of course $1$.
== Sources ==
* : Solutions: $215$. -- An Ingenious Match Puzzle
* : Answers: $498$. An Ingenious Match Puzzle
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/216 - Fifty-Seven to Nothing

== by : $216$ ==
;Fifty-Seven to Nothing
:Place $6$ matches as shown, so as to represent the number $57$ in Roman numerals.
:
:Remove and replace any $2$ of them (without disturbing the others) to make an expression representing zero ($0$).
:There are two completely different solutions.
's originally posted problem used cigarettes instead of matches.
=== Click here for solution $1$ ===
=== Click here for solution $2$ ===
== Sources ==
* : Unclassified Problems: $216$. -- Fifty-Seven to Nothing
* : Match Puzzles: $499$. Fifty-Seven to Nothing
Category:Henry Ernest Dudeney/Modern Puzzles
Category:Henry Ernest Dudeney/Modern Puzzles/216 - Fifty-Seven to Nothing
Henry Ernest Dudeney/Modern Puzzles/216 - Fifty-Seven to Nothing/Solution 1

== by : $216$ ==
;Fifty-Seven to Nothing
== Solution ==
The dotted lines are where the matches were moved from, and the red lines are where they were moved to.
:
That is: $\sqrt 1 - 1$ which is of course $0$.
== Sources ==
* : Solutions: $216$. -- Fifty-Seven to Nothing
* : Answers: $499$. Fifty-Seven to Nothing
Category:Henry Ernest Dudeney/Modern Puzzles/216 - Fifty-Seven to Nothing
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/216 - Fifty-Seven to Nothing/Solution 2

== by : $216$ ==
;Fifty-Seven to Nothing
== Solution ==
The dotted lines are where the matches were moved from, and the red lines are where they were moved to.
:
That is: $\text {NIL}$.
== Sources ==
* : Solutions: $216$. -- Fifty-Seven to Nothing
* : Answers: $499$. Fifty-Seven to Nothing
Category:Henry Ernest Dudeney/Modern Puzzles/216 - Fifty-Seven to Nothing
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/217 - The Five Squares

== by : $217$ ==
;The Five Squares
:$12$ matches are arranged to form $4$ squares.
:
:Rearrange the same matches (so they lie flat on the table) to make $5$ squares.
:All the squares must be empty.
=== Click here for solution ===
== Sources ==
* : Unclassified Problems: $217$. -- The Five Squares
* : Match Puzzles: $500$. The Five Squares
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/217 - The Five Squares/Solution

== by : $217$ ==
;The Five Squares
== Solution ==
Keep the same basic arrangement, but make it so that the $4$ matches that meet at the centre have their non-head ends together.
Then you can move them so their ends look like this when you zoom in really closely:
:
See, there is a square there, but it's really, really tiny.
== Sources ==
* : Solutions: $217$. -- The Five Squares
* : Answers: $500$. The Five Squares
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/218 - A Square with Four Pennies

== by : $218$ ==
;A Square with Four Pennies
:Can you place four (old) pennies together so as to show a square?
:They must all lie flat on the table.
=== Click here for solution ===
== Sources ==
* : Unclassified Problems: $218$. -- A Square with Four Pennies
* : Unclassified Puzzles: $514$. A Square with Four Pennies
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/218 - A Square with Four Pennies/Solution

== by : $218$ ==
;A Square with Four Pennies
== Solution ==
Heh! Bazinga.
:
The sides of the square are formed from the straight lines over the date.
== Sources ==
* : Solutions: $218$. -- A Square with Four Pennies
* : Answers: $514$. A Square with Four Pennies
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/219 - A Calendar Puzzle

== by : $219$ ==
;A Calendar Puzzle
:I have stated in my book, , that, under our present calendar rules,
::the first day of a century cannot fall on a Sunday or a Wednesday or a Friday.
:I am frequently asked the reason why.
:Try to explain the mystery in as simple a way as possible.
:Note that $1$st January $1901$ was the first day of the $20$th century, not $1900$.
=== Click here for solution ===
== Sources ==
* : Unclassified Problems: $219$. -- A Calendar Puzzle
* : Unclassified Puzzles: $516$. A Calendar Puzzle
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/219 - A Calendar Puzzle/Solution

== by : $219$ ==
;A Calendar Puzzle
== Solution ==
The Gregorian calendar is such that the pattern of leap years repeats every $400$ years.
So $400$ years contains:
:$400 \times 365$ days
plus:
:an extra $4 \times 25$ days for the leap years in each century
minus:
:$3$ days for the years divisible by $100$ and not $400$ which are not actually in fact leap years.
Thus there are $400 \times 365 + 4 \times 25 - 3 = 146 \, 097$ days in $400$ years.
We have that $146 \, 097 = 7 \times 20 \, 871$.
Hence the first day of the century in a given block of $400$ years is always the same.
We can look up the day on which $1$st January $2001$ falls, and see it is Monday.
Now, between $2001$ and $2100$, $2101$ and $2200$, and $2201$ and $2300$ there are $100 \times 365 + 24$ days, that is, $36 \, 524$ days.
This is congruent to $5$ modulo $7$.
So subsequent first days of the century after $2001$ are:
:$2101$: Saturday ($5$ days after Monday)
:$2201$: Thursday ($5$ days after Saturday)
:$2301$: Tuesday ($5$ days after Thursday)
and then, as we have seen, $2401$ starts on the same day as $2001$.
So the centuries begin on Monday, Saturday, Thursday, Tuesday, Monday, ... and so on.
That is, the century can never begin on Wednesday, Friday or Sunday.
== Sources ==
* : Solutions: $219$. -- A Calendar Puzzle
* : Answers: $516$. A Calendar Puzzle
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/21 - Their Ages

== by : $21$ ==
;Their Ages
:If you add the square of Tom's age to the age of Mary,
::the sum is $62$;
:but if you add the square of Mary's age to the age of Tom,
::the result is $176$.
:Can you say what are the ages of Tom and Mary?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Age and Kinship Puzzles: $21$. -- Their Ages
* : Arithmetical and Algebraical Problems: Age Puzzles: $32$. Their Ages
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/21 - Their Ages/Solution

== by : $21$ ==
;Their Ages
== Solution ==
Tom is $7$ and Mary is $13$.
== Proof ==
Let $T$ be the age of Tom and $M$ be the age of Mary.
Expressing one variable in terms of another and eliminating one of them results in a quartic:
Good luck with that.
So, let us assume that both $T$ and $M$ are natural numbers.
It is apparent that $T < 8$ and $M < 14$, otherwise the totals will be such that either $T$ or $M$ will be negative.
So:
:$176 - M^2 < 8$
which leads us to:
:$M^2 = 169$
giving us:
:$M = 13$
and:
:$T = 176 - 169 = 7$
We check the other equation:
:$7^2 + 13 = 49 + 13 = 62$
and all is well.
== Sources ==
* : Solutions: $21$. -- Their Ages
* : Answers: $32$. Their Ages
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/220 - The Fly's Tour

== by : $220$ ==
;The Fly's Tour
:I had a ribbon of paper, divided into squares on each side.
:I joined the ends together to make a ring, and tossed it down onto the table.
:Then I watched a fly land on the ring and walk in a line over every one of those squares on both sides,
::returning to the point where it started, without ever passing over the edge of the paper.
:Its course passed through the centre of the squares all the time.
:How was this possible?
=== Click here for solution ===
== Sources ==
* : Unclassified Problems: $220$. -- The Fly's Tour
* : Unclassified Puzzles: $517$. The Fly's Tour
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/220 - The Fly's Tour/Solution

== by : $220$ ==
;The Fly's Tour
== Solution ==
's answer is:
:Before you join the ends give one end of the ribbon a half-turn, so that there is a twist in the ring.
:Then the fly can walk over all the squares without going over the edge,
::for we have the curious paradox of a piece of paper with only one side and one edge!
 has of course described a Möbius strip.
The author of this page is not convinced by his exposition, because at least part of the paper will end up flat on the table, and the fly may have trouble negotiating that bit.
== Sources ==
* : Solutions: $220$. -- The Fly's Tour
* : Answers: $517$. The Fly's Tour
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/221 - A Musical Enigma

== by : $221$ ==
;A Musical Enigma
:Here is an old musical enigma that has been pretty well known in Germany for some years.
:
=== Click here for solution ===
== Sources ==
* : Unclassified Problems: $221$. -- A Musical Enigma
* : Unclassified Puzzles: $518$. A Musical Enigma
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/221 - A Musical Enigma/Solution

== by : $221$ ==
;A Musical Enigma
== Solution ==
:The solution is $\text {B A C H}$.
:If you turn the cross round, you get successively:
::$\text B \flat$ (treble clef)
::$\text A$ (tenor clef)
::$\text C$ (alto clef)
::$\text B \natural$ (treble clef).
:In the German tradition, $\text B \flat$ (that is, $\text B$ flat) is called $\text B$, and $\text B \natural$ (that is, $\text B$ natural) is called $\text H$,
::making it read $\text {B A C H}$.
 then goes on to discuss an organ fugue by , which begins as follows:
:
but fails to call to mind the last work of his father , who included that exact motif in the final (uncompleted) movement of his .
This sequence of four notes is known as the .
== Sources ==
* : Solutions: $221$. -- A Musical Enigma
* : Answers: $518$. A Musical Enigma
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/222 - A Mechanical Paradox

== by : $222$ ==
;A Mechanical Paradox
:A remarkable mechanical paradox, invented by about the year $1751$, ought to be known by everyone, but, unfortunately, it is not.
:It was contrived by him as a challenge to a sceptical watchmaker during a metaphysical controversy.
:"Suppose," said, "I make one wheel as thick as three others and cut teeth in them all,
::and then put the three wheels all loose upon one axis and set the thick wheel to turn them,
::so that its teeth may take into those of the three thin ones.
:Now, if I turn the thick wheel round, how must it turn the others?"
:The watchmaker replied that it was obvious that all three must be turned the contrary way.
:Then produced his simple machine, which anyone can make in a few hours,
::showing that, turning the thick wheel which way you would,
::one of the thin wheels revolved in the same way, the second the contrary way, and the third remained stationary.
::Although the watchmaker took the machine away for careful examination, he failed to detect the cause of the strange paradox.
=== Click here for solution ===
== Also see ==
* Ferguson's Paradox
== Sources ==
* : Unclassified Problems: $222$. -- A Mechanical Paradox
* : Geometrical Problems: Circle Puzzles: $296$. A Mechanical Paradox
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/222 - A Mechanical Paradox/Solution

== by : $222$ ==
;A Mechanical Paradox
:A remarkable mechanical paradox, invented by about the year $1751$, ought to be known by everyone, but, unfortunately, it is not.
:It was contrived by him as a challenge to a sceptical watchmaker during a metaphysical controversy.
:"Suppose," said, "I make one wheel as thick as three others and cut teeth in them all,
::and then put the three wheels all loose upon one axis and set the thick wheel to turn them,
::so that its teeth may take into those of the three thin ones.
:Now, if I turn the thick wheel round, how must it turn the others?"
:The watchmaker replied that it was obvious that all three must be turned the contrary way.
:Then produced his simple machine, which anyone can make in a few hours,
::showing that, turning the thick wheel which way you would,
::one of the thin wheels revolved in the same way, the second the contrary way, and the third remained stationary.
::Although the watchmaker took the machine away for careful examination, he failed to detect the cause of the strange paradox.
== Solution ==
:The machine shown in our diagram consist of two pieces of thin wood, $B$ and $C$, made into a frame by being joined at the corners.
:This frame, by means of the handle $n$, may be turned round an axle $a$, which pierces the frame and is fixed in a stationary board or table, $A$,
::and carries within the frame an immovable wheel.
:
:This first wheel, $D$, when the frame revolves, turns a second and thick wheel, $E$,
::which, like the remaining three wheels, $F$, $G$ and $H$, moves freely on its axis.
:The thin wheels, $F$, $G$ and $H$, are driven by the wheel $E$ in such a manner that when the frame revolves
::$H$ turns the same way as $E$ does,
::$G$ turns the contrary way,
::and $F$ remains stationary.
:The secret lies in the fact that though the wheels may be all of the same diameter,
::and $D$, $E$ and $F$ may ($D$ and $F$ must) have an equal number of teeth,
:yet $G$ must have at least one tooth fewer, and $H$ at least one tooth more, than $D$.
== Sources ==
* : Solutions: $222$. -- A Mechanical Paradox
* : Answers: $296$. A Mechanical Paradox
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/22 - Mrs. Wilson's Family

== by : $22$ ==
;Mrs. Wilson's Family
:Mrs. Wilson had three children, Edgar, James and John.
:Their combined ages were half of hers.
:Five years later, during which time Ethel was born, Mrs. Wilson's age equalled the total of all her children's ages.
:Ten years more have passed, Daisy appearing during that interval.
:At the latter event Edgar was as old as John and Ethel together.
:The combined ages of all the children are now double Mrs. Wilson's age, which is, in fact, only equal to that of Edgar and James together.
:Edgar's age also equals that of the two daughters.
:Can you find all their ages?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Age and Kinship Puzzles: $22$. -- Mrs. Wilson's Family
* : Arithmetical and Algebraical Problems: Age Puzzles: $33$. Mrs. Wilson's Family
Category:Henry Ernest Dudeney/Modern Puzzles
Category:Henry Ernest Dudeney/Modern Puzzles/22 - Mrs. Wilson's Family
Henry Ernest Dudeney/Modern Puzzles/22 - Mrs. Wilson's Family/Solution

== by : $22$ ==
;Mrs. Wilson's Family
== Solution ==
:Edgar is $21$
:James and John are both $18$ (they are twins)
:Ethel is $12$
:Daisy is $9$
:Mrs. Wilson herself is $39$.
== Proof ==
The wording is not completely clear, but it is assumed that "now" is the same time as "ten years more have passed".
So, let us define the following variables:
:Let $a$, $b$, $c$, $d$ and $e$ be the current ages of Edgar, James, John, Daisy and Ethel.
:Let $M$ be the current age of Mrs. Wilson.
:Let $x$ be the number of years before now that Daisy appeared.
We can set up a set of equations encoding the statements in the question as follows:
We set up this system of linear simultaneous equations in matrix form as:
:$\begin {pmatrix}
2 & 2 & 2 & 0 & 0 & -1 \\
1 & 1 & 1 & 0 & 1 & -1 \\
1 & 0 & -1 & 1 & -1 & 0 \\
1 & 1 & 1 & 1 & 1 & -2 \\
1 & 1 & 0 & 0 & 0 & -1 \\
1 & 0 & 0 & -1 & -1 & 0 \\
\end {pmatrix} \begin {pmatrix} a \\ b \\ c \\ d \\ e \\ M \end {pmatrix} = \begin {pmatrix} 75 \\ 30 \\ 0 \\ 0 \\ 0 \\ 0 \end {pmatrix}$
It remains to solve this matrix equation.
Rearranging into a more convenient order for conversion into reduced echelon form:
:$\begin {pmatrix}
1 & 0 & -1 & 1 & -1 & 0 \\
1 & 1 & 1 & 1 & 1 & -2 \\
1 & 1 & 0 & 0 & 0 & -1 \\
1 & 0 & 0 & -1 & -1 & 0 \\
1 & 1 & 1 & 0 & 1 & -1 \\
2 & 2 & 2 & 0 & 0 & -1 \\
\end {pmatrix} \begin {pmatrix} a \\ b \\ c \\ d \\ e \\ M \end {pmatrix} = \begin {pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 30 \\ 75 \\ \end {pmatrix}$
In reduced echelon form, this gives:
from which the ages can be read off directly.
== Sources ==
* : Solutions: $22$. -- Mrs. Wilson's Family
* : Answers: $33$. Mrs. Wilson's Family
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Category:Henry Ernest Dudeney/Modern Puzzles/22 - Mrs. Wilson's Family
Henry Ernest Dudeney/Modern Puzzles/22 - Mrs. Wilson's Family/Working

== Working for by : $22$ -- Mrs. Wilson's Family ==
The simultaneous equations in matrix form:
:$\begin {pmatrix}
1 & 0 & -1 & 1 & -1 & 0 \\
1 & 1 & 1 & 1 & 1 & -2 \\
1 & 1 & 0 & 0 & 0 & -1 \\
1 & 0 & 0 & -1 & -1 & 0 \\
1 & 1 & 1 & 0 & 1 & -1 \\
2 & 2 & 2 & 0 & 0 & -1 \\
\end {pmatrix} \begin {pmatrix} a \\ b \\ c \\ d \\ e \\ M \end {pmatrix} = \begin {pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 30 \\ 75 \\ \end {pmatrix}$ 
when converted to reduced echelon form, gives:
:$\begin {pmatrix} 
1 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 \\ \end {pmatrix} \begin {pmatrix} a \\ b \\ c \\ d \\ e \\ M \end {pmatrix} = \begin {pmatrix} 21 \\ 18 \\ 18 \\ 9 \\ 12 \\ 39 \\ \end {pmatrix}$ 
== Proof ==
Category:Henry Ernest Dudeney/Modern Puzzles/22 - Mrs. Wilson's Family
Henry Ernest Dudeney/Modern Puzzles/23 - De Morgan and Another

== by : $23$ ==
;De Morgan and Another
:, the mathematician, who died in $1871$, used to boast that he was $x$ years old in the year $x^2$.
:My living friend, Jasper Jenkins, wishing to improve on this, tells me he was $a^2 + b^2$ in $a^4 + b^4$;
:that he was $2 m$ in the year $2 m^2$;
:and that he was $3 n$ years old in the year $3 n^4$.
:Can you give the years in which and Jenkins were respectively born?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Age and Kinship Puzzles: $23$. -- De Morgan and Another
* : Arithmetical and Algebraical Problems: Age Puzzles: $34$. De Morgan and Another
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/23 - De Morgan and Another/Solution

== by : $23$ ==
;De Morgan and Another
== Solution ==
 was born in $1806$.
Jasper Jenkins was born in $1860$.
== Proof ==
Note that was writing this in the $1920$s.
The square numbers around the $18$th and $19$th century are:
:$42^2 = 1764$
:$43^2 = 1849$
:$44^2 = 1936$
of which only $1849 = 43^2$ can plausibly fit the parameters for .
Hence it is deduced that was born on $1849 - 43 = 1806$.
When we inspect his page, we see that indeed he was born on $18$th June $1806$.
As for Jasper, we need to inspect square numbers around the $900$ to $1000$ region:
:$2 \times 30^2 = 2 \times 900 = 1800$
:$2 \times 31^2 = 2 \times 961 = 1922$
:$2 \times 32^2 = 2 \times 1024 = 2048$
Clearly Jasper was $2 \times 31 = 62$ in $1922$.
Hence it appears Jasper was born in $1860$.
We check the $4$th powers over the range $600$ to $700$ and find:
:$3 \times 5^3 = 3 \times 625 = 1875$
which corroborates the above: Jasper was $3 \times 5 = 15$ in $1875$.
Continuing to explore the $4$th powers, we have this list:
:$1^4 = 1$
:$2^4 = 16$
:$3^4 = 81$
:$4^4 = 256$
:$5^4 = 625$
:$6^4 = 1296$
:$7^4 = 2401$
and we have gone high enough.
Inspecting these numbers, we have that:
:$625 + 1296 = 1921$
at which time Jasper was $5^2 + 6^2 = 25 + 36 = 61$.
== Sources ==
* : Solutions: $23$. -- De Morgan and Another
* : Answers: $34$. De Morgan and Another
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/24 - Simple Arithmetic

== by : $24$ ==
;"Simple" Arithmetic
:Two gentlemen with an eccentric approach to philosophy were pinned down by your investigative reporter. 
:They wished to riddle my mathematical understanding.
:"Our two ages combined," said the first, "is $44$."
:"Don't be silly," said the other, "it's $1280$."
:They looked at me and said, "You see, we didn't tell you how we were combining them."
:It was clear to me that the first number was their difference and the second was their product.
:Now, how old were these two gentlemen?
Note that the above is a contemporary paraphrase of the original statement of this problem, which used language and concepts which are likely to cause offence in today's society.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Age and Kinship Puzzles: $24$. -- "Simple" Arithmetic
* : Arithmetical and Algebraical Problems: Age Puzzles: $35$. "Simple" Arithmetic
Category:Henry Ernest Dudeney/Modern Puzzles