nyuuzyou/wikis huggingface dataset:
Henry Ernest Dudeney/Puzzles and Curious Problems/153 - A Military Puzzle/Solution

== 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?
== Solution ==
Place the men along the sides of a regular dodecahedron with the officer in its centre.
You will find that the men at the vertices of this dodecahedron will be in two rows of $11$ men at the same time.
Hence $120$ men are needed for this arrangement.
== Sources ==
* : Solutions: $153$. -- A Military Puzzle
* : Answers: $266$. A Military Puzzle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/154 - Marching an Army

== by : $154$ ==
;Marching an Army
:A body of soldiers was marching in regular column, with $5$ men more in depth than in front.
:When the enemy came in sight the front was increased by $845$ men,
:and the whole was drawn up in $5$ lines.
:How many men were there in all?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $154$. -- Marching an Army
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $225$. Marching an Army
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/154 - Marching an Army/Solution

== by : $154$ ==
;Marching an Army
:A body of soldiers was marching in regular column, with $5$ men more in depth than in front.
:When the enemy came in sight the front was increased by $845$ men,
:and the whole was drawn up in $5$ lines.
:How many men were there in all?
== Solution ==
There were $4550$ men in the army.
== Proof ==
Let $m$ be the total number of men.
Let $n$ be the number of men in front while marching.
We have:
== Sources ==
* : Solutions: $154$. -- Marching an Army
* : Answers: $225$. Marching an Army
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/155 - The Orchard Problem

== by : $155$ ==
;The Orchard Problem
:A market gardener was planting a new orchard.
:The young trees were arranged in rows so as to form a square,
::and it was found that there were $146$ trees unplanted.
:To enlarge the square by an extra row each way he had to buy $31$ additional trees.
:How many trees were there in the orchard when it was finished?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $155$. -- The Orchard Problem
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $228$. The Orchard Problem
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/155 - The Orchard Problem/Solution

== by : $155$ ==
;The Orchard Problem
:A market gardener was planting a new orchard.
:The young trees were arranged in rows so as to form a square,
::and it was found that there were $146$ trees unplanted.
:To enlarge the square by an extra row each way he had to buy $31$ additional trees.
:How many trees were there in the orchard when it was finished?
== Solution ==
:$7921$ trees.
== Proof ==
Let $n$ be the number of trees on each side of the square of already planted trees.
Then there are $n^2$ trees in that square.
From the Odd Number Theorem, to make a square of $\paren {n + 1}^2$ trees we need $2 n + 1$ more trees.
We are told we need $146 + 31 = 177$ more trees.
Then we have that:
:$177 = 2 \times 88 + 1$
and so:
:$n = 88$
and the total number of trees in the new orchard is $\paren {88 + 1}^2 = 89^2 = 7921$.
== Sources ==
* : Solutions: $155$. -- The Orchard Problem
* : Answers: $228$. The Orchard Problem
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/156 - Multiplying the Nine Digits

== by : $156$ ==
;Multiplying the Nine Digits
:They were discussing mental problems at the Crackham's breakfast-table,
::when George suddenly asked his sister Dora to multiply as quickly as possible:
::$1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9 \times 0$''
:How long would it have taken the reader?
It is suspected that mental problems means in this context problems in mental arithmetic rather than anything related to illness.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $156$. -- Multiplying the Nine Digits
* : Arithmetical and Algebraical Problems: Digital Puzzles: $141$. Multiplying the Nine Digits
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/156 - Multiplying the Nine Digits/Solution

== by : $156$ ==
;Multiplying the Nine Digits
== Solution ==
As long as it would take me to notice there is a $0$ amongst the multiplicands.
Wake up at the back.
== Sources ==
* : Solutions: $156$. -- Multiplying the Nine Digits
* : Answers: $141$. Multiplying the Nine Digits
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/157 - Counting the Matches

== by : $157$ ==
;Counting the Matches
:A friend bought a box of midget matches, each one inch in length.
:He found he could arrange them all in the form of a triangle whose area was just as many square inches as there were matches.
:He then used up $6$ of the matches,
::and found that with the remainder he could again construct a triangle whose area was just as many square inches as there were matches.
:And using another $6$ matches he could again do precisely the same.
:How many matches were there in the box originally?
:The number is less than $40$.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $157$. -- Counting the Matches
* : Match Puzzles: $510$. Counting the Matches
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/157 - Counting the Matches/Solution

== by : $157$ ==
;Counting the Matches
== Solution ==
First we recall Heron's Formula for the area of a triangle with sides equal to $a$, $b$ and $c$:
:$\AA = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$
where $s = \dfrac {a + b + c} 2$ is the semiperimeter of $\triangle ABC$.
With $36$ matches you can make a triangle with sides $17$, $10$, $9$ whose area $\AA$ is calculated by Heron's Formula to be:
With $30$ matches you can make a triangle with sides $13$, $12$, $5$, which is the $5$-$12$-$13$ Pythagorean Triangle whose area $\AA$ is calculated by Area of Triangle in Terms of Side and Altitude to be:
With $24$ matches you can make a triangle with sides $10$, $8$, $6$, which is the $3$-$4$-$5$ Pythagorean Triangle scaled up by a factor of $2$, whose area $\AA$ is calculated by Area of Triangle in Terms of Side and Altitude to be:
== Sources ==
* : Solutions: $157$. -- Counting the Matches
* : Answers: $510$. Counting the Matches
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/158 - Newsboys

== by : $158$ ==
;Newsboys
:A contest took place amongst some newspaper boys.
:Tom Smith sold one paper more than a quarter of the whole lot they had secured;
::Billy Jones disposed of one paper more than a quarter of the remainder;
::Ned Smith sold one paper more than a quarter of what was left;
::and Charlie Jones disposed of just one paper more than a quarter of the remainder.
:At this stage it was found that the Smiths were exactly $100$ papers ahead,
:but little Jimmy Jones, the youngest kid of the bunch, sold all that were left,
::so that in this friendly encounter the Joneses won by how many papers do you think?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $158$. -- Newsboys
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/158 - Newsboys
Henry Ernest Dudeney/Puzzles and Curious Problems/158 - Newsboys/Solution

== by : $158$ ==
;Newsboys
== Solution ==
The Jones team won, by $620$ papers to $400$.
== Proof ==
Let $n$ be the total number of papers sold
Let $S_T$, $J_B$, $S_N$, $J_C$ and $J_J$ be the number of papers sold by Tom Smith, Billy Jones, Ned Smith, Charlie Jones and Jimmy Jones respectively.
We have:
These can be expressed more usefully as:
This set of simultaneous linear equations can be expressed conveniently in matrix form as:
:$\begin {pmatrix}
-1 & 0 & 0 & 0 & 0 & 4 \\
-1 & 0 & 0 & 0 & 4 & 1 \\
-1 & 0 & 0 & 4 & 1 & 1 \\
-1 & 0 & 4 & 1 & 1 & 1 \\
-1 & 1 & 1 & 1 & 1 & 1 \\
 0 & 0 & -1 & 1 & -1 & 1 \\
\end {pmatrix} \begin {pmatrix} n \\ J_J \\ J_C \\ S_N \\ J_B \\ S_T \end {pmatrix} = \begin {pmatrix} 4 \\ 4 \\ 4 \\ 4 \\ 0 \\ 100 \end {pmatrix}$
It remains to solve this matrix equation.
In reduced echelon form, this gives:
The result can be read off directly.
There were $1020$ newspapers.
The Smith brothers sold $256$ and $144$ between them, making $400$.
The two older Jones brothers sold $192$ and $108$ between them, making $300$.
But then Jimmy Jones sold the remaining $320$ and the Jones team won by a margin of $220$.
== Sources ==
* : Solutions: $158$. -- Newsboys
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/158 - Newsboys
Henry Ernest Dudeney/Puzzles and Curious Problems/158 - Newsboys/Working

== Working for by : $158$ -- Newsboys ==
The simultaneous equations in matrix form:
:$\begin {pmatrix}
-1 & 0 & 0 & 0 & 0 & 4 \\
-1 & 0 & 0 & 0 & 4 & 1 \\
-1 & 0 & 0 & 4 & 1 & 1 \\
-1 & 0 & 4 & 1 & 1 & 1 \\
-1 & 1 & 1 & 1 & 1 & 1 \\
 0 & 0 & -1 & 1 & -1 & 1 \\
\end {pmatrix} \begin {pmatrix} n \\ J_J \\ J_C \\ S_N \\ J_B \\ S_T \end {pmatrix} = \begin {pmatrix} 4 \\ 4 \\ 4 \\ 4 \\ 0 \\ 100 \end {pmatrix}$
when converted to reduced echelon form, gives:
:$\begin {pmatrix} 
 1 & 0 & 0 & 0 & 0 & 0 \\
 0 & 1 & 0 & 0 & 0 & 0 \\
 0 & 0 & 1 & 0 & 0 & 0 \\
 0 & 0 & 0 & 1 & 0 & 0 \\
 0 & 0 & 0 & 0 & 1 & 0 \\
 0 & 0 & 0 & 0 & 0 & 1 \\
\end {pmatrix} \begin {pmatrix} n \\ J_J \\ J_C \\ S_N \\ J_B \\ S_T \end {pmatrix} = \begin {pmatrix} 1020 \\ 320 \\ 108 \\ 144 \\ 192 \\ 256 \end {pmatrix}$
== Proof ==
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/158 - Newsboys
Henry Ernest Dudeney/Puzzles and Curious Problems/159 - The Year 1927

== by : $159$ ==
;The Year $1927$
:Can you find values for $p$ and $q$ such that $p^q - q^p = 1927$?
:To make it perfectly clear, we will give an example for the year $1844$, where $p = 3$ and $q = 7$:
:::$3^7 - 7^3 = 1844$
:Can you express $1927$ in the same curious way?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $159$. -- The Year $1927$
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $226$. The Year $1927$
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/159 - The Year 1927/Solution

== by : $159$ ==
;The Year $1927$
== Solution ==
:$2^{11} - 11^2 = 2048 - 121 = 1927$
== Sources ==
* : Solutions: $159$. -- The Year $1927$
* : Answers: $226$. The Year $1927$
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/15 - Buying Turkeys

== by : $15$ ==
;Buying Turkeys
:A man bought a number of turkeys at a cost of $\pounds 60$,
::and after reserving fifteen of the birds he sold the remainder for $\pounds 54$,
::thus gaining $2 \shillings$ a head by these.
:How may turkeys did he buy?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Money Puzzles: $15$. -- Buying Turkeys
* : Arithmetical and Algebraical Problems: Money Puzzles: $21$. Buying Turkeys
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/15 - Buying Turkeys/Solution

== by : $15$ ==
;Buying Turkeys
== Solution ==
$75$ turkeys for $16 \shillings$ each.
He sold $60$ turkeys for $18 \shillings$ each.
== Proof ==
Let $n$ be the number of turkeys bought.
Let $t_1 \shillings$ be the price he paid per turkey for the original $n$ turkeys.
Let $t_2 \shillings$ be the price he sold the remaining $n - 15$ turkeys for.
All monetary values will be expressed in shillings.
We have:
Clearly only the positive solution makes sense.
== Sources ==
* : Solutions: $15$. -- Buying Turkeys
* : Answers: $21$. Buying Turkeys
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/160 - Boxes of Cordite

== by : $160$ ==
;Boxes of Cordite
:Cordite charges for $6$-inch howitzers were served out from ammunition dumps in boxes of $15$, $18$ and $20$.
:"Why the three different sizes of boxes?" I asked the officer on the dump.
:He answered: "So that we can give any battery the number of charges it needs without breaking a box.
:This was an excellent system for the delivery of a large number of boxes,
::but failed in small cases, like $5$, $10$, $25$ and $61$.
:Now, what is the biggest number of charges that cannot be served out in whole boxes of $15$, $18$ and $20$?
:It is not a very large number.
=== Click here for solution ===
== Historical Note ==
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $160$. -- Boxes of Cordite
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $227$. Boxes of Cordite
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/160 - Boxes of Cordite/Historical Note

== Historical Note on by : $160$ - Boxes of Cordite ==
 reports that this puzzle is the work of one W. H.-J., but it is not apparent who that person is.
When the collection was edited for its $1968$ re-publication, the presentation of these initials was changed to W. H. J.
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $160$. -- Boxes of Cordite
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $227$. Boxes of Cordite
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/160 - Boxes of Cordite/Solution

== by : $160$ ==
;Boxes of Cordite
:Cordite charges for $6$-inch howitzers were served out from ammunition dumps in boxes of $15$, $18$ and $20$.
:"Why the three different sizes of boxes?" I asked the officer on the dump.
:He answered: "So that we can give any battery the number of charges it needs without breaking a box.
:This was an excellent system for the delivery of a large number of boxes,
::but failed in small cases, like $5$, $10$, $25$ and $61$.
:Now, what is the biggest number of charges that cannot be served out in whole boxes of $15$, $18$ and $20$?
:It is not a very large number.
== Solution ==
:$97$
== Proof ==
The dump officer gives boxes of $18$ until the remainder is a multiple of $5$.
Then, unless this is $5$, $10$ or $25$, the remainder is given in $15$s or $20$s.
The biggest number for which this breaks down is $72 + 25 = 97$.
Take the case of a higher numbers, such as $133$.
$6$ boxes of $18$ makes $108$, leaving $25$.
So you give one box of $18$, leaving $115$, which can be delivered in one box of $15$ and five of $20$.
But in the case of $97$, $72$ is the first and only case leaving a multiple of $5$, that is, $25$.
== Sources ==
* : Solutions: $160$. -- Boxes of Cordite
* : Answers: $227$. Boxes of Cordite
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/161 - Blocks and Squares

== by : $161$ ==
;Blocks and Squares
:Here is a curious but not easy puzzle that appeared, we believe, some ten years ago, though the author is not traced.
:Three children each possess a box containing similar cubic blocks, the same number of blocks in every box.
::
:The first girl was able, using all her blocks, to make a hollow square, as indicated by $A$.
:The second girl made a still larger square, as $B$.
:The third girl made a still larger square, as $C$ but had four blocks left over for the corners, as shown.
:What is the smallest number of blocks that each box could have contained?
:The diagram must not be taken to truly represent the proportion of the various squares.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $161$. -- Blocks and Squares
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $229$. Blocks and Squares
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/161 - Blocks and Squares/Solution

== by : $161$ ==
;Blocks and Squares
== Solution ==
The smallest number of blocks in each box appears to be $1344$.
The innermost hollow square is $34^2$.
Square $A$ is $50^2$.
Square $B$ is $62^2$.
Square $C$ is $72^2$, with four blocks left over at the corners.
We have:
:$50^2 - 34^2 = 2500 - 1156 = 1344$
:$62^2 - 50^2 = 3844 - 2500 = 1344$
:$72^2 - 62^2 = 5184 - 3844 = 1344 - 4$
== Proof ==
== Sources ==
* : Solutions: $161$. -- Blocks and Squares
* : Answers: $229$. Blocks and Squares
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/162 - Find the Triangle

== by : $162$ ==
;Find the Triangle
:The sides and height of a triangle are four consecutive whole numbers.
:What is the area of the triangle?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $162$. -- Find the Triangle
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $230$. Find the Triangle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Category:Heronian Triangles
Henry Ernest Dudeney/Puzzles and Curious Problems/163 - Domino Fractions

== by : $163$ ==
;Domino Fractions
:Here is a new puzzle with dominoes.
:Taking an ordinary box, discard all doubles and blanks.
:Then, substituting figures for the pips, regard the remaining $15$ dominoes as fractions.
:Arrange these $15$ dominoes in $3$ rows of $5$ dominoes so that each row adds up to $10$.
:You are allowed to use as many improper fractions as you like.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $163$. -- Domino Fractions
* : Domino Puzzles: $480$. Domino Fractions
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Category:Dominoes
Henry Ernest Dudeney/Puzzles and Curious Problems/163 - Domino Fractions/Solution

== by : $163$ ==
;Domino Fractions
== Solution ==
:
This gives us the equations:
== Sources ==
* : Solutions: $163$. -- Domino Fractions
* : Answers: $480$. Domino Fractions
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/164 - Cow, Goat and Goose

== by : $164$ ==
;Cow, Goat and Goose
:A farmer found
::that his cow and goat would eat all the grass in a certain field in $45$ days,
::that the cow and the goose would eat it in $60$ days,
::but that it would take the goat and the goose $90$ days to eat it down.
:Now, if he had turned cow, goat and goose into the field together, how long would it have taken them to eat all the grass?
: showed us how to solve a puzzle of this kind with the grass growing all the time;
::but, for the sake of greater simplicity, we will assume that the season and conditions were such that the grass was not growing.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $164$. -- Cow, Goat and Goose
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $231$. Cow, Goat and Goose
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Category:Cistern Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/164 - Cow, Goat and Goose/Solution

== by : $164$ ==
;Cow, Goat and Goose
== Solution ==
:$40$ days.
== Proof ==
Let $t$ be the number of days it takes for them all to eat all the grass.
Let $a, b, c$ be the eating rate in numbers of fields per day of (respectively) cow, goat and goose.
In $t$ days, the various contributions of each of the animals to the eating process is $a t$, $b t$ and $c t$ respectively.
So for the total contribution to be $1$ field, we have:
We have:
and so:
So the field will be eaten bare in $40$ days.
== Sources ==
* : Solutions: $164$. -- Cow, Goat and Goose
* : Answers: $231$. Cow, Goat and Goose
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/165 - The Postage-Stamps Puzzle

== by : $165$ ==
;The Postage-Stamps Puzzle
:A youth who collects postage stamps was asked how many he had in his collection, and he replied:
:"The number, if divided by $2$, will give a remainder $1$;
::divided by $3$, a remainder $2$;
::divided by $4$, a remainder $3$;
::divided by $5$, a remainder $4$;
::divided by $6$, a remainder $5$;
::divided by $7$, a remainder $6$;
::divided by $8$, a remainder $7$;
::divided by $9$, a remainder $8$;
::divided by $10$, a remainder $9$.
:But there are fewer than $3000$."
:Can you tell how many stamps there were in the album?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $165$. -- The Postage-Stamps Puzzle
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $232$. The Postage Stamps Puzzle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/165 - The Postage-Stamps Puzzle/Solution

== by : $165$ ==
;The Postage-Stamps Puzzle
== Solution ==
:$2519$.
== Proof ==
Let $n$ be the number of stamps in the collection.
We have that $n + 1$ is divisible by $2$, $3$, $4$, $5$, $6$, $7$, $8$, $9$ and $10$.
The smallest such $n + 1$ is $\lcm \set {2, 3, 4, 5, 6, 7, 8, 9, 10}$ which is $8 \times 9 \times 5 \times 7 = 2520$.
No other number smaller than $3000$ fits the conditions.
Hence $n = 2519$.
== Sources ==
* : Solutions: $165$. -- The Postage-Stamps Puzzle
* : Answers: $232$. The Postage Stamps Puzzle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/166 - Hens and Tens

== by : $166$ ==
;Hens and Tens
:If ten hen-pens cost ten and tenpence (that is, $10 \shillings 10 \oldpence$),
::and ten hens and one hen-pen cost ten and tenpence,
::what will ten hens without any hen-pens cost?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $166$. -- Hens and Tens
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/166 - Hens and Tens/Solution

== by : $166$ ==
;Hens and Tens
== Solution ==
:$9 \shillings 9 \oldpence$
== Proof ==
Let $h$ pence be the cost of one hen.
Let $p$ pence be the cost of one hen-pen.
We have that:
:$10 \shillings 10 \oldpence = 130 \oldpence$
Hence:
Converting from pence into shillings and pence is all that remains to be done.
== Sources ==
* : Solutions: $166$. -- Hens and Tens
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions