nyuuzyou/wikis huggingface dataset:
Henry Ernest Dudeney/Modern Puzzles/70 - The Solitary Seven

== by : $70$ ==
;The Solitary Seven
 *7***
 ---------
 ***)********
 ****
 ----
 ***
 ***
 ---
 ****
 ***
 ----
 ****
 ****
 ----
:It is the first example I have seen of one of these missing-figure puzzles in which only one figure is given,
::and there appears to be only one possible solution.
:And, curiously enough, it is not difficult to construct this simple division sum.
:For example, as the divisor when multiplied by $7$ produces only $3$ figures we know that the first figure in the divisor must be $1$.
:We can then prove that the first figure in the dividend must be $1$,
::that, in consequence of bringing down together the last $2$ figures of the dividend,
::the last but one figure in the quotient must be $0$,
::that the first and last figures in the quotient must be greater than $7$,
::because they each produce $4$ figures in the sum,
::and so on.
=== Click here for solution ===
== Historical Note ==
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $70$. -- The Solitary Seven
* : Arithmetical and Algebraical Problems: Skeleton Puzzles: $144$. The Solitary Seven
Category:Henry Ernest Dudeney/Modern Puzzles
Category:Skeleton Puzzles
Henry Ernest Dudeney/Modern Puzzles/70 - The Solitary Seven/Historical Note

== Historical Note on by : $70$ - The Solitary Seven ==
This puzzle was originally submitted to by a person identified merely as the Rev. E. F. O.
 then published it in .
Research is needed to determine which issue.
The uniqueness of the solution was proved at the time by .
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $70$. -- The Solitary Seven
* : Arithmetical and Algebraical Problems: Skeleton Puzzles: $144$. The Solitary Seven
* : Answers: $144$. The Solitary Seven
Category:Historical Notes
Henry Ernest Dudeney/Modern Puzzles/70 - The Solitary Seven/Solution

== by : $70$ ==
;The Solitary Seven
== Solution ==
 97809
 ---------
 124)12128316
 1116
 ----
 968
 868
 ---
 1003
 992
 ----
 1116
 1116
 ----
== Proof ==
Let $d$ denote the divisor.
Let $q$ denote the quotient.
Let $n$ denote the dividend.
Let $n_1$ to $n_4$ denote the partial dividends which are subject to the $1$st to $4$th division operations respectively.
Let $p_1$ to $p_4$ denote the partial products generated by the $1$st to $4$th division operations respectively.
As suggested by :
:$(1): \quad$ Because $7 d$ has $3$ digits, the first digit in $d$ is $1$.
:$(2): \quad$ $p_1$ is a $4$-digit number, which can be no greater than $199 \times 9 = 1791$. Hence the first digit in $n$ is $1$, as is that of $p_1$.
:$(3): \quad$ The last $2$ digits of $n$ come down together, which means the last but one digit of $q$ is $0$.
So far:
 *7*0*
 ---------
 1**)1*******
 1***
 ----
 ***
 ***
 ---
 ****
 ***
 ----
 ****
 ****
 ----
:$(4): \quad$ The first and last digits of $q$ must be greater than $7$, because its product with $d$ has $4$ digits.
:$(5): \quad$ $n_3$, $n_4$ and $p_4$ all have $4$ digits, and so begin with $1$, for the same reason as $(2)$.
:$(6): \quad$ The $3$-digit $p_3$ must begin with $9$, as its difference from the $4$-digit $n_3$ has $2$ digits.
:$(7): \quad$ Because $n_4$ begins with $1$:
::::the second digit of $n_3$ must be $0$.
::::the third digit of $n_3$ must be $0$ or $1$.
::::the second digit of $p_3$ must must be $8$ or $9$.
:$(8): \quad$ $d \le 142$, otherwise $7 d$ would have $4$ digits.
:$(9): \quad$ $d \ge 112$, otherwise $9 d$ would have $3$ digits.
:$(10): \quad$ As $142 \times 9 = 1278$, the second digit of $n$ can be no greater than $2$.
So far:
 *7*0*
 ---------
 1**)1*******
 1***
 ----
 ***
 ***
 ---
 10**
 9**
 ----
 1***
 1***
 ----
As $142 \times 9 = 1278$, the difference $n_3 - p_3$ cannot exceed $12$.
Since $n_3 \ge 1000$, we have:
:$988 \le p_3 \le 999$
We also have $p_3 \ge 7 d$ as $p_3$ must be the largest $3$-digit multiple of $d$,
and that $p_3 < 9 d$ since the latter has $4$ digits.
If $p_3 = 7 d$, then $p_2$ would also start with a $9$.
Then $n_2$, being at least $100$ larger, cannot be a $3$-digit number.
Hence we must have $p_3 = 8 d$.
The only number between $988$ and $999$ that is divisible by $8$ is $992$:
:$992 = 8 \times 124$
Hence we arrive at $q = 97 \, 809$ and we can fill in all other numbers accordingly.
== Historical Note ==
== Sources ==
* : Solutions: $70$. -- The Solitary Seven
* : Answers: $144$. The Solitary Seven
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/71 - A Complete Skeleton

== by : $71$ ==
;A Complete Skeleton
:Here is an arrangement without any figures at all.
:Note the decimal dot in the quotient.
:the extension to four places of decimals makes it curiously easy to solve.
 ****.****
 ---------
 ***)******
 ***
 ---
 ***
 ***
 ---
 ***
 ***
 ---
 ** *
 ** *
 ----
 ****
 ****
 ----
=== Click here for solution ===
== Historical Note ==
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $71$. -- A Complete Skeleton
* : Arithmetical and Algebraical Problems: Skeleton Puzzles: $145$. A Complete Skeleton
Category:Henry Ernest Dudeney/Modern Puzzles
Category:Skeleton Puzzles
Henry Ernest Dudeney/Modern Puzzles/71 - A Complete Skeleton/Historical Note

== Historical Note on by : $71$ - A Complete Skeleton ==
According to , this puzzle was the work of Mr. , whose identity has not been determined.
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $71$. -- A Complete Skeleton
* : Arithmetical and Algebraical Problems: Skeleton Puzzles: $145$. A Complete Skeleton
Category:Historical Notes
Henry Ernest Dudeney/Modern Puzzles/71 - A Complete Skeleton/Solution

== by : $71$ ==
;A Complete Skeleton
== Solution ==
 1011.1008
 ---------
 625)631938
 625
 ---
 693
 625
 ---
 688
 625
 ---
 63 0
 62 5
 ----
 5000
 5000
 ----
== Proof ==
Let $d$ denote the divisor.
Let $q$ denote the quotient.
Let $n$ denote the dividend.
Let $n_1$ to $n_5$ denote the partial dividends which are subect to the $1$st to $5$th division operations respectively.
Let $p_1$ to $p_5$ denote the partial products generated by the $1$st to $5$th division operations respectively.
As suggested, the $3$ zeroes are dropped down into $n_5$ and $p_5$.
Another zero is dropped into $n_4$.
This also gives us the last two but one digits of $q$.
Thus $d$ is a divisor of a multiple of $1000$.
The divisors of $d$ can only be $5$ or $2$ up to a multiplicity of $3$ each, or $x$ where $x < 10$.
Because $d$ has $3$ digits, one divisor must be $5$, and the last digit must therefore be $5$ or $0$.
The subtraction from the final $0$ in $n_4$ giving a digit in the of the difference means it must be $5$.
$d$ cannot have $2$ as a divisor or $d$ will end in $0$.
It follows immediately that $d = 5^4 = 625$, and that all of $p_1$ to $p_4$ are also $625$.
The remaining digits of $q$ are now obviously either $1$ or $0$:
 1011.1008
 ---------
 625)******
 625
 ---
 ***
 625
 ---
 ***
 625
 ---
 ** 0
 62 5
 ----
 5000
 5000
 ----
and $n$ follows by multiplying $d$ by $q$.
== Sources ==
* : Solutions: $71$. -- A Complete Skeleton
* : Answers: $145$. A Complete Skeleton
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/72 - Alphabetical Sums

== by : $72$ ==
;Alphabetical Sums
:There is a family resemblance between puzzles where an arithmetical working has to be reconstructed from a few figures and a number of asterisks,
::and those in which every digit is represented by a letter,
::but they are really quite different.
:The resemblance lies in the similarity of the process of solving.
:Here is a little example of the latter class.
:It can hardly be called difficult.
 RSR
 ------
 PR)MTVVR
 MVR
 ---
 KKV
 KMD
 ---
 MVR
 MVR
 ---
:Can you reconstruct this simple division sum?
:Every digit is represented by a different letter.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $72$. -- Alphabetical Sums
* : Arithmetical and Algebraical Problems: Cryptarithm Puzzles: $151$. Alphabetical Sums
Category:Henry Ernest Dudeney/Modern Puzzles
Category:Cryptarithms
Henry Ernest Dudeney/Modern Puzzles/72 - Alphabetical Sums/Solution

== by : $72$ ==
;Alphabetical Sums
== Solution ==
 565
 ------
 35)19775
 175
 ---
 227
 210
 ---
 175
 175
 ---
== Proof ==
We have immediately that $R \times R = R \pmod {10}$.
However, because $R \times PR = MTV$ it is clear $R \ne 1$ and (obviously) $R \ne 0$.
So $R = 5$ or $R = 6$.
To get $V$ in the fifth line it is clear that $D = 0$.
Thus we have that $K = 2 M$.
In order for $D$ to be $0$:
:if $R = 5$ then $S$ must be even.
:if $R = 6$ then $S = 5$.
Note that $KMD > MVR$, which implies that $S > R$.
Hence it cannot be the case that $R = 6$ and $S = 5$.
And thus we conclude $R = 5$ and $S = 6$ or $8$.
The divisor $PR$ is a factor of $KMD$, which can only be $210, 420, 630$ or $840$.
Therefore $PR$ is a factor of the lowest common multiple of these numbers, $2520$.
Of the factors of $2520$, only $15$, $35$ and $45$ are $2$-digit numbers ending in $5$.
However $15 \times RSR < 15 \times 600 = 9000 < 10 \, 000$, so $PR \ne 15$.
Finally we check the remaining $4$ possibilities:
:$35 \times 565 = 19 \, 775$
:$45 \times 565 = 25 \, 425$
:$35 \times 585 = 20 \, 475$
:$45 \times 585 = 26 \, 325$
and only $19 \, 775$ matches the pattern $MTVVR$, so it must be our dividend.
The rest of the puzzle can be filled in accordingly.
== Sources ==
* : Solutions: $72$. -- Alphabetical Sums
* : Answers: $151$. Alphabetical Sums
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/73 - Alphabetical Arithmetic

== by : $73$ ==
;Alphabetical Arithmetic
:Here is a puzzle that will keep the reader agreeably employed for several minutes.
 F G
 Less A B multiplied by C = D E
 ---
 Leaving H I
 ---
:Each letter stands for a different figure ($1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$ or $9$) and $0$ is not allowed.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $73$. -- Alphabetical Arithmetic
* : Arithmetical and Algebraical Problems: Cryptarithm Puzzles: $152$. Alphabetical Arithmetic
Category:Henry Ernest Dudeney/Modern Puzzles
Category:Cryptarithms
Henry Ernest Dudeney/Modern Puzzles/73 - Alphabetical Arithmetic/Solution

== by : $73$ ==
;Alphabetical Arithmetic
== Solution ==
 9 3
 Less 1 7 multiplied by 4 = 6 8
 ---
 Leaving 2 5
 ---
== Proof ==
Immediately: 
:$F > D$
:$A \le 4$
:$B \ne 1, 5$
:$C \ne 1, 5$
Suppose $A = 4$.
Since $3 \times 40 > 100$, $C = 2$ and $D \ge 8$.
Since $9 \ge F > D$, we must have $D = 8$, leaving $H = 1$.
$E$ is even, so $E = 6$ and $B = 3$.
So far we have:
 9 G
 Less 4 3 multiplied by 2 = 8 6
 ---
 Leaving 1 I
 ---
The two remaining digits are $5$ and $7$, and cannot fit in the positions of $G$ and $I$.
Therefore $A \ne 4$.
Suppose $A = 3$.
Since $4 \times 30 > 100$, $C = 2$ and $D \ge 6$.
Since $F \le 9$, $H \le F - D \le 9 - 6 = 3$.
This forces $H = 1$.
We check the case $B = 4$.
So far we have:
 F G
 Less 3 4 multiplied by 2 = 6 8
 ---
 Leaving 1 I
 ---
The remaining digits are $5, 7$ and $9$.
This forces $F = 7$.
But then $5$ and $9$ cannot fit in the positions of $G$ and $I$.
Thus $B \ne 4$.
For $B > 5$, we have $D = 7$.
$B \ne 6$ since $E \ne 2$.
We check the cases $B = 8, 9$.
For $B = 8$ we have:
 F G
 Less 3 8 multiplied by 2 = 7 6
 ---
 Leaving 1 I
 ---
The remaining digits are $4, 5$ and $9$.
This forces $F = 9$.
But then $4$ and $5$ cannot fit in the positions of $G$ and $I$.
Thus $B \ne 8$.
For $B = 9$ we have:
 F G
 Less 3 9 multiplied by 2 = 7 8
 ---
 Leaving 1 I
 ---
The remaining digits are $4, 5$ and $6$.
None of the digits can fit $F$.
Therefore $B \ne 9$, and thus $A \ne 3$.
Suppose $A = 2$.
We have $C = 3$ or $4$.
If $C = 4$, since $B \ge 3$:
:$D E \ge 23 \times 4 = 92$
But then $F > 9$ is impossible.
Thus $C = 3$.
Similar to the above, we have $H = 1$.
If $B = 9$, $DE = 87$.
But $FG > 87$ is impossible as $9$ is taken.
If $B = 8$, $DE = 84$ which repeats $8$.
If $B = 7$, $DE = 81$ which repeats $1$.
If $B = 4$, $DE = 72$ which repeats $2$.
Thus $B = 6$.
So far we have:
 F G
 Less 2 6 multiplied by 3 = 7 8
 ---
 Leaving 1 I
 ---
The remaining digits are $4, 5$ and $9$.
Thus $F = 9$.
But then $4$ and $5$ cannot fit in the positions of $G$ and $I$.
Therefore $A \ne 2$.
We arrive at $A = 1$.
Then $DE = FG - HI \le 98 - 23 = 75$.
This leaves the following cases for $AB \times C = DE$ to check:
:$13 \times 4 = 52$
:$13 \times 2 = 26$ repeats $2$
:$14 \times 3 = 42$ repeats $4$
:$14 \times 2 = 28$ repeats $2$
:$16 \times 4 = 64$ repeats $4$ and $6$
:$16 \times 3 = 48$
:$16 \times 2 = 32$ repeats $2$
:$17 \times 4 = 68$
:$17 \times 3 = 51$ repeats $1$
:$17 \times 2 = 34$
:$18 \times 4 = 72$
:$18 \times 3 = 54$
:$18 \times 2 = 36$
:$19 \times 4 = 76$
:$19 \times 3 = 57$
:$19 \times 2 = 38$
 
== Sources ==
* : Solutions: $73$. -- Alphabetical Arithmetic
* : Answers: $152$. Alphabetical Arithmetic
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/74 - Queer Division

== by : $74$ ==
;Queer Division
:The following is a rather curious puzzle in which we are restricted to the use of only $2$ digits.
:Find the smallest number which when divided successively by $45$, $454$, $4545$, and $45454$
::leaves the remainders $4$, $45$, $454$, and $4545$ respectively.
:This is perhaps not very easy but it affords a good arithmetical exercise.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $74$. -- Queer Division
* : Arithmetical and Algebraical Problems: Digital Puzzles: $121$. Queer Division
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/74 - Queer Division/Solution

== by : $74$ ==
;Queer Division
== Solution ==
:$35 \, 641 \, 667 \, 749$
== Proof ==
We are to solve the following system of simultaneous linear congruences:
First note that $(3)$ implies $(1)$.
Now we split $(2)$ using the Chinese Remainder Theorem.
We have:
Note that $(5)$ is implied by $(4)$, so we just need to simutaneously solve $(3), (4), (6)$.
As $227, 4545$ and $45454$ are pairwise coprime, the Chinese Remainder Theorem guarantees a unique solution modulo $227 \times 4545 \times 45454 = 46 \, 895 \, 573 \, 610$.
To find this solution, we first need to find, in each modulo, the inverse of the product of the other modulos.
One way to find them is to use the Euclidean Algorithm.
We have:
Next we multiply each inverse with the remainder in that modulo from our equations, and then to the product of the other modulos and sum them up.
This will give us the solution modulo $46 \, 895 \, 573 \, 610$:
which is the smallest positive solution to our congruences.
== Sources ==
* : Solutions: $74$. -- Queer Division
* : Answers: $121$. Queer Division
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/75 - A Teasing Legacy

== by : $75$ ==
;A Teasing Legacy
:Professor Rackbrain left his typist what he called a trifle of a legacy if she was able to claim it.
:The legacy was the largest amount that she could find in an addition sum,
::where pounds, shillings and pence were all represented and no digit used more than once.
:Every digit must be used once, a single $0$ may or may not appear, as in the examples below, and the dash may be employed in the manner shown.
 £ s. d. £ s. d.
 - 3 7 4 2 5
 - 4 8 6 7 3
 - 5 9 --------
 1 6 - £10 9 8
--------
£2 - -
:The young lady was cleverer than he thought.
:What was the largest amount that she could claim?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $75$. -- Queer Division
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/75 - A Teasing Legacy/Solution

== by : $75$ ==
;A Teasing Legacy
== Solution ==
In the words of :
:The best answer, as far as I have been able to discover, is the following:
 £ s. d.
 296 - 5
 8 - 7
 ---------
£304 1 -
== Sources ==
* : Solutions: $75$. -- A Teasing Legacy
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/76 - The Nine Volumes

== by : $76$ ==
;The Nine Volumes
:In a small bookcase were arranged $9$ volumes of some big work,
::numbered from $1$ to $9$ inclusive,
::on $3$ shelves exactly as shown:
 26 5 9
 7
 184 3
:The $9$ digits express money value.
:You will see that they are so arranged that $\pounds 26 \ 5 \shillings \ 9 \oldpence$ multiplied by $7$ will produce $\pounds 184 \ 0 \shillings \ 3 \oldpence$
:Every digit represented once, and yet they form a correct sum in the multiplication of money.
:But in the blank shillings space in the bottom row is a slight defect, and I want to correct it.
:The puzzle is to use the multiplier $3$, instead of $7$, and get a correct result with the $9$ volumes, without any blank space;
::with pounds, shillings and pence all represented in both the top and bottom line.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $76$. -- The Nine Volumes
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/76 - The Nine Volumes/Solution

== by : $76$ ==
;The Nine Volumes
== Solution ==
 offers up $2$ solutions:
 21 8 7
 3
 --------
 64 5 9
 8 5 7
 3
 --------
 24 16 9
== Sources ==
* : Solutions: $76$. -- The Nine Volumes
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/77 - The Ten Volumes

== by : $77$ ==
;The Ten Volumes
:As an extension of the last puzzle, let us introduce a $10$th volume marked $0$.
:If we arrange the $10$ volumes as follows, we get a sum of money correctly multiplied by $2$.
 54 3 9
 2
 108 7 6
:Can you do the same thing with the multiplier $4$ so that the $9$ digits and $0$ are all represented,
::once and once only?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $77$. -- The Ten Volumes
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/78 - The Miller's Toll

== by : $78$ ==
;The Miller's Toll
:Here is a very simple puzzle, yet I have seen people perplexed by it for a few minutes.
:A miller was accustomed to take as toll one-tenth of the flour that he ground for his customers.
:How much did he grind for a man who had just one bushel after the toll had been taken?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $78$. -- The Miller's Toll
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $160$. The Miller's Toll
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/78 - The Miller's Toll/Solution

== by : $78$ ==
;The Miller's Toll
== Solution ==
:$1 \tfrac 1 9$ bushels
== Proof ==
Let $x$ bushels be the total quantity of flour ground.
Then:
:$x - \dfrac x {10} = 1$
from which:
:$x = \dfrac {10} 9$
The result follows.
== Sources ==
* : Solutions: $78$. -- The Miller's Toll
* : Answers: $160$. The Miller's Toll
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/79 - Egg Laying

== by : $79$ ==
;Egg Laying
:The following is a new variation of an old friend.
:Though it looks rather complicated and difficult, it is absurdly easy if properly attacked.
:If a hen and a half lays an egg and a half in a day and a half,
::how many and a half who lay better by half will lay half a score and a half in a week and a half?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $79$. -- Egg Laying
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $161$. Egg Laying
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/79 - Egg Laying/Solution

== by : $79$ ==
;Egg Laying
== Solution ==
One hen.
== Proof ==
Half a score and a half is $10 \tfrac 1 2$.
A week and a half is $10 \tfrac 1 2$ days.
We have that $1 \tfrac 1 2$ hens lay $1 \tfrac 1 2$ eggs in $1 \tfrac 1 2$ days.
Hence $1$ hen lays $1$ egg in $1 \tfrac 1 2$ days.
A hen who lays better by half will lay $1 \tfrac 1 2$ eggs in $1 \tfrac 1 2$ days.
That is, $1$ egg in $1$ day.
Let $n + \dfrac 1 2$ be the number of better-by-half laying hens who lay $10 \tfrac 1 2$ eggs in $10 \tfrac 1 2$ days.
The same number of hens lay $1$ egg in $1$ day.
That is, half a hen and half a hen, or one hen.
== Sources ==
* : Solutions: $79$. -- Egg Laying
* : Answers: $161$. Egg Laying
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/7 - Selling Eggs

== by : $7$ ==
;Selling Eggs
:A woman took a certain number of eggs to market and sold some of them.
:The next day, through the industry of her hens, the number left over had been doubled, and she sold the same number as the previous day.
:On the third day the new remainder was trebled, and she sold the same number as before.
:On the fourth day the remainder was quadrupled, and her sales the same as before.
:On the fifth day what had been left over were quintupled, yet she sold exactly the same as on all the previous occasions, and so disposed of her entire stock.
:What is the smallest number of eggs she could have taken to market the first day, and how many did she sell daily?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Money Puzzles: $7$. -- Selling Eggs
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $163$. Selling Eggs
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/7 - Selling Eggs/Solution

== by : $7$ ==
;Selling Eggs
== Solution ==
She came to market on the first day with $103$ eggs, and sold $60$ every day.
== Proof ==
Let $n$ be the original number of eggs taken to market.
Let $d$ be the number of eggs sold each day.
Let $r_1$, $r_2$, $r_3$ and $r_4$ be the remaining eggs after days $1$, $2$, $3$ and $4$.
On the first day, she sold $d = n - r_1$ eggs.
On the second day, she sold $d = 2 r_1 - r_2$ eggs.
On the third day, she sold $d = 3 r_2 - r_3$ eggs.
On the fourth day, she sold $d = 4 r_3 - r_4$ eggs.
On the fifth day, she sold $d = 5 r_4$ eggs.
So:
The smallest $n$ and $r_1$ that fit this equation are:
So:
:She came to market on the first day with $103$ eggs.
:She sold $60$ eggs per day.
:At the end of the first day she had $43$ eggs remaining.
:On the second day she started with $2 \times 43 = 86$ eggs, sold $60$ and had $26$ remaining.
:On the third day she started with $3 \times 26 = 78$ eggs, sold $60$ and had $18$ remaining.
:On the fourth day she started with $4 \times 18 = 72$ eggs, sold $60$ and had $12$ remaining.
:On the fifth day she started with $5 \times 12 = 60$ eggs and sold them all.
== Sources ==
* : Solutions: $7$. -- Selling Eggs
* : Answers: $163$. Selling Eggs
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/80 - The Flocks of Sheep

== by : $80$ ==
;The Flocks of Sheep
:Four brothers were comparing the number of sheep that they owned.
:It was found that Claude had ten more sheep than Dan.
:If Claude gave a quarter of his sheep to Ben,
::then Claude and Adam would together have the same number as Ben and Dan together.
:If, then, Adam gave one-third to Ben,
::and Ben gave a quarter of what he then held to Claude,
::who then passed on a fifth of his holding to Dan,
::and then Ben divided one-quarter of the number he then possessed equally among Adam, Claude and Dan,
::they would all have an equal number of sheep.
:How many sheep did each possess?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $80$. -- The Flocks of Sheep
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $162$. The Flocks of Sheep
Category:Henry Ernest Dudeney/Modern Puzzles