nyuuzyou/wikis huggingface dataset:
Henry Ernest Dudeney/Puzzles and Curious Problems/129 - Squares and Cubes

== by : $129$ ==
;Squares and Cubes
:Find two different numbers such that the sum of their squares shall equal a cube, and the sum of their cubes equals a square.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $129$. -- Squares and Cubes
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $202$. Squares and Cubes
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/129 - Squares and Cubes/Solution

== by : $129$ ==
;Squares and Cubes
:Find two different numbers such that the sum of their squares shall equal a cube, and the sum of their cubes equals a square.
== Solution ==
We have:
:$625^2 + 1250^2 = 125^3$
while:
:$625^3 + 1250^3 = 46 \, 875^2$
== Proof ==
We wish to find integral solutions to:
 states that if one number is $625 m^6$ and the other number double the first, you can get any number of solutions of a particular series.
The answer given is for $m 1$.
He probably exploited the fact that $1 + 2^3 = 3^2$.
Therefore for any $k$:
:$k^3 + \paren {2 k}^3 = 9 k^3$
$9 k^3$ would be a square if $k$ is itself a square.
So we write $k = n^2$:
:$\paren {n^2}^3 + \paren {2 n^2}^3 = \paren {3 n^3}^2$
and we see that $(2)$ is satisfied with $x = n^2, y = 2 n^2, w = 3 n^3$.
Substituting this to $(1)$, we have:
:$\paren {n^2}^2 + \paren {2 n^2}^2 = 5 n^4 = z^3$
Since $5 n^4$ is a cube, $n$ must be divisible by $5$.
By writing $n = 5 a$, we still require $5^5 a^4$ to be a cube.
Hence $a$ must again be divisible by $5$.
By writing $a = 5 b$, we only require $5^9 b^4$ to be a cube.
Thus $b$ must itself be a cube.
Finally we write $b = m^3$.
:$k = n^2 = 25 a^2 = 625 b^2 = 625 m^6$
and we see that:
:$\paren {625 m^6}^2 + \paren {1250 m^6}^2 = 5^9 m^{12} = \paren {125 m^4}^3$
which is a cube, so $(1)$ is also satisfied.
Substituting $x = 625 m^6, y = 1250 m^6$ back to $(2)$, we have:
:$\paren {625 m^6}^3 + \paren {1250 m^6}^3 = 2 \, 197 \, 265 \, 625 m^{18} = \paren {46 \, 875 m^9}^2$
and thus we have obtained a general solution.
== Sources ==
* : Solutions: $129$. -- Squares and Cubes
* : Answers: $202$. Squares and Cubes
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/12 - Reductions in Price

== by : $12$ ==
;Reductions in Price
:"I have often been mystified," said Colonel Crackham, "at the startling reductions some people make in their prices,
::and wondered on what principles they went to work.
:For example, a man offered me a motor-car two years ago for $\pounds 512$;
::a year later his price was $\pounds 320$;
::a little while after he asked a level $\pounds 200$;
::and last week he was willing to sell for $\pounds 125$.
:The next time he reduces I shall buy.
:At what price shall I purchase if he makes a consistent reduction?"
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Money Puzzles: $12$. -- Reductions in Price
* : Arithmetical and Algebraical Problems: Money Puzzles: $19$. Reductions in Price
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/12 - Reductions in Price/Solution

== by : $12$ ==
;Reductions in Price
== Solution ==
:$\pounds 78 \tfrac 1 8$, or $\pounds 78, 2 \shillings 6 \oldpence$
== Proof ==
The numbers:
:$512, 320, 200, 125$ form a geometric sequence whose common ratio is $\dfrac 5 8$.
Hence the next reduction is expected to be $125 \times \dfrac 5 8 = \dfrac {625} 8 = 78 \tfrac 1 8$.
== Sources ==
* : Solutions: $12$. -- Reductions in Price
* : Answers: $19$. Reductions in Price
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/130 - Milk and Cream

== by : $130$ ==
;Milk and Cream
:A dairyman found that the milk supplied by his cows was $5$ per cent cream and $95$ per cent skimmed milk.
:He wanted to know how much skimmed milk he must add to a quart of whole milk to reduce the percentage of cream to $4$ per cent.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $130$. -- Milk and Cream
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $203$. Milk and Cream
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/130 - Milk and Cream/Solution

== by : $130$ ==
;Milk and Cream
== Solution ==
Half a pint.
== Proof ==
Let $v_c$ be the volume of cream in a volume $v$ of whole milk.
Let $w$ be the quantity of skimmed milk needed to reduce the volume of cream to $4 \%$.
We have:
As we are told that $v$ is $1$ quart, it follows that the amount of skimmed milk to be added is half a pint.
== Sources ==
* : Solutions: $130$. -- Milk and Cream
* : Answers: $203$. Milk and Cream
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/131 - Feeding the Monkeys

== by : $131$ ==
;Feeding the Monkeys
:A man went to the zoo with a bag of nuts to feed the monkeys.
:He found that if he divided them equally amongst the $11$ monkeys in the first cage he would have $1$ nut over;
::if he divided them equally amongst the $13$ monkeys in the second cage there would be $8$ left;
::if he divided them amongst the $17$ monkeys in the last cage $3$ nuts would remain.
:He also found that if he divided them equally amongst the $41$ monkeys in all $3$ cages,
::or amongst the monkeys in any $2$ cages,
::there would always be some left over.
:What is the smallest number of nuts that the man could have in his bag?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $131$. -- Feeding the Monkeys
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $204$. Feeding the Monkeys
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/131 - Feeding the Monkeys/Solution

== by : $131$ ==
;Feeding the Monkeys
== Solution ==
:$2179$ nuts.
== Proof ==
Let $n$ be the number of nuts he had in his bag.
We have the following congruences:
From $(1)$, we have that:
:$n \in \set {1, 12, 23, 34, 45, 56, \ldots}$
From $(2)$, we have that:
:$n \in \set {8, 21, 34, 47, 60, 73, \ldots}$
Thus:
:$n \equiv 34 \pmod {11 \times 13}$
that is:
:$n \equiv 34 \pmod {143}$
By definition of modulo arithmetic, there is some integer $k$ such that:
so we wish to find $k$ such that:
:$n \equiv 7 k \equiv 3 \pmod {17}$
Since:
:$7 \times 5 = 35 \equiv 1 \pmod {17}$
we have:
so the smallest positive integer $k$ that satisfies the equation is $15$.
In other words:
:$n \equiv 34 + 143 \times 15 \equiv 2179 \pmod {11 \times 13 \times 17}$
We have:
:$11 \times 13 \times 17 = 2431 > 2179$
so there is no smaller positive integer solution to the congruence above.
We also see that $2179$ satisfies all other conditions in the puzzle:
:$2179 = 41 \times 53 + 6$
:$2179$ is odd, so it cannot be a multiple of $24, 28$ or $30$
and hence $2179$ is the smallest possible solution to the puzzle.
== Sources ==
* : Solutions: $131$. -- Feeding the Monkeys
* : Answers: $204$. Feeding the Monkeys
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/132 - Sharing the Apples

== by : $132$ ==
;Sharing the Apples
:If $3$ boys had $169$ apples which they shared in the ratio of one-half, one-third and one-fourth, how many apples did each receive?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $132$. -- Sharing the Apples
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $205$. Sharing the Apples
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/132 - Sharing the Apples/Solution

== by : $132$ ==
;Sharing the Apples
:If $3$ boys had $169$ apples which they shared in the ratio of one-half, one-third and one-fourth, how many apples did each receive?
== Solution ==
The first boy got $78$ apples.
The second boy got $52$ apples.
The third boy got $39$ apples.
== Proof ==
We have that:
:$\dfrac 1 2 + \dfrac 1 3 + \dfrac 1 4 = \dfrac {13} {12}$
We have that
:$169 = \dfrac {13} {12} 156$
Hence:
:the first boy got $\dfrac 1 2 156 = 78$ apples
:the second boy got $\dfrac 1 3 156 = 52$ apples
:the first boy got $\dfrac 1 4 156 = 39$ apples
We note that $78 + 52 + 39 = 169$ and all is well with the world.
== Sources ==
* : Solutions: $132$. -- Sharing the Apples
* : Answers: $205$. Sharing the Apples
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/133 - Sawing and Splitting

== by : $133$ ==
;Sawing and Splitting
:Two men can saw $5$ cords of wood per day,
::or they can split $8$ cords of wood when sawed.
:How many cords must they saw in order that they may be occupied for the rest of the day in splitting it?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $133$. -- Sawing and Splitting
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $206$. Sawing and Splitting
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/133 - Sawing and Splitting/Solution

== by : $133$ ==
;Sawing and Splitting
== Solution ==
:$3 \tfrac 1 {13}$ cords
== Proof ==
Let $n$ be the number of cords they are to work on.
We have that:
:$\dfrac n 5 + \dfrac n 8 = 1$
from which we get:
:$8 n + 5 n = 40$
That is:
:$n = \dfrac {40} {13} = 3 \tfrac 1 {13}$
== Sources ==
* : Solutions: $133$. -- Sawing and Splitting
* : Answers: $206$. Sawing and Splitting
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/134 - The Bag of Nuts

== by : $134$ ==
;The Bag of Nuts
:There are $100$ nuts distributed between $5$ bags.
:In the first and second there are altogether $52$ nuts;
::in the second and third there are $43$;
::in the third and fourth there are $34$;
::in the fourth and fifth, $30$.
:How many nuts are there in each bag?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $134$. -- The Bag of Nuts
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $207$. The Bag of Nuts
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/134 - The Bag of Nuts
Henry Ernest Dudeney/Puzzles and Curious Problems/134 - The Bag of Nuts/Solution

== by : $134$ ==
;The Bag of Nuts
== Solution ==
In the first to fifth bags in order, the number of nuts is:
:$27, 25, 18, 16, 14$
== Proof ==
Let $a$, $b$, $c$, $d$ and $e$ be the number of nuts in each of the first, second, third, fourth and fifth bags respectively.
We have:
We set up this system of linear simultaneous equations in matrix form as:
:$\begin {pmatrix}
 1 & 1 & 1 & 1 & 1 \\
 1 & 1 & 0 & 0 & 0 \\
 0 & 1 & 1 & 0 & 0 \\
 0 & 0 & 1 & 1 & 0 \\
 0 & 0 & 0 & 1 & 1 \\
\end {pmatrix} \begin {pmatrix} a \\ b \\ c \\ d \\ e \end {pmatrix} = \begin {pmatrix} 100 \\ 52 \\ 43 \\ 34 \\ 30 \end {pmatrix}$
It remains to solve this matrix equation.
In reduced echelon form, this gives:
from which the contents of the bags can be read off directly.
== Sources ==
* : Solutions: $134$. -- The Bag of Nuts
* : Answers: $207$. The Bag of Nuts
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/134 - The Bag of Nuts
Henry Ernest Dudeney/Puzzles and Curious Problems/134 - The Bag of Nuts/Working

== Working for by : $134$ -- The Bag of Nuts ==
The simultaneous equations in matrix form:
:$\begin {pmatrix}
 1 & 1 & 1 & 1 & 1 \\
 1 & 1 & 0 & 0 & 0 \\
 0 & 1 & 1 & 0 & 0 \\
 0 & 0 & 1 & 1 & 0 \\
 0 & 0 & 0 & 1 & 1 \\
\end {pmatrix} \begin {pmatrix} a \\ b \\ c \\ d \\ e \end {pmatrix} = \begin {pmatrix} 100 \\ 52 \\ 43 \\ 34 \\ 30 \end {pmatrix}$
when converted to reduced echelon form, gives:
:$\begin {pmatrix} 
 1 & 0 & 0 & 0 & 0 \\
 0 & 1 & 0 & 0 & 0 \\
 0 & 0 & 1 & 0 & 0 \\
 0 & 0 & 0 & 1 & 0 \\
 0 & 0 & 0 & 0 & 1 \end {pmatrix} \begin {pmatrix} a \\ b \\ c \\ d \\ e \end {pmatrix} = \begin {pmatrix} 27 \\ 25 \\ 18 \\ 16 \\ 14 \end {pmatrix}$
== Proof ==
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/134 - The Bag of Nuts
Henry Ernest Dudeney/Puzzles and Curious Problems/135 - Distributing Nuts

== by : $135$ ==
;Distributing Nuts
:Aunt Martha bought some nuts.
:She gave Tommy one nut and a quarter of the remainder;
::Bessie then received one nut and a quarter of what were left;
::Bob, one nut and a quarter of the remainder;
::and, finally, Jessie received one nut and a quarter of the remainder.
:It was then noticed that the boys had received exactly $100$ nuts more than the girls.
:How many nuts had Aunt Martha retained for her own use?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $135$. -- Distributing Nuts
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $208$. Distributing Nuts
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/135 - Distributing Nuts/Solution

== by : $135$ ==
;Distributing Nuts
== Solution ==
Aunt Martha kept $321$ nuts for herself.
== Proof ==
Let $n$ be the total number of nuts.
Let $a$, $b$, $c$, $d$ and $m$ be the number of nuts received or retained by Tommy, Bessie, Bob, Jessie and Aunt Martha respectively.
We have:
Rewriting $(1)$ to $(5)$ in a more convenient form:
At this stage we recognise that $a$, $b$, $c$ and $c$ must be positive integers, which gives us the obvious Ansatz:
We note that:
:$a + c = 100$
:$b + d = 75$
giving a difference between $a + c$ and $b + d$ of $25$.
So we multiply our Ansatz by $4$ to get:
We see that:
giving the appropriate difference.
Thus:
== Sources ==
* : Solutions: $135$. -- Distributing Nuts
* : Answers: $208$. Distributing Nuts
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/136 - Juvenile Highwaymen

== by : $136$ ==
;Juvenile Highwaymen
:Three juvenile highwaymen called upon an apple-woman to "stand and deliver."
:Tom seized half of the apples, but returned $10$ to the basket;
:Ben took one-third of what were left, but returned $2$ that he did not fancy;
:Jim took half of the remainder, but threw back one that was worm-eaten.
:The woman was then left with only $12$ in her basket.
:How many had she before the raid was made?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $136$. -- Juvenile Highwaymen
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $209$. Juvenile Highwaymen
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/136 - Juvenile Highwaymen/Solution

== by : $136$ ==
;Juvenile Highwaymen
== Solution ==
The apple-woman had $40$ apples of which Tom had $10$, Ben had $8$ and Jim had $10$.
== Proof ==
Let $n$ be the number of apples the apple-woman started with.
Let $T$, $B$ and $J$ be the number of apples stolen by Tom, Ben and Jim respectively.
We have:
These can be more conveniently written as:
== Sources ==
* : Solutions: $136$. -- Juvenile Highwaymen
* : Answers: $209$. Juvenile Highwaymen
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/137 - Buying Dog Biscuits

== by : $137$ ==
;Buying Dog Biscuits
:A salesman packs his dog biscuits (all of one quality) in boxes containing $16$, $17$, $23$, $24$, $39$ and $40 \ \mathrm{lbs.}$ (that is, pounds weight),
::and he will not sell them in any other way, or break into a box.
:A customer asked to be supplied with $100 \ \mathrm{lbs.}$ of the biscuits.
:Could you have carried out the order?
:If not, now near could you have got to making up the $100 \ \mathrm{lbs.}$ supply?
:Of course, he has an ample supply of boxes of each size.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $137$. -- Buying Dog Biscuits
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $210$. Buying Dog Biscuits
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/137 - Buying Dog Biscuits/Solution

== by : $137$ ==
;Buying Dog Biscuits
== Solution ==
:$2$ of the $16 \ \mathrm{lbs.}$ boxes and $4$ of the $17 \ \mathrm{lbs.}$ boxes make $100 \ \mathrm{lbs.}$
== Sources ==
* : Solutions: $137$. -- Buying Dog Biscuits
* : Answers: $210$. Buying Dog Biscuits
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/138 - The Three Workmen

== by : $138$ ==
;The Three Workmen
:"Me and Bill," said Casey, "can do the job for you in ten days,
::but give me Alec instead of Bill, and we can get it done in nine days."
:"I can do better than that," said Alec. "Let me take Bill as a partner, and we will do the job for you in eight days."
:Then how long would each man take over the job alone?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $138$. -- The Three Workmen
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $211$. The Three Workmen
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Category:Cistern Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/138 - The Three Workmen/Solution

== by : $138$ ==
;The Three Workmen
== Solution ==
Alfred, Bill and Casey would take respectively $14 \tfrac {34} {49}$ days, $17 \tfrac {23} {41}$ days, and $23 \tfrac 7 {31}$ days.
== Proof ==
Let $a, b, c$ be the rate of working in jobs per day of (respectively) Alec, Bill and Casey.
Let $t_a, t_b, t_c$ be the number of days it would take (respectively) Alec, Bill and Casey to do the job alone.
We have:
and so:
Hence and so.
== Also see ==
* Henry Ernest Dudeney: Modern Puzzles: $84$ - The Three Brothers, exactly the same problem but with different names.
== Sources ==
* : Solutions: $138$. -- The Three Workmen
* : Answers: $211$. The Three Workmen
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/139 - Working Alone

== by : $139$ ==
;Working Alone
:Alfred and Bill together can do a job of work in $24$ days.
:If Alfred can do two-thirds as much as Bill, how long will it take each of them to do the work alone?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $139$. -- Working Alone
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $212$. Working Alone
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Category:Cistern Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/139 - Working Alone/Solution

== by : $139$ ==
;Working Alone
== Solution ==
Alfred would take $60$ days, while Bill would take $40$ days.
== Proof ==
Let $t_a$ and $t_b$ be the time taken for Alfred and Bill to do the job.
Let $a$ and $b$ be the rate in days at which Alfred and Bill can do the job.
In $24$ days, the contributions of Alfred and Bill is $24 a$ and $24 b$ respectively.
So for the total contribution to the job, we have:
We have:
and so:
So Alfred would take $60$ days, while Bill would take $40$ days.
== Sources ==
* : Solutions: $139$. -- Working Alone
* : Answers: $212$. Working Alone
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/13 - The Three Hospitals

== by : $13$ ==
;The Three Hospitals
:Colonel Crackham said that a hospital collection brought in the following contributions:
::A cheque for $\pounds 2, 10 \shillings$,
::two cheques for $\pounds 1, 5 \shillings$ each,
::three $\pounds 1$ Treasury notes,
::three $10 \shillings$ Treasury notes,
::two crowns,
::two postal orders for $3 \shillings$ each,
::two florins,
::and three shillings.
:As this money had to be divided amongst three hospitals, just as it stood,
::since nobody happened to have any change in his pocket,
::how was it to be done?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Money Puzzles: $13$. -- The Three Hospitals
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/13 - The Three Hospitals/Solution

== by : $13$ ==
;The Three Hospitals
== Solution ==
The total take was $213 \shillings$.
Hence each hospital is to receive $71 \shillings$.
Hence we need to find a way of splitting the above accordingly:
:$50 \shillings$, $10 \shillings$, $5 \shillings$, $3 \shillings$, $2 \shillings$, $1 \shillings$
:$25 \shillings$, $20 \shillings$, $20 \shillings$, $3 \shillings$, $2 \shillings$, $1 \shillings$
:$25 \shillings$, $20 \shillings$, $10 \shillings$, $10 \shillings$, $5 \shillings$, $1 \shillings$
== Sources ==
* : Solutions: $13$. -- The Three Hospitals
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/140 - A Curious Progression

== by : $140$ ==
;A Curious Progression
:A correspondent sent this:
:"An arithmetical progression is $10, 20, 30, 40, 50$, the five terms of which sum is $150$.
:Find another progression of five terms, without fractions, which sum to $153$."
:We noted at once the wily omission of a word in the last sentence,
::because such an arithmetical progression is not possible.
:We therefore suggested, by way of jest, this queer solution:
::a progression of $5$ [then] current silver coins: $3 \oldpence$, $1 \shillings$, $2 \shillings$, $2 \shillings 6 \oldpence$, $4 \shillings$, $5 \shillings$,
::which sum to $153$ pence.
:But this is not his own answer, which is quite satisfying -- no algebraic complexities.
:What is it?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $140$. -- A Curious Progression
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/140 - A Curious Progression/Solution

== by : $140$ ==
;A Curious Progression
== Solution ==
The solution given by 's correspondent, which is as good a solution as any, is:
:$153 = 1! + 2! + 3! + 4! + 5!$
where $n!$ denotes the factorial of $n$.
We at came to the same conclusion by clicking on 153, on which this fact is presented.
== Also see ==
* 153
* Sum of Sequence of Factorials
== Sources ==
* : Solutions: $140$. -- A Curious Progression
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Category:153
Henry Ernest Dudeney/Puzzles and Curious Problems/141 - The First Boomerang Puzzle

== by : $141$ ==
;The First "Boomerang" Puzzle
:You ask someone to think of any whole number between $1$ and $100$,
::and then divide it successively by $3$, $5$ and $7$,
::telling you the remainder in each case.
:You then immediately announce the number that was thought of.
:Can the reader discover a simple method of mentally performing this feat?
:If not, he will perhaps be interested in seeing how the ancient mathematicians did it.
=== Click here for solution ===
== Historical Note ==
== 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