nyuuzyou/wikis huggingface dataset:
Henry Ernest Dudeney/Modern Puzzles/57 - A Misunderstanding/Solution

== by : $57$ ==
;A Misunderstanding
== Solution ==
:$\dfrac {857142} 3 = 285714$
== Proof ==
We are being asked to find a number:
:$N = \sqbrk {a_k a_{k - 1} \ldots a_2 a_1}_{10}$
such that:
:$\dfrac N 3 = \sqbrk {a_1 a_k a_{k - 1} \ldots a_2}_{10}$
where $a_1$ is required to be $2$.
We extract the general case below:
Let $q$ be the rational number which can be expressed as:
:$q = 0 \cdotp \dot a_k a_{k - 1} \ldots a_2 \dot a_1$
such that:
:$\dfrac q 3 = 0 \cdotp \dot a_1 a_k a_{k - 1} \ldots \dot a_2$
Then:
From the construction of the problem, the possible values for $a_1$ are $1$ and $2$, giving:
as possible answers.
From 7 is Cyclic Number:
:$q = 6 \times 0 \cdotp \dot 14285 \dot 7 = 0 \cdotp \dot 85714 \dot 2$
which leads us to:
:$N = 857142$
So the specific answer required was:
:$\dfrac {857142} 3 = 285714$
while we also have:
:$\dfrac {428571} 3 = 142857$
== Sources ==
* : Solutions: $57$. -- A Misunderstanding
* : Answers: $108$. A Misunderstanding
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Category:Cyclic Numbers
Henry Ernest Dudeney/Modern Puzzles/58 - The Two Fours

== by : $58$ ==
;The Two Fours
:The point [of the Four Fours puzzle] is to express all possible whole numbers with four fours (no more and no fewer), using the various arithmetical signs.
:Thus:
:
:and:
:
:All numbers up to $112$ inclusive may be solved, using only the signs for addition, subtraction, multiplication, division, square root, decimal points, and the factorial sign $4!$ which means $1 \times 2 \times 3 \times 4$, or $24$, but $113$ is impossible.
:It is necessary to discover which numbers can be formed with one four, with two fours, and with three fours, and to record these for combination as required.
:It is the failure to find some of these that leads to so much difficulty.
:For example, I think very few discover that $64$ can be expressed with only two fours.
:Can the reader do it?
=== Click here for solution ===
== Historical Note ==
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $58$. -- The Two Fours
* : Arithmetical and Algebraical Problems: Digital Puzzles: $109$. The Two Fours
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/58 - The Two Fours/Historical Note

== Historical Note on $58$. - The Two Fours ==
 reports:
:I am perpetually receiving inquiries about the old "Four Fours" puzzle.
:I published it in $1899$, but have since found that it first appeared in the first volume of ($1881$).
:It has since been dealt with at some length by various writers.
 locates that original article in as being the December $30$th issue.
He then goes on to cite a number of more recent discussions on the subject, including his exposition in his own column in for January $1964$.
He finishes with a reference to an article by in which it is proved that all positive integers up to $208$ can be expressed with nothing but one $4$, instances of the square root sign, the factorial sign, and the floor function.
Because it is possible to express $4$ using four $4$s, it is hence possible to represent $113$ using four $4$s, although this representation may be somewhat complicated.
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $58$. -- The Two Fours
* 
* : Arithmetical and Algebraical Problems: Digital Puzzles: $109$. The Two Fours
* : Answers: $109$. The Two Fours
Category:Historical Notes
Henry Ernest Dudeney/Modern Puzzles/58 - The Two Fours/Solution

== by : $58$ ==
;The Two Fours
== Solution ==
If you allow the $\Gamma$ function:
:$\paren {\sqrt 4}^{\map \Gamma 4} = 2^6 = 64$
Otherwise, there is this ingenious creation that was supplied by himself, and may be original to him:
:$\sqrt {\paren {\sqrt {\sqrt 4} }^{4 !} } = \sqrt {\paren {\sqrt 2}^{24} } = \sqrt {2^{12} } = \sqrt {4096} = 64$
== Historical Note ==
== Sources ==
* : Solutions: $58$. -- The Two Fours
* : Answers: $109$. The Two Fours
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/59 - The Two Digits

== by : $59$ ==
;The Two Digits
:Write down any $2$-figure number (different figures and no $0$)
::and then express that number by writing the same figures in reverse order,
::with or without arithmetical signs.
:For example, $45 = 5 \times 9$ would be correct if only the $9$ had happened to be a $4$.
:Or $81 = \paren {1 + 8}^2$ would do, except for the fact that it introduces a third figure: the $2$.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $59$. -- The Two Digits
* : Arithmetical and Algebraical Problems: Digital Puzzles: $110$. The Two Digits
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/59 - The Two Digits/Solution

== by : $59$ ==
;The Two Digits
== Solution ==
Examples given by :
It is worth pointing out that the second inadequate example quoted by himself:
:$81 = \paren {1 + 8}^2$
which is ineligible because of the $2$, leads us on neatly into:
:$\sqrt {81} = 1 + 8$
which unaccountably missed.
There may be more.
 offers up:
:$94 \equiv 4 \pmod 9$
which may or may not be eligible.
And indeed:
and so on.
== Sources ==
* : Solutions: $59$. -- The Two Digits
* : Answers: $110$. The Two Digits
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/5 - Doubling the Value

== by : $5$ ==
;Doubling the Value
:It is a curious fact that if you double $\pounds 6 \ 13 \shillings$, you get $\pounds 13 \ 6 \shillings$, which is merely changing the shillings and the pounds.
:Can you find another sum of money that has the same peculiarity that, when multiplied by any number you may choose to select, will merely exchange the shillings and the pounds?
:There is only one other multiplier and sum of money, besides the case shown, that will work.
:What is it?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Money Puzzles: $5$. -- Doubling the Value
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/5 - Doubling the Value/Solution

== by : $5$ ==
;Doubling the Value
== Solution ==
:$\pounds 2 \ 17 \shillings$
which, when multiplied by $6$, becomes:
:$\pounds 17 \ 2 \shillings$
== Proof ==
It is assumed that when says number he means natural number
Let $k$ denote the shilling value of the original sum of money.
Recall there are $20$ shillings to the pound.
Let $s$ and pounds $p$ denote the number of shillings and pounds that $k$ consists of such that $s < 20$.
That is:
:$k = \pounds p \ s \shillings$
where:
:$0 \le s < 20$
and of course trivially $s \ne 0$.
Let $n$ be a number which switches the shillings and pounds when you multiply $k$ by it.
Hence we want to find values for $n$ that will result in:
:$k = 20 p + s$
and:
:$n k = 20 s + p$
where both $p < 20$ and $s < 20$.
We have:
It remains to substitute values for $n$ and see where that gets us.
Clearly $n < 20$ in order for $\paren {20 - n} s > 0$.
$n = 2$:
which is the answer we have been provided.
$n = 3$:
$n = 4$:
$n = 5$:
$n = 6$:
Hence we have that:
:$k = \pounds 2 \ 17 \shillings$
and we check that:
It remains to be shown that this is the only solution.
$n = 7$:
$n = 8$:
$n = 9$:
$n = 10$:
$n = 10$:
and at this stage $\dfrac p s > 20$ and so there can be no solutions such that $p < 20$.
That exhausts all possibilities for $n$.
It is interesting to see what happens when $n = 1$ is entered:
which is of course obvious.
== Sources ==
* : Solutions: $5$. -- Doubling the Value
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/60 - Digital Coincidences

== by : $60$ ==
;Digital Coincidences
:If I multiply, and also add, $9$ and $9$, I get $81$ and $18$, which contain the same figures.
:If I multiply and add $2$ and $47$ I get $94$ and $49$ -- the same figures.
:If I multiply and add $3$ and $24$ I get the same figures -- $72$ and $27$.
:Can you find two numbers that, when multiplied and added will, in this simple manner, produce the same three figures?
=== Click here for solution ===
== Also see ==
* Numbers the Sum of whose Digits is Reversal of Product of Digits
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $60$. -- Digital Coincidences
* : Arithmetical and Algebraical Problems: Digital Puzzles: $111$. Digital Coincidences
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/60 - Digital Coincidences/Solution

== by : $60$ ==
;Digital Coincidences
== Solution ==
 offers up:
and:
== Also see ==
* Numbers which Multiplied by 2 are the Reverse of when Added to 2
== Historical Note ==
== Sources ==
* : Solutions: $60$. -- Digital Coincidences
* : Answers: $111$. Digital Coincidences
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/61 - Palindromic Square Numbers

== by : $61$ ==
;Palindromic Square Numbers
:This is a curious subject for investigation -- the search for square numbers the figures of which read backwards and forwards alike.
:Some of them are very easily found.
:For example, the squares of $1$, $11$, $111$ and $1111$ are respectively $1$, $121$, $12321$, and $1234321$, all palindromes,
::and the rule applies for any number of $1$'s provided the number does not contain more than nine.
:But there are other cases that we may call irregular, such as the square of $264 = 69696$ and the square of $2285 = 5221225$.
:Now, all the examples I have given contain an odd number of digits.
:Can the reader find a case where the square palindrome contains an even number of figures?
=== Click here for solution ===
== Also see ==
* Square of Small-Digit Palindromic Number is Palindromic
* Palindromic Squares with Non-Palindromic Roots
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $61$. -- Palindromic Square Numbers
* : Arithmetical and Algebraical Problems: Digital Puzzles: $112$. Palindromic Square Numbers
Category:Henry Ernest Dudeney/Modern Puzzles
Category:Palindromic Numbers
Category:Square Numbers
Henry Ernest Dudeney/Modern Puzzles/61 - Palindromic Square Numbers/Solution

== by : $61$ ==
;Palindromic Square Numbers
== Solution ==
 gives:
:$836^2 = 698896$
This is now known to be the smallest.
There are larger ones, but they are fairly rare.
== Also see ==
* Numbers whose Square is Palindromic with Even Number of Digits
== Sources ==
* : Solutions: $61$. -- Palindromic Square Numbers
* : Answers: $112$. Palindromic Square Numbers
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/62 - Factorizing

== by : $62$ ==
;Factorizing
:What are the factors (the numbers that will divide it without any remainder) of this number -- $1000000000001$?
:This is easily done if you happen to know something about numbers of this peculiar form.
:In fact, it is just as easy for me to give two factors if you insert, say $101$ noughts, instead of $11$, between the two ones.
:There is a curious, easy, and beautiful rule for these cases.
:Can you find it?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $62$. -- Factorizing
* : Arithmetical and Algebraical Problems: Digital Puzzles: $113$. Factorizing
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/62 - Factorizing/Solution

== by : $62$ ==
;Factorizing
== Solution ==
== Proof ==
This specific result is an instance of Divisors of One More than Power of 10: Number of Zero Digits Congruent to 2 Modulo 3.
== Also see ==
* Divisors of One More than Power of 10
* Divisors of One More than Power of 10: Number of Zero Digits Even
* Prime Factors of One More than Power of 10
== Sources ==
* : Solutions: $62$. -- Factorizing
* : Answers: $113$. Factorizing
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/63 - Find the Factors

== by : $63$ ==
;Find the Factors
:Find $2$ whole numbers with the smallest possible difference between them
::which, when multiplied together, will produce $1234567890$.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $63$. -- Find the Factors
* : Arithmetical and Algebraical Problems: Digital Puzzles: $114$. Find the Factors
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/63 - Find the Factors/Solution

== by : $63$ ==
;Find the Factors
== Solution ==
:$36070 \times 34227$
== Proof ==
We have:
:$1234567890 = 2 \times 3^2 \times 5 \times 3607 \times 3803$
It is safe to say that one of the factors will contain $3607$ and the other $3803$.
It remains a matter of putting these together with the smaller ones to get the smallest difference.
The remaining factors can be collected as $3^2 = 9$ and $2 \times 5 = 10$
and so to the result.
== Sources ==
* : Solutions: $63$. -- Find the Factors
* : Answers: $114$. Find the Factors
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/64 - Dividing by Eleven

== by : $64$ ==
;Dividing by Eleven
:If the $9$ digits are written at haphazard in any order,
::for example $4 \ 1 \ 2 \ 5 \ 3 \ 9 \ 7 \ 6 \ 8$, what are the chances that the number that happens to be produced will be divisible by $11$ without remainder?
:The number I have written at random is not, I see, so divisible, but if I had happened to make the $1$ and the $8$ change places it would be.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $64$. -- Dividing by Eleven
* : Arithmetical and Algebraical Problems: Digital Puzzles: $115$. Dividing by Eleven
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/64 - Dividing by Eleven/Solution

== by : $64$ ==
;Dividing by Eleven
== Solution ==
:$115$ to $11$ against a haphazard arrangement being divisible by $11$.
== Proof ==
Let $\sqbrk {abcdefghi}$ be a haphazard arrangement of the $9$ digits.
We set:
:$m = a + c + e + g + i$
:$n = b + d + f + h$
From Divisibility by 11, $\sqbrk {abcdefghi}$ is divisible by $11$ :
:$m - n = 11k$
for some integer $k$.
We have:
Moreover, since $m + n = 45$ is odd, so is $m - n = \paren {m + n} - 2n$.
Hence we can only have $m - n = \pm 11$, which give the following cases when solved:
:Case $1$: $m = 28, n = 17$
:Case $2$: $m = 17, n = 28$
For Case $1$, the following $9$ quadruples of distinct digits give $n = 17$:
:$1, 2, 5, 9$
:$1, 2, 6, 8$
:$1, 3, 4, 9$
:$1, 3, 5, 8$
:$1, 3, 6, 7$
:$1, 4, 5, 7$
:$2, 3, 4, 8$
:$2, 3, 5, 7$
:$2, 4, 5, 6$
For Case $2$, the following $2$ quintuples of distinct digits give $m = 17$:
:$1, 2, 3, 4, 7$
:$1, 2, 3, 5, 6$
Hence there are exactly:
:$\paren {9 + 2} \times 4! \times 5! = 31680$
numbers, out of $9! = 362880$ possibilities that our number is divisible by $11$.
This gives a probability of:
:$\dfrac {31680} {362880} = \dfrac {11} {126}$
which is exactly $115$ to $11$.
== Sources ==
* : Solutions: $64$. -- Dividing by Eleven
* : Answers: Digital Puzzles: $115$. Dividing by Eleven
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/65 - Dividing by 37

== by : $65$ ==
;Dividing by $37$
:I want to know whether the number $49,129,308,213$ is exactly divisible by $37$,
::or if not, what is the remainder when so divided.
:How may I do this quite easily without any process of actual division whatever?
:It can be done by inspection in a few seconds -- if you know how.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $65$. -- Dividing by $37$
* : Arithmetical and Algebraical Problems: Digital Puzzles: $116$. Dividing by $37$
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/65 - Dividing by 37/Solution

== by : $65$ ==
;Dividing by $37$
== Solution ==
:$49,129,308,213 = 37 \times 1,327,819,140 + 33$
Hence the remainder is $33$.
== Proof ==
We proceed by Divisibility by 37.
In each group of $3$ digits of $49,129,308,213$, counting from the :
:$\color {blue} {49}, \color {red} 1 \color {blue} {29}, \color {red} 3 \color {blue} {08}, \color {red} 2 \color {blue} {13}$
Then we add up the numbers:
 49
 29
 08
+13
---
 99
and subtract $11$ times the sum of the numbers:
11 x (1 + 3 + 2) = 66
99 - 66 = 33
Hence the remainder is $33$.
To check, we perform the calculation using an online calculator or otherwise:
:$49,129,308,213 = 37 \times 1,327,819,140 + 33$
== Sources ==
* : Solutions: $65$. -- Dividing by $37$
* : Answers: Digital Puzzles: $116$. Dividing by $37$
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/66 - Another 37 Division

== by : $66$ ==
;Another $37$ Division
:Here is an interesting extension of the last puzzle.
:If the $9$ digits are written at haphazard in any order, for example $412539768$,
::what are the chances that the number that happens to be produced will be divisible by $37$ without remainder?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $66$. -- Another $37$ Division
* : Arithmetical and Algebraical Problems: Digital Puzzles: $117$. Another $37$ Division
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/67 - A Digital Difficulty

== by : $67$ ==
;A Digital Difficulty
:Arrange the $10$ digits, $1 \ 2 \ 3 \ 4 \ 5 \ 6 \ 7 \ 8 \ 9 \ 0$, in such order that they shall form a number
::that may be divided by every number from $2$ to $18$ without in any case a remainder.
:As an example, if I arrange them thus, $1 \ 2 \ 7 \ 4 \ 9 \ 5 \ 3 \ 6 \ 8 \ 0$, this number can be divided by $2$, $3$, $4$, $5$ and so on up to $16$, without any remainder,
::but it breaks down at $17$.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $67$. -- A Digital Difficulty
* : Arithmetical and Algebraical Problems: Digital Puzzles: $118$. A Digital Difficulty
Category:Henry Ernest Dudeney/Modern Puzzles
Category:Pandigital Integers
Category:Pandigital Numbers Divisible by All Integers up to 18
Henry Ernest Dudeney/Modern Puzzles/67 - A Digital Difficulty/Solution

== by : $67$ ==
;A Digital Difficulty
== Solution ==
There are $4$ solutions:
:$2 \, 438 \, 195 \, 760$
:$3 \, 785 \, 942 \, 160$
:$4 \, 753 \, 869 \, 120$
:$4 \, 876 \, 391 \, 520$
== Proof ==
Let $N$ be such a number.
The sum of all the digits of $N$ is $45$.
Hence all numbers made with all the digits of $N$ is divisible by $3$ and $9$.
By making the final digit $0$, $N$ will also be divisible by both $2$ and $5$, hence also $6$, $10$, $15$ and $18$.
Making the last but one digit even makes $N$ divisible by $4$, and therefore also $12$.
It remains to consider $7$, $11$, $13$, $16$ and $17$, from which $8$ and $14$ will follow.
By Divisibility by 11, either the digits in the odd positions must add to $17$ and those in the even positions must add to $28$ or vice versa.
Leaving the $0$ off the end of $N$ does not affect its divisibility by $7$, $11$ or $13$.
So, let $M$ be $N$ without the $0$ on the end.
By Divisibility Test for 7, 11 and 13, the numbers formed from $M$ of the first three and last three digits must add up to the number formed by the middle three digits.
It remains to try them out.
Note that the third case is actually $474$ and $1386$ and $912$, with the $1$ carried forward and added to the last $4$ of $474$.
== Also see ==
* Pandigital Numbers Divisible by All Integers up to 18
== Sources ==
* : Solutions: $67$. -- A Digital Difficulty
* : Answers: $118$. A Digital Difficulty
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/68 - Threes and Sevens

== by : $68$ ==
;Threes and Sevens
:What is the smallest number composed only of the digits $3$ and $7$ that may be divided by $3$ and $7$,
::and also the sum of its digits by $3$ and $7$, without any remainder.
:Thus, for example, $7 \ 7 \ 3 \ 3 \ 7 \ 3 \ 3$ is divisible by $3$ and $7$ without remainder,
::but the sum of its digits ($33$), while divisible by $3$, is not divisible by $7$ without remainder.
:Therefore it is not a solution.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $68$. -- Threes and Sevens
* : Arithmetical and Algebraical Problems: Digital Puzzles: $119$. Threes and Sevens
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/68 - Threes and Sevens/Solution

== by : $68$ ==
;Threes and Sevens
== Solution ==
 offers up $3,333,377,733$.
This is divisible by $3$ and $7$, and the sum of its digits is $42$, also divisible by $3$ and $7$.
In his words:
:There must be at fewest three $7$'s and seven $3$'s, and the $7$'s must be placed as far to the right as possible.
However, we also offer up $777$ which equally well fulfils the conditions of the question.
== Sources ==
* : Solutions: $68$. -- Threes and Sevens
* : Answers: $119$. Threes and Sevens
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/69 - Root Extraction

== by : $69$ ==
;Root Extraction
:In a conversation I had with Professor Simon Greathead, the eminent mathematician, ...
::the extraction of the cube root.
:"Ah," said the professor, "it is astounding what ignorance prevails ...
::... the simple fact that, to extract the cube root of a number, all you have to do is to add together the digits.
:Thus, ignoring the obvious case of the number $1$, if we want the cube root of $512$, add the digits -- $8$, and there you are!"
:I suggested that that was a special case.
:"Not at all," he replied. "Take another number at random -- $4913$ -- and the digits add to $17$, the cube of which is $4913$."
:I did not presume to argue the point with the learned man,
::but I will just ask the reader to discover all the other numbers whose cube root is the same as the sum of their digits.
:They are so few that they can be counted on the fingers of one hand.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $69$. -- Root Extraction
* : Arithmetical and Algebraical Problems: Digital Puzzles: $120$. Root Extraction
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/69 - Root Extraction/Solution

== by : $69$ ==
;Root Extraction
== Solution ==
The other numbers are:
== Proof ==
See Sequence of Dudeney Numbers.
== Sources ==
* : Solutions: $69$. -- Root Extraction
* : Answers: $120$. Root Extraction
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Category:Dudeney Numbers
Henry Ernest Dudeney/Modern Puzzles/6 - Generous Gifts

== by : $6$ ==
;Generous Gifts
:A generous man set aside a certain sum of money for equal distribution weekly to the needy of his acquaintance.
:One day he remarked:
::"If there are five fewer applicants next week, you will each receive $2$ shillings more."
:Unfortunately, instead of there being fewer there were actually four more persons applying for the gift.
::"This means," he pointed out, "that you will each receive one shilling less."
:Now, how much did each person receive at that last distribution?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Money Puzzles: $6$. -- Generous Gifts
* : Arithmetical and Algebraical Problems: Money Puzzles: $4$. Generous Gifts
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/6 - Generous Gifts/Solution

== by : $6$ ==
;Generous Gifts
== Solution ==
$20$ people each received a dole of $6$ shillings.
There is a total available weekly amount of $240$ shillings.
With $5$ fewer people, we have $120$ shillings divided between $15$ people, that is $8$ shillings each.
However, with $4$ more people, we have $120$ shillings divided between $24$ people, that is $5$ shillings each.
== Proof ==
Let $N$ denote the total number of shillings to be disbursed.
Let $m$ denote the number of shillings disbursed per person today.
Let $p$ denote the total number of people who are receiving a donation today.
Then we have:
== Sources ==
* : Solutions: $6$. -- Generous Gifts
* : Answers: $4$. Generous Gifts
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions