nyuuzyou/wikis huggingface dataset:
Henry Ernest Dudeney/Puzzles and Curious Problems/4 - Puzzling Legacies

== by : $4$ ==
;Puzzling Legacies
:A man bequeathed a sum of money, a little less than $\pounds 1500$, to be divided as follows:
:The five children and the lawyer received such sums that
::the square root of the eldest son's share,
::the second son's share divided by two,
::the third son's share minus $\pounds 2$,
::the fourth son's share plus $\pounds 2$,
::the daughter's share multiplied by two,
::and the square of the lawyer's fee
::all worked out at exactly the same sum of money.
:No pounds were divided, and no money was left over after the division.
:What was the total amount bequeathed?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Money Puzzles: $4$. -- Puzzling Legacies
* : Arithmetical and Algebraical Problems: Money Puzzles: $14$. Puzzling Legacies
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/4 - Puzzling Legacies/Solution

== by : $4$ ==
;Puzzling Legacies
:A man bequeathed a sum of money, a little less than $\pounds 1500$, to be divided as follows:
:The five children and the lawyer received such sums that
::the square root of the eldest son's share,
::the second son's share divided by two,
::the third son's share minus $\pounds 2$,
::the fourth son's share plus $\pounds 2$,
::the daughter's share multiplied by two,
::and the square of the lawyer's fee
::all worked out at exactly the same sum of money.
:No pounds were divided, and no money was left over after the division.
:What was the total amount bequeathed?
== Solution ==
:The oldest son gets $\pounds 1296$.
:The second son gets $\pounds 72$.
:The third son gets $\pounds 38$.
:The fourth son gets $\pounds 34$.
:The daughter gets $\pounds 18$.
:The lawyer's fee is $\pounds 6$.
Hence the total amount bequeathed is $\pounds 1464$.
== Proof ==
Let $a$, $b$, $c$, $d$, $e$ and $l$ be the shares received by the eldest son, second son, third son, fourth son, daughter and lawyer respectively.
We have:
We can either attempt to solve this quartic formally, or try some integer values of $l$.
There are not many such that $l^4 < 1500$, so this is a feasible approach.
and we have gone far enough.
$5$ is immediately eliminated because $\dfrac 9 2 5^2$ is not an integer.
$4$ looks too small for the numbers under discussion.
So, we try $6$:
:$6^4 + \dfrac 9 2 6^2 + 6 = 1296 + 162 + 6 = 1464$
which fits the bill perfectly.
The result follows.
== Sources ==
* : Solutions: $4$. -- Puzzling Legacies
* : Answers: $14$. Puzzling Legacies
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/50 - Buying Ribbon

== by : $50$ ==
;Buying Ribbon
:Here is a puzzle that appears to bear a strong family resemblance to others given in the past.
:But it really requires an entirely different method of working.
:The author is unknown.
:Four mothers, each with one daughter, went into a shop to buy ribbon.
:Each mother bought twice as many yards as her daughter,
::and each person bought as many yards of ribbon as the number of farthings she paid for each yard.
:Mrs. Jones spent $1 \shillings 7 \oldpence$ more than Mrs. White;
::Nora bought three yards less than Mrs. Brown;
::Gladys bought two yards more than Hilda,
::who spent $1 \shillings$ less than Mrs. Smith.
:What is the name of Mary's mother?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Age and Kinship Puzzles: $50$. -- Buying Ribbon
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $186$. Buying Ribbon
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/51 - Sharing the Apples

== by : $51$ ==
;Sharing the Apples
:Eight children had a basket containing $32$ apples.
:They divided them amongst themselves as follows:
:Anne got one apple,
::Mary two,
::Jane three,
::and Kate four.
:Ned Smith took as many as his sister,
::Tom Brown twice as many as his sister,
::Bill Jones three times as many as his sister,
::and Jack Robinson four times as many as his sister.
:What are the full names of the girls?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Age and Kinship Puzzles: $51$. -- Sharing the Apples
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $185$. Sharing the Apples
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/38 - The Picnic
Henry Ernest Dudeney/Puzzles and Curious Problems/51 - Sharing the Apples/Solution

== by : $51$ ==
;Sharing the Apples
== Solution ==
:Ned Smith and his sister Jane took $3$ and $3$ respectively.
:Tom and Kate Brown took $8$ and $4$ respectively.
:Bill and Anne Jones took $3$ and $1$ respectively.
:Jack and Mary Robinson took $8$ and $2$ respectively.
== Proof ==
This is exactly the same problem as $38$ - The Picnic from the same collection.
== Sources ==
* : Solutions: $51$. -- Sharing the Apples
* : Answers: $185$. Sharing the Apples
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/52 - In the Year 1900

== by : $52$ ==
;In the Year $1900$
:A correspondent proposes the following question.
:The reader may think, at first sight, that there is insufficient data for an answer, but he will be wrong:
:A man's age at death was one-twenty-ninth of the year of his birth.
:How old was he in the year $1900$?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Age and Kinship Puzzles: $52$. -- In the Year $1900$
* : Arithmetical and Algebraical Problems: Age Puzzles: $42$. In the Year $1900$
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/52 - In the Year 1900/Solution

== by : $52$ ==
;In the Year $1900$
== Solution ==
:$44$.
== Proof ==
This puzzle was presented in $1932$.
Years immediately before then which are divisible by $29$ are:
:$1827 = 63 \times 29$
:$1856 = 64 \times 29$
:$1885 = 65 \times 29$
:$1914 = 66 \times 29$
The years of death for each of these can then be calculated as:
:$1827 + 63 = 1890$
which is too early, as it is implied he was still alive in $1900$.
:$1856 + 64 = 1920$
which produces the valid solution that he was $44$ in $1900$.
:$1885 + 65 = 1950$
which, in $1932$, is still in the future, so he has not died.
== Sources ==
* : Solutions: $52$. -- In the Year $1900$
* : Answers: $42$. In the Year $1900$
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/53 - Finding a Birthday

== by : $53$ ==
;Finding a Birthday
:A correspondent informs us incidentally that on Armistice Day (Nov. 11, 1928)
::he would have lived as long in the $20$th century as he lived in the $19$th.
:This tempted us to work out the day of his birth.
:Perhaps the reader may like to do the same.
:We will assume he was born at midday.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Age and Kinship Puzzles: $53$. -- Finding a Birthday
* : Arithmetical and Algebraical Problems: Age Puzzles: $43$. Finding a Birthday
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/53 - Finding a Birthday/Solution

== by : $53$ ==
;Finding a Birthday
== Solution ==
:Midday on .
== Proof ==
We calculate which day of the year (that is, which by index from $1$ to $366$) is , which is a leap year: $28 = 7 \times 4$:
:$31 + 29 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31 + 10 + \tfrac 1 2 = 315 \tfrac 1 2$
the $\tfrac 1 2$ because at noon we are halfway through the day.
Between $1901$ and $1927$ there are:
:$6$ leap years of $366$ days
leaving:
:$21$ years of $365$ days.
This adds up to $10 \, 176 \tfrac 1 2$ days since , which is where the century started.
It remains to determine the date which is the day $10 \, 176 \tfrac 1 2$ days before this date.
Some $28$ years before $1901$ is somewhere in February $1873$.
We need to narrow that down.
Between $1874$ and $1900$ there are again $6$ leap years of $366$ days, and $21$ years of $365$ days, so:
:the number of days between the start of $1874$ and the end of $1900$
is exactly the same as:
:the number of days between the start of $1901$ and the end of $1927$
so we do not have to adjust for that.
Instead we know that we want the date of day $315 \tfrac 1 2$ counting backwards from .
:$365 - 315 \tfrac 1 2 = 49 \tfrac 1 2$
Day $49$ of $1873$ is .
Half a day after that brings us to midday on .
== Sources ==
* : Solutions: $53$. -- Finding a Birthday
* : Answers: $43$. Finding a Birthday
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/54 - The Birth of Boadicea

== by : $54$ ==
;The Birth of Boadicea
: died $129$ years after was born.
:Their united ages (that is, the combined years of their complete lives) were $100$ years.
:Cleopatra died .
:When was Boadicea born?
=== Click here for solution ===
=== Click here for linguistic note on Boadicea ===
== Historical Note ==
== Sources ==
* : Arithmetical and Algebraical Problems: Age and Kinship Puzzles: $54$. -- The Birth of Boadicea
* : Arithmetical and Algebraical Problems: Age Puzzles: $44$. The Birth of Boadicea
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/54 - The Birth of Boadicea/Historical Note

== Historical Note on by : $54$ -- The Birth of Boadicea ==
 attributes this puzzle to a correspondent he identifies as R.D., but no record remains of who this may have been.
== Sources ==
* : Arithmetical and Algebraical Problems: Age and Kinship Puzzles: $54$. -- The Birth of Boadicea
* : Arithmetical and Algebraical Problems: Age Puzzles: $44$. The Birth of Boadicea
Category:Historical Notes
Henry Ernest Dudeney/Puzzles and Curious Problems/54 - The Birth of Boadicea/Solution

== by : $54$ ==
;The Birth of Boadicea
== Solution ==
She was born in .
== Proof ==
Let $C_b$, $C_d$, $B_b$, $B_d$ denote the years of birth and death of Cleopatra and Boadicea respectively.
We have:
== Sources ==
* : Solutions: $54$. -- The Birth of Boadicea
* : Answers: $44$. The Birth of Boadicea
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/55 - Eliza's Surname

== by : $55$ ==
;Eliza's Surname
:Readers will remember the old puzzle of the "Dutchmen's Wives," nearly two hundred years old.
:Here is a new extension of it (sent by C. C. P. S.):
:Smith, Brown and Robinson have provided themselves with a penny pencil each,
::and took their wives to a stockbroker's office to buy shares.
:Mary bought $50$ more shares than Brown,
::and Robinson $120$ more than Jane.
:Each man paid as many shillings per share as he bought shares,
::and each wife as many pence per share as she bought shares,
::and every man spent one guinea more than his wife.
:What was Eliza's surname?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Age and Kinship Puzzles: $55$. -- Eliza's Surname
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/56 - The Ambiguous Clock

== by : $56$ ==
;The Ambiguous Clock
:A man had a clock with an hour hand and minute hand of the same length and indistinguishable.''
:If it was set going at noon, what would be the first time that it would be impossible, by reason of the similarity of the hands, to be sure of the correct time?
:Readers will remember with these clock puzzles there is the convention that we may assume it possible to indicate fractions of seconds.
:On this assumption an exact answer can be given.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Clock Puzzles: $56$. -- The Ambiguous Clock
* : Arithmetical and Algebraical Problems: Clock Puzzles: $48$. The Ambiguous Clock
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Category:Clock Puzzles
Category:Ambiguous Times
Henry Ernest Dudeney/Puzzles and Curious Problems/56 - The Ambiguous Clock/Solution

== by : $56$ ==
;The Ambiguous Clock
== Solution ==
:$12:05 \tfrac 5 {143}$, where it cannot be distinguished from $01:00 \tfrac {60} {143}$.
:
== Proof ==
See Ambiguous Times for full analysis.
== Sources ==
* : Solutions: $56$. -- The Ambiguous Clock
* : Answers: $48$. The Ambiguous Clock
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/57 - The Broken Clock Face

== by : $57$ ==
;The Broken Clock Face
:How may a clock dial with Roman numerals be broken into four parts
::so that the numerals on each part add up in every case to $20$?
:Here is an example of an attempt at a solution where the parts do not quite add to $20$:
:
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Clock Puzzles: $57$. -- The Broken Clock Face
* : Arithmetical and Algebraical Problems: Clock Puzzles: $49$. The Broken Clock Face
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Category:Clock Puzzles
Henry Ernest Dudeney/Puzzles and Curious Problems/57 - The Broken Clock Face/Historical Note

== Historical Note on by : $57$ - The Broken Clock Face ==
 points out, in his $1968$ repackaging , that there are a total of $13$ possible solutions here.
They are all listed in his columns in for May and June $1966$, reprinted in his $1975$ collection .
He also notes that the solution given in the first edition of was inferior, requiring that the $\text {IX}$ be viewed upside down and interpreted as $\text {XI}$
== Sources ==
* : Answers: $49$. The Broken Clock Face
* : $17$. Cooks and Quibble-Cooks
Category:Historical Notes
Henry Ernest Dudeney/Puzzles and Curious Problems/57 - The Broken Clock Face/Solution

== by : $57$ ==
;The Broken Clock Face
== Solution ==
There are $13$ solutions in total, exemplified by the one given by himself:
:
== Historical Note ==
== Sources ==
* : Solutions: $57$. -- The Broken Clock Face
* : Answers: $49$. The Broken Clock Face
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/58 - When did the Dancing Begin

== by : $58$ ==
;When did the Dancing Begin?
:"The guests at that ball the other night," said Dora at the breakfast-table,
::"thought that the clock had stopped,
::because the hands appeared in exactly the same position as when the dancing began.
:But it was found that they had really only changed places.
:As you know, the dancing commenced between ten and eleven o'clock.
:What was the exact time of the start?"
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Clock Puzzles: $58$. -- When did the Dancing Begin?
* : Arithmetical and Algebraical Problems: Clock Puzzles: $50$. When did the Dancing Begin?
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Category:Clock Puzzles
Category:Ambiguous Times
Henry Ernest Dudeney/Puzzles and Curious Problems/58 - When did the Dancing Begin/Solution

== by : $58$ ==
;When did the Dancing Begin?
== Solution ==
The answer given by is:
:The dancing must have begun at $59 \tfrac {83} {143}$ minutes past ten,
::and the hands were noticed to have changed places at $54 \tfrac {138} {143}$ minutes past eleven.
However, this is merely one of $11$ possible solutions, although some are less feasible than others.
This is a list of all the ambiguous times between $10:00$ and $11:00$:
:
$10 : 04 \tfrac {28} {143}$	$12 : 50 \tfrac {50} {143}$
$10 : 09 \tfrac {33} {143}$	$01 : 50 \tfrac {110} {143}$
$10 : 14 \tfrac {38} {143}$	$02 : 51 \tfrac {27} {143}$
$10 : 19 \tfrac {43} {143}$	$03 : 51 \tfrac {87} {143}$
$10 : 24 \tfrac {48} {143}$	$04 : 52 \tfrac 4 {143}$
$10 : 29 \tfrac {53} {143}$	$05 : 52 \tfrac {64} {143}$
$10 : 34 \tfrac {58} {143}$	$06 : 52 \tfrac {124} {143}$
$10 : 39 \tfrac {63} {143}$	$07 : 53 \tfrac {41} {143}$
$10 : 44 \tfrac {68} {143}$	$08 : 53 \tfrac {101} {143}$
$10 : 49 \tfrac {73} {143}$	$09 : 54 \tfrac {18} {143}$
$10 : 59 \tfrac {83} {143}$	$11 : 54 \tfrac {138} {143}$
Of all of these, the period of time between $10 : 59 \tfrac {83} {143}$ and $11 : 54 \tfrac {138} {143}$ is the shortest:
:
== Proof ==
See Ambiguous Times for the analysis of this.
== Sources ==
* : Solutions: $58$. -- When did the Dancing Begin?
* : Answers: $50$. When did the Dancing Begin?
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/59 - Mistaking the Hands

== by : $59$ ==
;Mistaking the Hands
:"Between two and three o'clock yesterday," said Colonel Crackham,
::"I looked at the clock and mistook the minute hand for the hour hand,
::and consequently the time appeared to be fifty-five minutes earlier than it actually was.
:What was the correct time?"
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Clock Puzzles: $59$. -- Mistaking the Hands
* : Arithmetical and Algebraical Problems: Clock Puzzles: $51$. Mistaking the Hands
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Category:Clock Puzzles
Henry Ernest Dudeney/Puzzles and Curious Problems/59 - Mistaking the Hands/Solution

== by : $59$ ==
;Mistaking the Hands
== Solution ==
The time was $02:05 \tfrac 5 {11}$, but it appeared to be $01 : 10 \tfrac 5 {11}$.
:
== Proof ==
Let $T$ be the time in question.
Let $m$ be the number of minutes after $2:00$ at $T$.
The time mistaken is $T - 55$ minutes, that is $m + 5$ minutes after $1:00$.
Let $\theta \degrees$ be the angle made by the minute hand with respect to twelve o'clock at time $T$.
Let $\phi \degrees$ be the angle made by the hour hand with respect to twelve o'clock at time $T$.
At $T - 55$ minutes, the angle made by the minute hand with respect to twelve o'clock is $\theta + 30 \degrees$.
But this is the angle of the hour hand at $T$.
Hence from Condition for Valid Time Indication, we have that:
:$12 \paren {\phi - 60} = \theta$
as $T$ is between $2:00$ and $3:00$.
Hence:
Hence the time was $02:05 \tfrac 5 {11}$.
== Sources ==
* : Solutions: $59$. -- Mistaking the Hands
* : Answers: $51$. Mistaking the Hands
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/5 - Dividing the Legacy

== by : $5$ ==
;Dividing the Legacy
:A man left $\pounds 100$ to be divided between his two sons Alfred and Benjamin.
:If one-third of Alfred's legacy be taken from one-fourth of Benjamin's, the remainder would be $\pounds 11$.
:What was the amount of each legacy?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Money Puzzles: $5$. -- Dividing the Legacy
* : Arithmetical and Algebraical Problems: Money Puzzles: $15$. Dividing the Legacy
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/5 - Dividing the Legacy/Solution

== by : $5$ ==
;Dividing the Legacy
:A man left $\pounds 100$ to be divided between his two sons Alfred and Benjamin.
:If one-third of Alfred's legacy be taken from one-fourth of Benjamin's, the remainder would be $\pounds 11$.
:What was the amount of each legacy?
== Solution ==
Alfred gets $\pounds 24$ and Benjamin gets $\pounds 76$.
== Proof ==
Let $A$ and $B$ be the amount of Alfred's and Benjamin's legacies respectively.
We have:
== Sources ==
* : Solutions: $5$. -- Dividing the Legacy
* : Answers: $15$. Dividing the Legacy
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/60 - Equal Distances

== by : $60$ ==
;Equal Distances
:At what time between three and four o'clock is the minute hand the same distance from $\text {VIII}$ as the hour hand is from $\text {XII}$?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Clock Puzzles: $60$. -- Equal Distances
* : Arithmetical and Algebraical Problems: Clock Puzzles: $52$. Equal Distances
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Category:Clock Puzzles
Henry Ernest Dudeney/Puzzles and Curious Problems/60 - Equal Distances/Solution

== by : $60$ ==
;Equal Distances
== Solution ==
:$23 \tfrac 1 {13}$ past $3$.
:
== Proof ==
Let $T$ be the time in question.
Let $m$ be the number of minutes past $3:00$ that $T$ is.
Let $\theta \degrees$ be the angle made by the minute hand with respect to twelve o'clock at time $T$.
Let $\phi \degrees$ be the angle made by the hour hand with respect to twelve o'clock at time $T$.
Between $3:00$ and $4:00$ we have that $90 < \phi < 120$.
Hence from Condition for Valid Time Indication, we have that:
:$12 \paren {\phi - 90} = \theta$
Then we are told that:
:$\size {240 - \theta} = \phi$
This can mean that:
:$240 - \theta = \phi$
in which case:
Or it can mean that:
:$\theta - 240 = \phi$
in which case:
But in this case $T = 4:00$, which is not (strictly) between $3:00$ and $4:00$.
Hence the result.
== Sources ==
* : Solutions: $60$. -- Equal Distances
* : Answers: $52$. Equal Distances
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/61 - Right and Left

== by : $61$ ==
;Right and Left
:At what time between three and four o'clock will the minute hand be as far from $12$ on the left side of the dial plate as the hour hand is from $12$ on the right side of the dial plate?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Clock Puzzles: $61$. -- Right and Left
* : Arithmetical and Algebraical Problems: Clock Puzzles: $53$. Right and Left
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Category:Clock Puzzles
Henry Ernest Dudeney/Puzzles and Curious Problems/61 - Right and Left/Solution

== by : $61$ ==
;Right and Left
== Solution ==
:$41 \tfrac 7 {13}$ past $3$.
:
== Proof ==
Let $T$ be the time in question.
Let $m$ be the number of minutes past $3:00$ that $T$ is.
Let $\theta \degrees$ be the angle made by the minute hand with respect to twelve o'clock at time $T$.
Let $\phi \degrees$ be the angle made by the hour hand with respect to twelve o'clock at time $T$.
Between $3:00$ and $4:00$ we have that $90 < \phi < 120$.
Hence from Condition for Valid Time Indication, we have that:
:$12 \paren {\phi - 90} = \theta$
Then we are told that:
:$\theta = 360 - \phi$
Hence:
Hence the result.
== Sources ==
* : Solutions: $61$. -- Right and Left
* : Answers: $53$. Right and Left
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/62 - At Right Angles

== by : $62$ ==
;At Right Angles
:How soon between the hours of five and six will the hour and minute hands of a clock be exactly at right angles?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Clock Puzzles: $62$. -- At Right Angles
* : Arithmetical and Algebraical Problems: Clock Puzzles: $54$. At Right Angles
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/62 - At Right Angles
Category:Clock Puzzles
Henry Ernest Dudeney/Puzzles and Curious Problems/62 - At Right Angles/Solution

== by : $62$ ==
;At Right Angles
== Solution ==
:$10 \tfrac {10} {11}$ past $5$.
:
== Proof $1$ ==
== Proof $2$ ==
== Sources ==
* : Solutions: $62$. -- At Right Angles
* : Answers: $54$. At Right Angles
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/62 - At Right Angles