nyuuzyou/wikis huggingface dataset:
Henry Ernest Dudeney/Modern Puzzles/46 - Baxter's Dog

== by : $46$ ==
;Baxter's Dog
:This is an interesting companion to the "Man and Dog" puzzle.
:Anderson set off from an hotel at San Remo at nine o'clock and had been walking an hour when Baxter went after him along the same road.
:Baxter's dog started at the same time as his master and ran uniformly forwards and backwards between him and Anderson until the two men were together.
:Anderson's speed is $2$, Baxter's $4$, and the dog's $10$ miles an hour.
:How far had the dog run when Baxter overtook Anderson?
:My correspondent in Italy who sends me this is an exact man, and he says,
::"Neglect the length of dog and time spent in turning."
:I will merely add, neglect also the dog's name and the day of the month.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Locomotion and Speed Puzzles: $46$. -- Baxter's Dog
* : Arithmetical and Algebraical Problems: Speed & Distance Puzzles: $74$. Baxter's Dog
Category:Henry Ernest Dudeney/Modern Puzzles
Category:The Bulldozers and the Bee
Henry Ernest Dudeney/Modern Puzzles/46 - Baxter's Dog/Solution

== by : $46$ ==
;Baxter's Dog
== Solution ==
:$10$ miles.
== Proof ==
This is a variant on The Bulldozers and the Bee.
During the first hour, Anderson walks $2$ miles.
Baxter walks twice as fast, so catches up with Anderson after the latter has walked another $2$ miles, taking another hour to do so.
During that hour, the dog travels at $10$ miles an hour.
Thus the dog runs $10$ miles.
== Sources ==
* : Solutions: $46$. -- Baxter's Dog
* : Answers: $74$. Baxter's Dog
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/47 - The Runner's Refreshment

== by : $47$ ==
;The Runner's Refreshment
:A man runs $n$ times round a circular track whose radius is $t$ miles.
:He drinks $s$ quarts of beer for every mile that he runs.
:Prove that he will only need one quart!
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Locomotion and Speed Puzzles: $47$. -- The Runner's Refreshment
* : Arithmetical and Algebraical Problems: Speed & Distance Puzzles: $75$. The Runner's Refreshment
Category:Henry Ernest Dudeney/Modern Puzzles
Category:Trick Puzzles
Henry Ernest Dudeney/Modern Puzzles/47 - The Runner's Refreshment/Solution

== by : $47$ ==
;The Runner's Refreshment
== Proof ==
A lame variant of the Second Pizza Theorem that the author of this page did not believe was worth adding to the jokes page.
From Perimeter of Circle, the distance run is $n \times 2 \pi t$.
During that run, he drinks $n \times 2 \pi t \times s$ quarts of beer.
That is, rearranging the variables:
:$2 \, \text {pi} n t s$
The result follows from the definition of a quart.
Ba-dum tsh.
== Sources ==
* : Solutions: $47$. -- The Runner's Refreshment
* : Answers: $75$. The Runner's Refreshment
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/48 - Railway Shunting

== by : $48$ ==
;Railway Shunting
:How are the trains in our illustration to pass one another, and proceed with their engines in front?
:
:The small side track is large enough to hold one engine or one carriage at a time, and no tricks, such as ropes and flying-switches, are allowed.
:Every reversal -- that is, change of direction -- of an engine is counted as a move in the solution.
:What is the smallest number of moves necessary?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Locomotion and Speed Puzzles: $48$. -- Railway Shunting
* : Geometrical Problems: Moving Counter Puzzles: $374$. Railway Shunting
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/48 - Railway Shunting/Solution

== by : $48$ ==
;Railway Shunting
== Solution ==
Refer to the following diagram, in which:
:the engines are $L$ and $R$
:$L$'s carriages, never separated in the solution, are denoted by $X$
:$R$'s carriages are $A$ and $B$
:$S$ is the siding.
:
Make the following moves:
== Sources ==
* : Solutions: $48$. -- Railway Shunting
* : Answers: $374$. Railway Shunting
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/49 - Exploring the Desert

== by : $49$ ==
;Exploring the Desert
:Nine travellers, each possessing a motor-car, meet on the eastern edge of a desert.
:They wish to explore the interior, always going due west.
:Each car can travel $40$ miles on the contents of the engine tank,
::which holds a gallon of petrol,
::and each can carry $9$ extra gallon tins of petrol and no more.
:Unopened tins can alone be transferred from car to car.
:What is the greatest distance at which they can enter the desert without making any depots of petrol for the return journey?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Locomotion and Speed Puzzles: $49$. -- Exploring the Desert
* : Arithmetical and Algebraical Problems: Speed & Distance Puzzles: $76$. Exploring the Desert
Category:Henry Ernest Dudeney/Modern Puzzles
Category:Desert Crossing Problems
Henry Ernest Dudeney/Modern Puzzles/49 - Exploring the Desert/Solution

== by : $49$ ==
;Exploring the Desert
== Solution ==
:$360$ miles out and home.
== Working ==
Let the $9$ men be denoted $A$, $B$, $C$, $D$, $E$, $F$, $G$, $H$ and $J$.
All $9$ men travel $40$ miles into the desert, carrying $9$ gallons each, using the gallon they have in the tank.
$A$ transfers $1$ gallon to each of the other $8$ drivers, and uses his other gallon to get home.
Those $8$ then travel $40$ miles further into the desert on the gallon they have in the tank.
$B$ transfers $1$ gallon to each of the remaining $7$ drivers, and uses his other $2$ gallons to get home.
Those $7$ then travel $40$ miles further into the desert on the gallon they have in the tank.
$C$ transfers $1$ gallon to each of the remaining $6$ drivers, and uses his other $3$ gallons to get home.
Those $6$ then travel $40$ miles further into the desert on the gallon they have in the tank.
$D$ transfers $1$ gallon to each of the remaining $5$ drivers, and uses his other $4$ gallons to get home.
Those $5$ then travel $40$ miles further into the desert on the gallon they have in the tank.
$E$ transfers $1$ gallon to each of the remaining $4$ drivers, and uses his other $5$ gallons to get home.
Those $4$ then travel $40$ miles further into the desert on the gallon they have in the tank.
$F$ transfers $1$ gallon to each of the remaining $3$ drivers, and uses his other $6$ gallons to get home.
Those $3$ then travel $40$ miles further into the desert on the gallon they have in the tank.
$G$ transfers $1$ gallon to each of the remaining $2$ drivers, and uses his other $7$ gallons to get home.
Those $2$ then travel $40$ miles further into the desert on the gallon they have in the tank.
$H$ transfers $1$ gallon to the remaining driver $J$, and uses his other $8$ gallons to get home.
$J$ then travels $40$ miles further into the desert on the gallon he has in the tank.
Then using the $9$ gallons he is carrying, $J$ travels the $9 \times 40 = 360$ miles home.
== Sources ==
* : Solutions: $49$. -- Exploring the Desert
* : Answers: $76$. Exploring the Desert
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/4 - Loose Cash

== by : $4$ ==
;Loose Cash
:What is the largest sum of money -- all in current silver coins and no four-shilling piece -- that I could have in my pocket without being able to give change for half a sovereign?
=== Click here for solution ===
=== Variant ===
== Sources ==
* : Arithmetical and Algebraical Problems: Money Puzzles: $4$. -- Loose Cash
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/4 - Loose Cash/Solution

== by : $4$ ==
;Loose Cash
== Solution ==
The largest sum possible is $15 \shillings 9 \oldpence$, composed of:
:one crown and one half-crown (or $3$ half-crowns)
:$4$ florins
:one threepenny bit.
== Proof ==
Recall the list of silver coins in the time of , expressed in shillings:
It is of course to be noted that these coins were actually an alloy consisting of only $50\%$ pure silver, but were still called "silver coins".
:The threepenny bit: $\tfrac 1 4 \shillings$
:The sixpence: $\tfrac 1 2 \shillings$
:The shilling
:The florin: $2 \shillings$
:The half-crown: $2 \tfrac 1 2 \shillings$
:The crown: $5 \shillings$
:The half sovereign: $10 \shillings$
's mention of the four-shilling piece refers to a coin in circulation for a short time in the late $19$th century, also known as the double florin.
Hence the question is seen to be the same as asking:
:What is the highest total of numbers from $\set {\tfrac 1 4, \tfrac 1 2, 1, 2, 2 \tfrac 1 2, 5}$ of which no subset adds to $10$?
We may take only one crown and only one half-crown, or $3$ half-crowns, as one more of either allow us to make $10$.
Then we see that $4$ florins take us to $5 + 2 \tfrac 1 2 + 4 \times 2 = 15 \tfrac 1 2$, while no combination of $5, 2 \tfrac 1 2, 2, 2, 2, 2$ can make $10$.
Adding another florin brings us to $16 \tfrac 1 2$, but then $2 + 2 + 2 + 2 + 2 = 10$.
Adding a shilling allows $5 + 2 + 2 + 1$, so we can add no shilling to our total.
Adding a sixpence allows $5 + 2 \tfrac 1 2 + 2 + \tfrac 1 2 = 10$, so we can add no sixpence to our total.
Adding a threepenny bit takes our total to $5 + 2 \tfrac 1 2 + 4 \times 2 + \tfrac 1 4 = 15 \tfrac 3 4$.
Adding another threepenny bit allows $5 + 2 \tfrac 1 2 + 2 + \tfrac 1 4 + \tfrac 1 4 = 10$, so we can add no further threepenny bit to our total.
We cannot remove larger coins and replace them with combinations of smaller ones, as this will always allow $10$ to be made.
Hence the highest total we can make with these coins so as not to make $10 \shillings$ is $15 \tfrac 3 4 \shillings$ or $15 \shillings 9 \oldpence$
== Sources ==
* : Solutions: $4$. -- Loose Cash
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/50 - Exploring Mount Neverest

== by : $50$ ==
;Exploring Mount Neverest
:Professor Walkingholme, one of the exploring party, was allotted the special task of making a complete circuit of the base of the mountain at a certain level.
:The circuit was exactly $100$ miles in length and he had to do it all alone on foot.
:He could walk $20$ miles a day, but he could only carry rations for $2$ days at a time,
::the rations for each day being packed in sealed boxes for convenience in dumping.
:He walked his full $20$ miles every day and consumed $1$ day's ration as he walked.
:What is the shortest time in which he could complete the circuit?
:This simple question will be found to form one of the most fascinating puzzles that we have considered for some time.
:It made a considerable demand on Professor Walkingholme's well-known ingenuity.
=== Click here for solution ===
== Historical Note ==
== Sources ==
* : Arithmetical and Algebraical Problems: Locomotion and Speed Puzzles: $50$. -- Exploring Mount Neverest
* : Arithmetical and Algebraical Problems: Speed & Distance Puzzles: $77$. Exploring Mount Neverest
Category:Henry Ernest Dudeney/Modern Puzzles
Category:Desert Crossing Problems
Henry Ernest Dudeney/Modern Puzzles/50 - Exploring Mount Neverest/Historical Note

== Historical Note on Henry Ernest Dudeney: Modern Puzzles: $50$ - Exploring Mount Neverest ==
In the words of himself:
:The idea was suggested to me by Mr. .
== Sources ==
* : Arithmetical and Algebraical Problems: Locomotion and Speed Puzzles: $50$. -- Exploring Mount Neverest
* : Arithmetical and Algebraical Problems: Speed & Distance Puzzles: $77$. Exploring Mount Neverest
Category:Historical Notes
Henry Ernest Dudeney/Modern Puzzles/50 - Exploring Mount Neverest/Solution

== by : $50$ ==
;Exploring Mount Neverest
== Solution ==
:$23 \tfrac 1 2$ days.
== Working ==
If the route is a straight march across a desert, that is, without being able to start at the end and work backwards, the minimum time is $86$ days, as follows:
== Sources ==
* : Solutions: $50$. -- Exploring Mount Neverest
* : Answers: $77$. Exploring Mount Neverest
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/51 - An Exceptional Number

== by : $51$ ==
;An Exceptional Number
:A number is formed of $5$ successive digits (not necessarily in regular order)
::so that the number formed by the first $2$ multiplied by the central digit will produce the number expressed by the last $2$.
:Thus if it were $1 \, 2 \,8 \, 9 \, 6$, then $12$ multiplied by $9$ produces $96$.
:But, unfortunately, $1, 2, 6, 8, 9$ are not successive numbers, so it will not do.
=== Click here for solution ===
=== Variant ===
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $51$. -- An Exceptional Number
* : Arithmetical and Algebraical Problems: Digital Puzzles: $102$. An Exceptional Number
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/51 - An Exceptional Number/Solution

== by : $51$ ==
;An Exceptional Number
== Solution ==
We have:
:$1 \, 3 \, 4 \, 5 \, 2$
as:
:$13 \times 4 = 52$
It remains to be shown that there are no more.
This could be implemented by means of a computer search.
== Sources ==
* : Solutions: $51$. -- An Exceptional Number
* : Answers: $102$. An Exceptional Number
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/52 - The Five Cards

== by : $52$ ==
;The Five Cards
:I have $5$ cards bearing the figures $1$, $3$, $5$, $7$ and $9$.
:How can I arrange them in a row so that the number formed by the $1$st pair multipied by the number formed with the last pair,
::with the central number subtracted,
::will produce a number composed of repetitions of one figure?
:::$\boxed 3 \boxed 1 \ \boxed 5 \ \boxed 7 \boxed 9$
:Thus, in the example I have shown, $31$ multiplied by $79$ and $5$ subtracted will produce $2444$,
::which would have been all right if that $2$ had happened to be another $4$.
:Of course, there must be $2$ solutions, for the pairs are clearly interchangeable.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $52$. -- The Five Cards
* : Arithmetical and Algebraical Problems: Digital Puzzles: $103$. The Five Cards
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/52 - The Five Cards/Solution

== by : $52$ ==
;The Five Cards
== Solution ==
:$\boxed 3 \boxed 9 \ \boxed 1 \ \boxed 5 \boxed 7$
or:
:$\boxed 5 \boxed 7 \ \boxed 1 \ \boxed 3 \boxed 9$
== Proof ==
Let $d_1$ and $d_2$ be the two $2$-f numbers at either end.
Let $s$ be the single-digit subtrahend.
Let $n$ be the repdigit that results from $d_1 \times d_2 - s$.
Suppose $n$ is odd.
Then $n + s = d_1 \times d_2$ is even.
But because $d_1$ and $d_2$ are both odd, this is impossible.
So $n$ is even.
Notice that:
:$15 \times 37 \le d_1 d_2 \le 93 \times 75$
that is:
:$555 \le d_1 d_2 \le 6975$
Hence we must have that $n$ is one of:
:$666, 888, 2222, 4444, 6666$
This leaves a small enough domain to perform an exhaustive search.
We factorize all $30$ possibilities of $n + s$, and filter the results using these criteria:
:$(1)$ Those with a prime factor greater than $100$ cannot be expressed as $d_1 \times d_2$.
:$(2)$ Any semiprime must be removed if their prime factorization involves even digits.
:$(3)$ Numbers divisible by $89$ should also be removed as no multiple of $89$ is expressible as $d_1$ or $d_2$.
We have:
This leaves:
:$675, 891, 897, 2223, 6669$
For $675$, it is divisible by $25$.
Since there is only one $5$ available, only one of $d_1, d_2$ can be divisible by $5$, and thus $25$.
The only multiple of $25$ not involving even digits, $75$, would result in:
:$675 = 75 \times 9$
so $675$ has no such expression.
For $891$, it is divisible by $11$.
All $2$-digit multiples of $11$ have repeated digits.
Therefore $891$ has no such expression.
For $897$, both $23$ and $23 \times 3 = 69$ have even digits.
Therefore $897$ has no such expression.
For $2223$, since:
:$13 \times 19 > 3^2 \times 19 > 3^2 \times 13 > 100$
we are forced to use:
:$\set {d_1, d_2} = \set {3 \times 13, 3 \times 19} = \set {39, 57}$
from which we derive our solutions:
:$\boxed 3 \boxed 9 \ \boxed 1 \ \boxed 5 \boxed 7$
or:
:$\boxed 5 \boxed 7 \ \boxed 1 \ \boxed 3 \boxed 9$
For $6669$, $13$ and $19$ cannot be multiplied together as their product is larger than $100$.
By Pigeonhole Principle, one of $13$ or $19$ must receive at least $3^2$ during multiplication.
But as shown before:
:$3^2 \times 19 > 3^2 \times 13 > 100$
so this is not possible.
Therefore $6667$ has no such expression.
We have exhausted all possibilities, so there are no more solutions.
== Sources ==
* : Solutions: $52$. -- The Five Cards
* : Answers: $103$. The Five Cards
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/53 - Squares and Digits

== by : $53$ ==
;Squares and Digits
:What is the smallest square number that terminates with the greatest possible number of similar digits?
:Thus the greatest possible number might be $5$ and the smallest square number with $5$ similar digits at the end might be $24677777$.
:But this is certainly not a square number.
:Of course, $0$ is not to be regarded as a digit.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $53$. -- Squares and Digits
* : Arithmetical and Algebraical Problems: Digital Puzzles: $104$. Squares and Digits
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/53 - Squares and Digits/Solution

== by : $53$ ==
;Squares and Digits
:What is the smallest square number that terminates with the greatest possible number of similar digits?
:Thus the greatest possible number might be $5$ and the smallest square number with $5$ similar digits at the end might be $24677777$.
:But this is certainly not a square number.
:Of course, $0$ is not to be regarded as a digit.
== Solution ==
:$1444 = 38^2$
== Proof ==
See Squares Ending in Repeated Digits.
== Sources ==
* : Solutions: $53$. -- Squares and Digits
* : Answers: $104$. Squares and Digits
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/54 - The Two Additions

== by : $54$ ==
;The Two Additions
:Can you arrange the following figures into two groups of $4$ figures each so that each group shall add to the same sum?
:::$1 \ 2 \ 3 \ 4 \ 5 \ 7 \ 8 \ 9$
:If you were allowed to reverse the $9$ so as to change it into the missing $6$ it would be very easy.
:For example, $1, 2, 7, 8$ and $3, 4, 5, 6$ add up to $18$ in both cases.
:But you are not allowed to make any such reversal.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $54$. -- The Two Additions
* : Arithmetical and Algebraical Problems: Digital Puzzles: $105$. The Two Additions
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/54 - The Two Additions/Solution

== by : $54$ ==
;The Two Additions
== Solution ==
Trick puzzle.
:$12 + 4 + 8 = 3 + 5 + 7 + 9 = 24$
There are probably many such answers.
 himself offers:
 173 85
 4 92
 --- ---
 177 177
== Sources ==
* : Solutions: $54$. -- The Two Additions
* : Answers: $105$. The Two Additions
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Category:Trick Puzzles
Henry Ernest Dudeney/Modern Puzzles/55 - The Repeated Quartette

== by : $55$ ==
;The Repeated Quartette
:If we multiply $64253$ by $365$ we get the product $23452345$, where the first $4$ figures are repeated.
:What is the largest number that we can multiply by $365$ in order to produce a similar product of $8$ figures repeated in the same order?
:There is no objection to a repetition of figures -- that is, the $4$ that are repeated need not be all different, as in the case shown.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $55$. -- The Repeated Quartette
* : Arithmetical and Algebraical Problems: Digital Puzzles: $106$. The Repeated Quartette
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/55 - The Repeated Quartette/Solution

== by : $55$ ==
;The Repeated Quartette
== Solution ==
:$273863$
== Proof ==
The key point is that a number of the form $\sqbrk {abcdabcd}_{10}$ is equal to $10001 \times \sqbrk {abcd}_{10}$.
We also have that:
:$10001 = 73 \times 137$
and:
:$365 = 5 \times 73$
So the $8$ digit number we end up with is a multiple of $5 \times 73 \times 137 = 50005$.
All we need to understand is that it has $2$ repeated sets of $4$ digits and ends in $5$.
The largest one of those is:
:$99959995 = 273863 \times 365$
and the job is done.
== Sources ==
* : Solutions: $55$. -- The Repeated Quartette
* : Answers: $106$. The Repeated Quartette
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/56 - Easy Division

== by : $56$ ==
;Easy Division
:To divide the number $8 \, 101 \, 265 \, 822 \, 784$ by $8$, all we need to do is transfer the $8$ from the beginning to the end!
:Can you find a number beginning with $7$ that can be divided by $7$ in the same simple manner?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $56$. -- Easy Division
* : Arithmetical and Algebraical Problems: Digital Puzzles: $107$. Easy Division
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/56 - Easy Division/Solution

== by : $56$ ==
;Easy Division
== Solution ==
:$7 \, 101 \, 449 \, 275 \, 362 \, 318 \, 840 \, 579$
== Proof ==
Let $7 N$ be the number in question.
This is an example of:
:Integer which is Multiplied by Last Digit when moving Last Digit to First
except here we are moving the first digit to last and dividing by it.
Hence $N$ is the recurring part of the fraction:
:$q = \dfrac {a_1} {10 a_1 - 1}$
where in this case $a_1 = 7$, giving us:
:$g = \dfrac 7 {69}$
We perform long division and see what we get:
 0.101449275362318840579710...
 -------------------------------
69)7.000000000000000000000000000
 6.9 345 552
 --- --- ---
 100 250 280
 ​ 69 207 276
 ​ --- --- ---
 310 430 400
 276 414 345
 --- --- ---
 340 160 550
 276 138 483
 --- --- ---
 640 220 670
 621 207 621
 --- --- ---
 190 130 490
 138 69 483
 --- --- ---
 520 610 70
 483 552 69
 --- --- --
 370 580 100
 345 552 ...
and we see that:
:$N = 7 \, 101 \, 449 \, 275 \, 362 \, 318 \, 840 \, 579$
== Sources ==
* : Solutions: $56$. -- Easy Division
* : Answers: $107$. Easy Division
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/57 - A Misunderstanding

== by : $57$ ==
;A Misunderstanding
:An American correspondent asks me to find a number composed of any number of digits that may be correctly divided by $2$
::by simply transferring the last figure to the beginning.
:He has apparently come across our last puzzle with the conditions wrongly stated.
:If you are to transfer the first figure to the end it is solved by $315 \, 789 \, 473 \, 684 \, 210 \, 526$,
::and a solution may easily be found from this with any given figure at the beginning.
:But if the figure is to be moved from the end to the beginning, there is no possible solution for the divisor $2$.
:But there is a solution for the divisor $3$.
:Can you find it?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $57$. -- A Misunderstanding
* : Arithmetical and Algebraical Problems: Digital Puzzles: $108$. A Misunderstanding
Category:Henry Ernest Dudeney/Modern Puzzles