nyuuzyou/wikis huggingface dataset:
Henry Ernest Dudeney/Puzzles and Curious Problems/73 - The Tube Stairs/Solution

== by : $73$ ==
;The Tube Stairs
== Solution ==
:$839$ steps.
== Proof ==
Let $n$ be the number of steps in the tube station.
Let $1$ be added to $n$ to make $m$.
We have that:
:$m$ is divisible by $2$
:$m$ is divisible by $3$
and so on, until:
:$m$ is divisible by $7$
Thus $m$ is divisible by the lowest common multiple of $\set {2, 3, 4, 5, 6, 7}$.
Hence we calculate:
and we note in passing that $420 = 21 \times 20$ and so is also divisible by $20$
As we are told there are nearly $1000$ steps, it is clear that $420$ is too small for $m$, so we multiply it by $2$ to get $840$.
We then note that $3 \times 420$ is way over $1000$ so cannot be the value for $m$.
So the only possible value for $m$ is indeed $840$.
Hence:
:$n = 840 - 1 = 839$
== Sources ==
* : Solutions: $73$. -- The Tube Stairs
* : Answers: $87$. The Subway Stairs
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/74 - The Omnibus Ride

== by : $74$ ==
;The Omnibus Ride
:George treated his best girl to a ride on a motor omnibus,
::but on account of his limited resources it was necessary that they should walk back.
:Now, if the bus goes at the rate of nine miles an hour and they walk at the rate of three miles an hour,
::how far can they ride so they may be back in eight hours?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Locomotion and Speed Puzzles: $74$. -- The Omnibus Ride
* : Arithmetical and Algebraical Problems: Speed & Distance Puzzles: $88$. The Bus Ride
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/74 - The Omnibus Ride/Solution

== by : $74$ ==
;The Omnibus Ride
== Solution ==
:$18$ miles.
== Proof ==
Let $D$ be the distance they can travel.
Let $T_1$ and $T_2$ be the lengths of time taken for the outward and return journeys respectively.
We have:
== Sources ==
* : Solutions: $74$. -- The Omnibus Ride
* : Answers: $88$. The Bus Ride
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/75 - A Question of Transport

== by : $75$ ==
;A Question of Transport
:Twelve soldiers had to get to a place twenty miles distant with the quickest possible dispatch,
::and all had to arrive at the same time.
:They requisitioned the services of a man with a small motor-car.
:"I can do twenty miles an hour," he said, "but I cannot carry more than four men at a time.
::At what rate do you walk?"
:"All of us can do a steady four miles an hour," they replied.
:"Very well," exclaimed the driver, "then I will go ahead with four men,
::drop them somewhere on the road to walk,
::then return and pick up four more (who will then be somewhere on the road),
::drop them off also, and return for the last four.
:So all you have to do is to keep walking while you are on your feet, and I will do the rest."
:As they started at noon, what was the exact time that they all arrived together?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Locomotion and Speed Puzzles: $75$. -- A Question of Transport
* : Arithmetical and Algebraical Problems: Speed & Distance Puzzles: $89$. A Question of Transport
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/75 - A Question of Transport
Henry Ernest Dudeney/Puzzles and Curious Problems/75 - A Question of Transport/Solution

== by : $75$ ==
;A Question of Transport
== Solution ==
:$14:36$
== Proof ==
Let us refer to the start and finish points as $A$ and $B$ respectively.
Hence the object of the operation is to get all $12$ men from $A$ to $B$ as quickly as possible.
Let $d_1$ miles be the distance from $A$ to which the driver takes the first batch of soldiers.
Let $t_1$ hours past noon be the time he drops them off.
Let $d_2$ miles be the distance from $A$ to which the driver returns to pick up the second batch.
Let $t_2$ hours past noon be the time he picks them up.
Let $d_3$ miles be the distance from $A$ to which the driver takes the second batch of soldiers.
Let $t_3$ hours past noon be the time he drops them off.
Let $d_4$ miles be the distance from $A$ to which the driver returns to pick up the third batch.
Let $t_4$ hours past noon be the time he picks them up.
Let $t_5$ hours past noon be the time they all arrive at $B$.
The following diagram illustrates the situation.
The journey of the driver is shown in blue, while those of the walkers are shown in red.
:
We have:
These can be expressed more usefully as:
This set of simultaneous linear equations can be expressed conveniently in matrix form as:
:$\begin {pmatrix}
 1 & 0 & 0 & 0 & -20 & 0 & 0 & 0 & 0 \\
 0 & 1 & 0 & 0 & 0 & -4 & 0 & 0 & 0 \\
 1 & -1 & 0 & 0 & 20 & -20 & 0 & 0 & 0 \\
-1 & 0 & 1 & 0 & 4 & 0 & -4 & 0 & 0 \\
 0 & -1 & 1 & 0 & 0 & 20 & -20 & 0 & 0 \\
 0 & -1 & 0 & 1 & 0 & 4 & 0 & -4 & 0 \\
 0 & 0 & 1 & -1 & 0 & 0 & 20 & -20 & 0 \\
 0 & 0 & 1 & 0 & 0 & 0 & -4 & 0 & 4 \\
 0 & 0 & 0 & 1 & 0 & 0 & 0 & -20 & 20 \\
\end {pmatrix} \begin {pmatrix} d_1 \\ d_2 \\ d_3 \\ d_4 \\ t_1 \\ t_2 \\ t_3 \\ t_4 \\ t_5 \end {pmatrix} = \begin {pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 20 \\ 20 \end {pmatrix}$
It remains to solve this matrix equation.
In echelon form, this gives:
We are interested only in $t_5$, which we see is $\dfrac {13} 5$, that is, $2 \dfrac 3 5$ hours after noon.
== Sources ==
* : Solutions: $75$. -- A Question of Transport
* : Answers: $89$. A Question of Transport
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/75 - A Question of Transport
Henry Ernest Dudeney/Puzzles and Curious Problems/75 - A Question of Transport/Working

== Working for by : $75$ -- A Question of Transport ==
The system of simultaneous equations in matrix form:
:$\begin {pmatrix}
 1 & 0 & 0 & 0 & -20 & 0 & 0 & 0 & 0 \\
 0 & 1 & 0 & 0 & 0 & -4 & 0 & 0 & 0 \\
 1 & -1 & 0 & 0 & 20 & -20 & 0 & 0 & 0 \\
-1 & 0 & 1 & 0 & 4 & 0 & -4 & 0 & 0 \\
 0 & -1 & 1 & 0 & 0 & 20 & -20 & 0 & 0 \\
 0 & -1 & 0 & 1 & 0 & 4 & 0 & -4 & 0 \\
 0 & 0 & 1 & -1 & 0 & 0 & 20 & -20 & 0 \\
 0 & 0 & 1 & 0 & 0 & 0 & -4 & 0 & 4 \\
 0 & 0 & 0 & 1 & 0 & 0 & 0 & -20 & 20 \\
\end {pmatrix} \begin {pmatrix} d_1 \\ d_2 \\ d_3 \\ d_4 \\ t_1 \\ t_2 \\ t_3 \\ t_4 \\ t_5 \end {pmatrix} = \begin {pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 20 \\ 20 \end {pmatrix}$
when converted to echelon form, gives:
:$\begin {pmatrix} 
 1 & 0 & 0 & 0 & -20 & 0 & 0 & 0 & 0 \\
 0 & 1 & 0 & 0 & 0 & -4 & 0 & 0 & 0 \\
 0 & 0 & 1 & 0 & -16 & 0 & -4 & 0 & 0 \\
 0 & 0 & 0 & 1 & 0 & 0 & 0 & -4 & 0 \\
 0 & 0 & 0 & 0 & 1 & 1 & -1 & 0 & 0 \\
 0 & 0 & 0 & 0 & 0 & 1 & -5/2 & 3/2 & 0 \\
 0 & 0 & 0 & 0 & 0 & 0 & 1 & -1 & -1/6 \\
 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 5/6 \\
 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 
\end {pmatrix} \begin {pmatrix} d_1 \\ d_2 \\ d_3 \\ d_4 \\ t_1 \\ t_2 \\ t_3 \\ t_4 \\ t_5 \end {pmatrix} = \begin {pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ -5/6 \\ 25/6 \\ 13/5 \end {pmatrix}$
== Proof ==
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/75 - A Question of Transport
Henry Ernest Dudeney/Puzzles and Curious Problems/76 - How Far Was It

== by : $76$ ==
;How Far Was It?
:"The steamer," remarked one of our officers home from the East, "was able to go twenty miles an hour down-stream,
::but could only do fifteen miles an hour upstream.
:So, of course, she took five hours longer in coming up than in going down."
:One could not resist working out mentally the distance from point to point.
:What was it?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Locomotion and Speed Puzzles: $76$. -- How Far Was It?
* : Arithmetical and Algebraical Problems: Speed & Distance Puzzles: $90$. How Far Was It?
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/76 - How Far Was It/Solution

== by : $76$ ==
;How Far Was It?
== Solution ==
:$300$ miles.
== Proof ==
Let $D$ be the distance travelled.
Let $T$ be the time taken to travel $D$ downstream.
We have:
== Sources ==
* : Solutions: $76$. -- How Far Was It?
* : Answers: $90$. How Far Was It?
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/77 - Out and Home

== by : $77$ ==
;Out and Home
:Mr Wilkinson walks from his country house into the neighbouring town at the rate of five miles per hour,
::and, as he is a little tired, he makes the return journey at the rate of three miles per hour.
:As the double journey takes him exactly seven hours, can you tell me the distance from his house to the town?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Locomotion and Speed Puzzles: $77$. -- Out and Home
* : Arithmetical and Algebraical Problems: Speed & Distance Puzzles: $91$. Out and Home
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/77 - Out and Home/Solution

== by : $77$ ==
;Out and Home
== Solution ==
:$13 \tfrac 1 8$ miles.
== Proof ==
Let $D$ be the distance to town.
Let $t_1$ and $t_2$ be the times taken to walk to town and to walk back respectively.
We have:
== Sources ==
* : Solutions: $77$. -- Out and Home
* : Answers: $91$. Out and Home
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/78 - The Meeting Cars

== by : $78$ ==
;The Meeting Cars
:The Crackhams made their first stop at Bugleminster, where they were to spend the night at a friend's house.
:This friend was to leave home at the same time and ride to London to put up at the Crackhams' house.
:They took the same route, and each car went at its own uniform speed.
:They kept a look-out for one another, and met forty miles from Bugleminster.
:George that evening worked out the following little puzzle:
:"I find that if, on our respective arrivals, we had each at once proceeded on the return journey at the same speeds
::we should meet $48$ miles from London."
:If this were so, what is the distance from London to Bugleminster?''
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Locomotion and Speed Puzzles: $78$. -- The Meeting Cars
* : Arithmetical and Algebraical Problems: Speed & Distance Puzzles: $92$. The Meeting Cars
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/78 - The Meeting Cars/Solution

== by : $78$ ==
;The Meeting Cars
== Solution ==
:$72$ miles.
== Proof ==
let $D$ be the distance from Bugleminster to London.
Let $v_1$ and $v_2$ miles per hour be the speed of Crackham and his friend respectively.
Let $t_1$ hours be the time after setting out that they met.
Let $t_2$ hours be the time after originally setting out that they would have met for the second time.
We have:
== Sources ==
* : Solutions: $78$. -- The Meeting Cars
* : Answers: $92$. The Meeting Cars
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/79 - A Cycle Race

== by : $79$ ==
;A Cycle Race
:Two cyclists race on a circular track.
:Brown can ride once round the track in six minutes,
::and Robinson in four minutes.
:In how many minutes will Robinson overtake Brown?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Locomotion and Speed Puzzles: $79$. -- A Cycle Race
* : Arithmetical and Algebraical Problems: Speed & Distance Puzzles: $93$. A Bicycle Race
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/79 - A Cycle Race/Solution

== by : $79$ ==
;A Cycle Race
== Solution ==
:$12$ minutes.
== Proof ==
After $4$ minutes, Robinson has completed one circuit, while Brown is $\dfrac 1 3$ the way round his first circuit.
After $8$ minutes, Robinson has completed two circuits, while Brown has completed $1 \tfrac 1 3$ circuits, but Robinson has not yet caught up with Brown.
After $12$ minutes, Robinson has completed $3$ circuits, while Brown has just about completed $2$ circuits, which is the point Robinson overtakes Brown.
== Sources ==
* : Solutions: $79$. -- A Cycle Race
* : Answers: $93$. A Bicycle Race
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/7 - Squaring Pocket-Money

== by : $7$ ==
;Squaring Pocket-Money
:A man has four different English coins in his pocket,
::and their sum in pence was a square number.
:He spent one of the coins, and the sum of the remainder in shillings was a square number.
:He then spent one of the three, and the sum of the other two in pence was a square number.
:And when he deducted the number of farthings in one of them from the number of halfpennies in the other, the remainder was a square number.
:What were the coins?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Money Puzzles: $7$. -- Squaring Pocket-Money
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/7 - Squaring Pocket-Money/Solution

== by : $7$ ==
;Squaring Pocket-Money
== Solution ==
The solution given by :
:The coins were a half crown, a shilling, a sixpence and a penny ($49$ pence).
:He spent the penny, and the number of shillings was then $4$.
:He then spent the shilling, and the number of pence was $36$.
:And as the half crown equals $60$ halfpence, and the sixpence equals $24$ farthings, their difference is $36$ -- again a square.
One suggestion is that the coins were a groat, two threepenny bits and a sixpence.
However, it may be assumed that the groat is disallowed on account of it not being legal tender at the time of writing this puzzle.
== Sources ==
* : Solutions: $7$. -- Squaring Pocket-Money
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/80 - A Little Train Puzzle

== by : $80$ ==
;A Little Train Puzzle
:A non-stop express going sixty miles an hour starts from Bustletown for Ironchester,
::and another non-stop express going forty miles an hour starts at the same time from Ironchester for Bustletown.
:How far apart are they exactly an hour before they meet?
:As I have failed to find these cities on any map or in any gazetteer, I cannot state the distance between them,
::so we will just assume that it is somewhere over $250$ miles.
:If this little puzzle gives the reader much trouble he will certainly smile when he sees the answer.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Locomotion and Speed Puzzles: $80$. -- A Little Train Puzzle
* : Arithmetical and Algebraical Problems: Speed & Distance Puzzles: $94$. A Little Train Puzzle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/80 - A Little Train Puzzle/Solution

== by : $80$ ==
;A Little Train Puzzle
== Solution ==
:$100$ miles.
== Proof ==
The first train is $1$ hour away from the meeting place, which at $60$ miles per hour is $60$ miles away.
The second train is also $1$ hour away from the meeting place, which at $60$ miles per hour is $40$ miles away.
$40$ miles plus $60$ miles is $100$ miles.
== Sources ==
* : Solutions: $80$. -- A Little Train Puzzle
* : Answers: $94$. A Little Train Puzzle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/81 - An Irish Jaunt

== by : $81$ ==
;An Irish Jaunt
:Colonel Crackham was going from Boghooley to Ballyfoyne, using Pat Doyle's horse and cart,
::which moved at a steady rate, but more slowly than would normally be expected.
:After they had been on the road for $20$ minutes, they had travelled half as far from Boghooley than it was to Pigtown.
:They stopped for refreshment at Pigtown when they arrived there.
:Five miles further on, it was half as far to Ballyfoyne as it was from Pigtown.
:After another hour they had arrived in Ballyfoyne.
:What is the distance from Boghooley to Ballyfoyne?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Locomotion and Speed Puzzles: $81$. -- An Irish Jaunt
* : Arithmetical and Algebraical Problems: Speed & Distance Puzzles: $95$. An Irish Jaunt
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/81 - An Irish Jaunt/Solution

== by : $81$ ==
;An Irish Jaunt
== Solution ==
:$10$ miles.
== Proof ==
Let $d_1$ and $d_2$ miles be the distances from Boghooley to Pigtown and from Pigtown to Ballyfoyne respectively.
Let $v$ miles per hour be the speed of transport.
We have:
== Sources ==
* : Solutions: $81$. -- An Irish Jaunt
* : Answers: $95$. An Irish Jaunt
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/82 - A Walking Problem

== by : $82$ ==
;A Walking Problem
:A man taking a walk in the country on turning round saw a friend of his walking $400$ yards behind in his direction.
:They each walked $200$ yards in a direct line, with their faces towards each other,
::and you would suppose that they must have met.
:Yet they found that after their $200$ yards walk that they were still $400$ yards apart.
:Can you explain?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Locomotion and Speed Puzzles: $82$. -- A Walking Problem
* : Arithmetical and Algebraical Problems: Speed & Distance Puzzles: $96$. A Walking Problem
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/82 - A Walking Problem/Solution

== by : $82$ ==
;A Walking Problem
:A man taking a walk in the country on turning round saw a friend of his walking $400$ yards behind in his direction.
:They each walked $200$ yards in a direct line, with their faces towards each other,
::and you would suppose that they must have met.
:Yet they found that after their $200$ yards walk that they were still $400$ yards apart.
:Can you explain?
== Solution ==
After the first man had turned around to see his friend, he walked backwards for $200$ yards in the same direction he had been walking.
== Sources ==
* : Solutions: $82$. -- A Walking Problem
* : Answers: $96$. A Walking Problem
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/83 - Three Different Digits

== by : $83$ ==
;Three Different Digits
:Find all $3$-digit numbers with distinct digits that are divisible by the square of the sum of those digits.
:For example, $112$ is such that $\paren {1 + 1 + 2}^2 = 16$, and $16$ is a divisor of $112$.
:However, the digits of $112$ are not all different, so it does not count.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $83$. -- Three Different Digits
* : Arithmetical and Algebraical Problems: Digital Puzzles: $122$. Three Different Digits
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/83 - Three Different Digits
Henry Ernest Dudeney/Puzzles and Curious Problems/83 - Three Different Digits/Solution

== by : $83$ ==
;Three Different Digits
== Solution ==
:$162$, $243$, $324$, $392$, $405$, $512$, $605$, $648$, $810$ and $972$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Solutions: $83$. -- Three Different Digits
* : Answers: $122$. Three Different Digits
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/83 - Three Different Digits
Henry Ernest Dudeney/Puzzles and Curious Problems/83 - Three Different Digits/Solution/Proof 1

== by : $83$ ==
;Three Different Digits
== Solution ==
== Proof ==
From , we have A072081, which provides us with all the numbers divisible by the square of the sum of their digits in base $10$:
:$1, 10, 20, 50, 81, 100, 112, 162, 200, 243, 324, 392, 400, 405, 500, 512, 605, 648, 810, 972, 1000, \ldots$
from which we pick up those required.
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/83 - Three Different Digits
Henry Ernest Dudeney/Puzzles and Curious Problems/83 - Three Different Digits/Solution/Proof 2

== by : $83$ ==
;Three Different Digits
== Solution ==
== Proof ==
Here is the source code of a program in C which finds them all:
 #include <stdio.h>
 
 int main()
 {
 for (int i = 100; i < 1000; i++)
 {
 int a = i / 100;
 int b = (i / 10) % 10;
 int c = i % 10;
 if (a != b && b != c && c!= a && i % ((a + b + c)*(a + b + c)) == 0)
 printf("%d, ", i);
 }
 }
Output:
 162, 243, 324, 392, 405, 512, 605, 648, 810, 972, 
Try it online!
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/83 - Three Different Digits
Henry Ernest Dudeney/Puzzles and Curious Problems/84 - Find the Cube

== by : $84$ ==
;Find the Cube
:A number increased by its cube is $592 \, 788$.
:What is that number?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $84$. -- Find the Cube
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/84 - Find the Cube/Solution

== by : $84$ ==
;Find the Cube
== Solution ==
:$84$
== Proof ==
We are looking for $n$ such that:
:$n^3 + n = 592 \, 788$
First we see that $\sqrt [3] {592 \, 788}$ is approximately $84 \cdotp 00...$
We calculate $84^3$ and get:
:$84^3 = 592 \, 704$
and see that:
:$84 + 592 \, 704 = 592 \, 788$
and our task is complete.
== Sources ==
* : Solutions: $84$. -- Find the Cube
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/85 - Squares and Triangulars

== by : $85$ ==
;Squares and Triangulars
:What is the third lowest number that is both a triangular and a square?
:$1$ and $36$ are the two lowest which fulfil the conditions.
:What is the next number?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $85$. -- Squares and Triangulars
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $187$. Squares and Triangulars
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/85 - Squares and Triangulars/Solution

== by : $85$ ==
;Squares and Triangulars
== Solution ==
:$1225$
== Proof ==
See Triangular Numbers which are also Square.
== Sources ==
* : Solutions: $85$. -- Squares and Triangulars
* : Answers: $187$. Squares and Triangulars
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/86 - Digits and Cubes

== by : $86$ ==
;Digits and Cubes
:Find all $5$-digit squares such that:
::the number formed from the first $2$ digits added to that formed by the last $2$ digits form a cube.
:For example, $141^2 = 19 \, 881$, where $19 + 81 = 100$.
:But $100$ is a square not a cube.
:How many are there altogether?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $86$. -- Digits and Cubes
* : Arithmetical and Algebraical Problems: Digital Puzzles: $123$. Digits and Cubes
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/86 - Digits and Cubes/Solution

== by : $86$ ==
;Digits and Cubes
== Solution ==
There are $3$ solutions:
== Proof ==
A computer program can be used to perform an exhaustive search.
== Sources ==
* : Solutions: $86$. -- Digits and Cubes
* : Answers: $123$. Digits and Cubes
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/87 - Reversing the Digits

== by : $87$ ==
;Reversing the Digits
:What $9$-digit number, when multiplied by $123\, 456 \, 789$, gives a product ending in $987 \, 654 \, 321$?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $87$. -- Reversing the Digits
* : Arithmetical and Algebraical Problems: Digital Puzzles: $124$. Reversing the Digits
Category:Henry Ernest Dudeney/Puzzles and Curious Problems