nyuuzyou/wikis huggingface dataset:
Henry Ernest Dudeney/Puzzles and Curious Problems/114 - Elementary Arithmetic/Solution

== by : $114$ ==
;Elementary Arithmetic
== Solution ==
:$2 \tfrac 2 3$
== Proof ==
According to , this is:
:merely a sum in simple proportion:
If $5$ be $4$, then:
:$3 \tfrac 1 3$ is $3 \tfrac 1 3 \times \dfrac 4 5 = \dfrac {10 \times 4} {3 \times 5} = 2 \tfrac 2 3$
== Historical Note ==
== Sources ==
* : Solutions: $114$. -- Elementary Arithmetic
* : Answers: $189$. Elementary Arithmetic
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/115 - The Eight Cards

== by : $115$ ==
;The Eight Cards
:Rearrange these cards, moving as few as possible, so that the two columns add up alike.
:Can it be done?
:$\begin{array} {} \boxed 1 & \boxed 3 \\ \boxed 2 & \boxed 4 \\ \boxed 7 & \boxed 5 \\ \boxed 9 & \boxed 8 \\ \end{array}$
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $115$. -- The Eight Cards
* : Arithmetical and Algebraical Problems: Digital Puzzles: $139$. The Eight Cards
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/115 - The Eight Cards/Solution

== by : $115$ ==
;The Eight Cards
== Solution ==
:$\begin{array} {} \boxed 1 & \boxed 3 \\ \boxed 2 & \boxed 4 \\ \boxed 7 & \boxed 5 \\ \boxed 8 & \boxed 6 \\ \end{array}$
having inverted the $\boxed 9$ to make $\boxed 6$.
== Proof ==
As the numbers add up to $39$, it is clear this cannot be done by keeping the cards in the same orientation.
However, a veteran of these puzzles will instantly consider turning the $\boxed 9$ card round to make $\boxed 6$.
Then we see the columns add to $16$ and $20$ respectively.
So $2$ needs to be added to column $1$ and subtracted from $2$.
This can be done by exchanging the $\boxed 9$, turning it round to make $\boxed 6$, and the $\boxed 8$.
== Sources ==
* : Solutions: $115$. -- The Eight Cards
* : Answers: $139$. The Eight Cards
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/116 - Transferring the Figures

== by : $116$ ==
;Transferring the Figures
:If we wish to multiply $571 \, 428$ by $5$ and divide by $4$,
::we need only transfer the $5$ from the beginning to the end for the correct answer $714 \, 285$.
:Can you find a number that we can multiply by $4$ and then divide the product by $5$ in the same simple manner,
::by moving the first figure to the end?
:Of course $714 \, 285$, just given, would do if we were allowed to transfer from the end to the beginning.
:But it must be from the beginning to the end!
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $116$. -- Transferring the Figures
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $190$. Transferring the Figures
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/116 - Transferring the Figures/Solution

== by : $116$ ==
;Transferring the Figures
== Solution ==
We have that:
:$2 \, 173 \, 913 \, 043 \, 478 \, 260 \, 869 \, 565 \times \dfrac 4 5 = 1 \, 739 \, 130 \, 434 \, 782 \, 608 \, 695 \, 652$
== Proof ==
Suppose $N = \sqbrk {a_1 a_2 \dots a_k}$ satisfies the property above.
Consider the rational number represented by $q = 0.\dot {a_1} {a_2} \dots \dot {a_k}$.
By the property we have $\dfrac 4 5 q = 0.\dot {a_2} {a_3} \dots {a_k} \dot {a_1}$.
Also $10 q = a_1.\dot {a_2} {a_3} \dots {a_k} \dot {a_1}$.
Notice that the recurring part of $10 q$ is the same as the recurring part of $\dfrac 4 5 q$.
Subtracting, we get $9.2 q = a_1$, or $q = \dfrac {a_1} {9.2}$.
For the case $a_1 = 1$, we get:
:$q = \dfrac 1 {9.2} = 0.1 \dot 086956521739130434782 \dot 6$
which is not of the form $q = 0.\dot {a_1} {a_2} \dots \dot {a_k}$.
For the case $a_1 = 2$, we get:
:$q = \dfrac 2 {9.2} = 0. \dot 217391304347826086956 \dot 5$
which gives us $N = 2 \, 173 \, 913 \, 043 \, 478 \, 260 \, 869 \, 565$ as the smallest positive integer satisfying the property.
It may be interesting to note that $a_1 = 4, 6, 8$ also give solutions to the equation:
:$4 \, 347 \, 826 \, 086 \, 956 \, 521 \, 739 \, 130 \times \dfrac 4 5 = 3 \, 478 \, 260 \, 869 \, 565 \, 217 \, 391 \, 304$
:$6 \, 521 \, 739 \, 130 \, 434 \, 782 \, 608 \, 695 \times \dfrac 4 5 = 5 \, 217 \, 391 \, 304 \, 347 \, 826 \, 086 \, 956$
:$8 \, 695 \, 652 \, 173 \, 913 \, 043 \, 478 \, 260 \times \dfrac 4 5 = 6 \, 956 \, 521 \, 739 \, 130 \, 434 \, 782 \, 608$
where all these are all multiples of $2 \, 173 \, 913 \, 043 \, 478 \, 260 \, 869 \, 565$;
and all numbers of the form $\sqbrk {NN}, \sqbrk {NNN}, \sqbrk {NNNN}$ also satisfy the conditions.
== Sources ==
* : Solutions: $116$. -- Transferring the Figures
* : Answers: $190$. Transferring the Figures
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/117 - A Queer Addition

== by : $117$ ==
;A Queer Addition
:Write down $5$ odd figures so that they will add up to make $14$.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $117$. -- A Queer Addition
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $191$. A Queer Addition
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/117 - A Queer Addition/Solution

== by : $117$ ==
;A Queer Addition
== Solution ==
:$11 + 1 + 1 + 1 = 14$
== Sources ==
* : Solutions: $117$. -- A Queer Addition
* : Answers: $191$. A Queer Addition
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/118 - Six Simple Questions

== by : $118$ ==
;Six Simple Questions
:$(1)$ Deduct four thousand eleven hundred and a half from twelve thousand twelve hundred and twelve.
:$(2)$ Add $3$ to $182$, and make the total less than $20$.
:$(3)$ What two numbers multiplied together will produce seven?
:$(4)$ What three figures multiplied by five will make six?
:$(5)$ If five times four are $33$, what is the fourth of $20$?
:$(6)$ Find a fraction whose numerator is less than its denominator, but which, when reversed, shall remain of the same value.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $118$. -- Six Simple Questions
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $192$. Six Simple Questions
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/118 - Six Simple Questions/Solution

== by : $118$ ==
;Six Simple Questions
== Solution ==
The answers given by to these trick questions are as follows:
:$(1): \quad 8111 \tfrac 1 2$
:$(2): \quad 18 \tfrac 2 3$
:$(3): \quad 1$ and $7$
:$(4): \quad 1 \tfrac 1 5$
:$(5): \quad 8 \tfrac 1 4$
:$(6): \quad \tfrac 6 9$
but these are not necessarily the only solutions.
== Proof ==
:$(2): \quad$ Trick question.
:::There is a case for exploiting another trick: adding $3$ to $182$ and inserting a decimal point to make $18 \cdotp 5$.
:$(3): \quad 7$ is of course prime.
:$(4): \quad$ It does not say how those $3$ figures are to be combined.
:$(5): \quad$ If $5 \times 4 = 33$, then $20$ is identified with $33$.
:::Hence dividing $20$ by $4$ is dividing $33$ by $4$, which is $\tfrac {33} 4 = 8 \tfrac 1 4$.
:::Messy and unsatisfying question, as the initial concept is absurd and meaningless.
:$(6): \quad$ By "reversed", means "turned upside down".
:::But there is an argument for "reversed" meaning "reflected in the vertical axis", in which case $\tfrac 1 8$ does the job just as well.
== Sources ==
* : Solutions: $118$. -- Six Simple Questions
* : Answers: $192$. Six Simple Questions
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/119 - The Three Drovers

== by : $119$ ==
;The Three Drovers
:Three drovers with varied flocks met on the highway.
:Said Jack to Jim: "If I give you six pigs for a horse then you will have twice as many animals in your drove as I will have in mine."
:Said Dan to Jack: "If I give you fourteen sheep for a horse, then you'll have three times as many animals as I have got."
:Said Jim to Dan: "But if I give you four cows for a horse, then you'll have six times as many animals as I."
:There were no deals; but can you tell me how many animals there were in the three droves?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $119$. -- The Three Drovers
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $193$. The Three Drovers
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/119 - The Three Drovers/Solution

== by : $119$ ==
;The Three Drovers
== Solution ==
Jack had $11$ animals, Jim had $7$, and Dan had $21$.
== Proof ==
Let $a$, $b$ and $c$ be the numbers of animals in each of Jack's, Jim's and Dan's droves respectively.
We have:
== Sources ==
* : Solutions: $119$. -- The Three Drovers
* : Answers: $193$. The Three Drovers
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/11 - Distribution

== by : $11$ ==
;Distribution
:Nine persons in a party, $A$, $B$, $C$, $D$, $E$, $F$, $G$, $H$, $K$, did as follows:
::First $A$ gave each of the others as much money as he (the receiver) already held;
::then $B$ did the same; then $C$; and so on to the last,
::$K$ giving to each of the other eight persons the amount the receiver then held.
:Then it was found that each of the nine persons held the same amount.
:Can you find the smallest amount in pence that each person could have originally held?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Money Puzzles: $11$. -- Distribution
* : Arithmetical and Algebraical Problems: Money Puzzles: $18$. Distribution
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/11 - Distribution/Solution

== by : $11$ ==
;Distribution
== Solution ==
:$10$, $19$, $37$, $73$, $145$, $289$, $577$, $1153$, $2305$.
$A$ is the largest holder, progressing to $K$ being the one whose holding is $10$.
At the end of the game, all hold $2^9 = 512$ pence.
== Proof ==
The question does not state this, but it is implicit in the nature of the answer that everybody starts with a whole number of pence.
This is another instance of Henry Ernest Dudeney: Modern Puzzles 12: A Weird Game.
The smallest number originally held is $1$ more than the number of persons.
The others are obtained by doubling and subtracting $1$.
== Sources ==
* : Solutions: $11$. -- Distribution
* : Answers: $18$. Distribution
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/120 - Proportional Representation

== by : $120$ ==
;Proportional Representation
:In a local election, there were ten names of candidates on a proportional representation ballot paper.
:Voters should place No. $1$ against the candidate of their first choice.
:They might also place No. $2$ against the candidate of their second choice,
::and so on until all the ten candidates have numbers placed against their names.
:The voters must mark their first choice, and any others may be marked or not as they wish.
:How many different ways might the ballot paper be marked by the voter?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $120$. -- Proportional Representation
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $194$. Proportional Representation
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/120 - Proportional Representation/Solution

== by : $120$ ==
;Proportional Representation
:In a local election, there were ten names of candidates on a proportional representation ballot paper.
:Voters should place No. $1$ against the candidate of their first choice.
:They might also place No. $2$ against the candidate of their second choice,
::and so on until all the ten candidates have numbers placed against their names.
:The voters must mark their first choice, and any others may be marked or not as they wish.
:How many different ways might the ballot paper be marked by the voter?
== Solutions ==
:$9 \, 684 \, 100$
== Proof ==
Let $n$ denoted the number of different ways to vote.
Let the voter vote for $k$ candidates, where $1 \le k \le 10$.
Each way of making $k$ votes is a $k$-permutation on $10$ objects, denoted ${}^{10} P_k$.
Thus:
== Also see ==
* Count of All Permutations on $n$ Objects
== Sources ==
* : Solutions: $120$. -- Proportional Representation
* : Answers: $194$. Proportional Representation
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/121 - Find the Numbers

== by : $121$ ==
;Find the Numbers
:Can you find $2$ numbers composed only of ones which give the same result by addition and multiplication?
:Of course $1$ and $11$ are very near, but they will not quite do,
::because added they make $12$, and multiplied they make only $11$.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $121$. -- Find the Numbers
* : Arithmetical and Algebraical Problems: Digital Puzzles: $140$. Find the Numbers
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/121 - Find the Numbers/Solution

== by : $121$ ==
;Find the Numbers
== Solution ==
:$11$ and $1 \cdotp 1$
== Proof ==
 has already raised this question in his Modern Puzzles: $93$ - Sum Equals Product, where he shows that:
:$y = \dfrac x {x - 1}$
for any pair $x$ and $y$ such that $x y = x + y$.
The only time $x$ and $y$ are both integers is when $x = y = 2$.
So there's a trick to look out for.
Suppose $x = 11$.
We have:
:$y = \dfrac {11} {11 - 1} = \dfrac {11} {10}$
and we notice that:
:$\dfrac {11} {10} = 1 \cdotp 1$
We note that:
:$11 + 1 \cdotp 1 = 1 \cdotp 21 = 11 \times 1 \cdotp 1$
== Sources ==
* : Solutions: $121$. -- Find the Numbers
* : Answers: $140$. Find the Numbers
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/122 - A Question of Cubes

== by : $122$ ==
;A Question of Cubes
:From Sum of Sequence of Cubes, the cubes of successive numbers, starting from $1$, sum to a square number.
:Thus the cubes of $1$, $2$, $3$ (that is, $1$, $8$, $27$), add to $36$, which is the square of $6$.
:If you are forbidden to use the $1$, the lowest answer is the cubes of $23$, $24$ and $25$, which together equal $204^2$.
:What is the next lowest number, using more than three consecutive cubes and as many more as you like, but excluding $1$?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $122$. -- A Question of Cubes
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $195$. A Question of Cubes
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/122 - A Question of Cubes/Solution

== by : $122$ ==
;A Question of Cubes
== Solution ==
:$\ds \sum_{n \mathop = 14}^{25} n^3 = 14^3 + 15^3 + 16^3 + \cdots + 25^3 = 97 \, 344 = 312^2$
The next one after that is:
:$\ds \sum_{n \mathop = 25}^{29} n^3 = 25^3 + 26^3 + 27^3 + 28^3 + 29^3 = 99 \, 225 = 315^2$
== Proof ==
== Sources ==
* : Solutions: $122$. -- A Question of Cubes
* : Answers: $195$. A Question of Cubes
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/123 - Two Cubes

== by : $123$ ==
;Two Cubes
:Can you find two cube numbers in integers whose difference shall be a square number?
:Thus the cube of $3$ is $27$, and the cube of $2$ is $8$,
::but the difference, $19$, is not here a square number.
:What is the smallest possible case?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $123$. -- Two Cubes
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $196$. Two Cubes
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/123 - Two Cubes/Solution

== by : $123$ ==
;Two Cubes
== Solution ==
The answer given by is:
:$8^3 - 7^3 = 512 - 343 = 169 = 13^2$
but it can be noticed that:
:$2^3 - \paren {-2}^3 = 8 - \paren {-8} = 16 = 4^2$
which, because $8$ and $-8$ are both integers, is also a valid solution.
== Sources ==
* : Solutions: $123$. -- Two Cubes
* : Answers: $196$. Two Cubes
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/124 - Cube Differences

== by : $124$ ==
;Cube Differences
:If we wanted to find a way of making the number $1 \, 234 \, 567$ the difference between two squares,
::we could of course write down $517 \, 284$ and $617 \, 283$ --
::a half of the number plus $\tfrac 1 2$ and minus $\tfrac 1 2$ respectively to be squared.
:But it will be found a little more difficult to discover two cubes the difference of which is $1 \, 234 \, 567$.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $124$. -- Cube Differences
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $197$. Cube Differences
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/124 - Cube Differences/Solution

== by : $124$ ==
;Cube Differences
== Solution ==
:$642^3 - 641^3 = 1 \, 234 \, 567$
== Proof ==
We need to find $a$ and $b$ such that $a^3 - b^3 = 1 \, 234 \, 567$.
The assumption is that $a, b \in \Z_{>0}$ so we are not concerned about negative numbers.
First we use Difference of Two Cubes to obtain:
:$a^3 - b^3 = \paren {a - b} \paren {a^2 + a b + b^2}$
We have that:
:$1 \, 234 \, 567 = 127 \times 9721$
from which it follows either that:
:$a - b = 127$ and $a^2 + b^2 + a b = 9721$
or:
:$a - b = 1$ and $a^2 + b^2 + a b = 1 \, 234 \, 567$
Trying the first of these:
However, the above quadratic does not have integer solutions.
So we try:
It is the positive solution we need here, which leads to:
:$a = 642$
:$b = 641$
== Sources ==
* : Solutions: $124$. -- Cube Differences
* : Answers: $197$. Cube Differences
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/125 - Accommodating Squares

== by : $125$ ==
;Accommodating Squares
:Can you find two three-digit square numbers (no noughts) that, when put together, will form a six-digit square number?
:Thus, $324$ and $900$ (the squares of $18$ and $30$) make $324 \, 900$, the square of $570$, only there it happens there are two noughts.
:There is only one answer.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $125$. -- Accommodating Squares
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $198$. Accommodating Squares
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/125 - Accommodating Squares/Solution

== by : $125$ ==
;Accommodating Squares
== Solution ==
:$475^2 = 225 \, 625$, while $15^2 = 225$ and $25^2 = 625$.
== Also see ==
* Squares whose Digits can be Separated into 2 other Squares
* 475
* 
== Sources ==
* : Solutions: $125$. -- Accommodating Squares
* : Answers: $198$. Accommodating Squares
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/126 - Making Squares

== by : $126$ ==
;Making Squares
:Find three whole numbers in arithmetic progression,
::the sum of every two of which shall be a square.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $126$. -- Making Squares
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $199$. Making Squares
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/126 - Making Squares/Solution

== by : $126$ ==
;Making Squares
:Find three whole numbers in arithmetic progression,
::the sum of every two of which shall be a square.
== Solution ==
We have the set:
:$\set {482, 3362, 6242}$
which have a common difference of $2880$.
We see that:
== Proof ==
== Sources ==
* : Solutions: $126$. -- Making Squares
* : Answers: $199$. Making Squares
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/127 - Find the Squares

== by : $127$ ==
;Find the Squares
:What is the number which, when added separately to $100$ and $164$, make them both perfect square numbers?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $127$. -- Find the Squares
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $200$. Find the Squares
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/127 - Find the Squares/Solution

== by : $127$ ==
;Find the Squares
== Solution ==
:$125$
== Proof ==
In this context , who is writing primarily for a non-mathematical audience, is looking for a strictly positive integer.
In puzzles such as this, where his audience consists of readers of popular journals, the concept of negative numbers will not be considered unless specifically prompted and primed.
We are looking for two square numbers whose difference is $64$, that is, $31 + 33$.
We have that:
:$31 = 2 \times 15 + 1$
:$33 = 2 \times 16 + 1$
By the Odd Number Theorem, those two square numbers are $15^2$ and $17^2$.
Thus we have:
:$15^2 = 225$
:$17^2 = 289$
Hence the number that needs to be added to $100$ and $164$, as required, is $125$.
== Sources ==
* : Solutions: $127$. -- Find the Squares
* : Answers: $200$. Find the Squares
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/128 - Forming Squares

== by : $128$ ==
;Forming Squares
:An officer arranged his men in a solid square, and had $39$ men left over.
:He then started increasing the number of men on a side by one, but found that $50$ new men would be needed to complete the new square.
:Can you tell me how many men the officer had?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $128$. -- Forming Squares
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $201$. Forming Squares
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/128 - Forming Squares/Solution

== by : $128$ ==
;Forming Squares
:An officer arranged his men in a solid square, and had $39$ men left over.
:He then started increasing the number of men on a side by one, but found that $50$ new men would be needed to complete the new square.
:Can you tell me how many men the officer had?
== Solution ==
:$1975$ men.
== Proof ==
Let the original square have $n$ men on each side.
Then we have:
So the officer had $44^2 + 39 = 1975$ men.
== Sources ==
* : Solutions: $128$. -- Forming Squares
* : Answers: $201$. Forming Squares
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions