nyuuzyou/wikis huggingface dataset:
Henry Ernest Dudeney/Puzzles and Curious Problems/167 - The Cancelled Cheque

== by : $167$ ==
;The Cancelled Cheque
Bankers at a certain bank would cancel their paid cheques by punching star-shaped holes in them.
:There was a case in which they happened to punch out the $6$ figures that form the number of the cheque.
:The puzzle is to find out what those figures were.
:It was a square number multiplied by $113$, and when divided into three $2$-figure numbers,
::each of these three numbers was a square number.
:Can you find the number of the cheque?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $167$. -- The Cancelled Cheque
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/167 - The Cancelled Cheque/Solution

== by : $167$ ==
;The Cancelled Cheque
== Solution ==
:$81 \, 64 \, 25 = 113 \times 85^2$
== Proof ==
Let $n$ be the $6$-digit number on the cheque.
We have that $n = 113 m$ where $m$ is a square number.
Because $100 \, 000 \le n \le 999 \, 999$ we have that:
:$885 \le m \le 8849$.
From Square Modulo 5: Corollary, no square number ends with $2$, $3$, $7$ or $8$.
So $m$, being square, ends in $0$, $1$, $4$, $5$, $6$ or $9$.
If $m$ ends in $1$, then $n$ ends in $3$, so is not a square.
If $m$ ends in $4$, then $n$ ends in $2$, so is not a square.
If $m$ ends in $6$, then $n$ ends in $8$, so is not a square.
If $m$ ends in $9$, then $n$ ends in $7$, so is not a square.
Suppose $m$ ends in $0$.
Then $n$ also ends in $0$.
But then the last two digits form a square ending in $0$.
Hence the last two digits of $n$ would have to be $00$.
In order for $n$ to end in $00$, we find that $m$ must also end in $00$.
Suppose $m$ ends in $5$.
Then $n$ also ends in $5$, and then in fact must end in $25$.
In order for $n$ to end in $25$, we find that $m$ must also end in $25$.
Thus $m$ is a square number which lies between $900$ and $8825$ inclusive, and ends in either $00$ or $25$.
It remains to investigate all square numbers of this form.
Thus we need to investigate the squares of all numbers ending in $0$ and $5$ between $30$ and $90$, as $25^2 = 526$ and $95^2 = 9025$.
This is not arduous, so we proceed:
Note that 's analysis dispenses of square numbers ending in $00$ before he even starts, suggesting that he may not be comfortable with defining $00$ itself as a square.
He also jumps straight from the fact that $m$ lying between $885$ and $8845$ eliminates all possibilities except $1225$ and $7225$ without considering the other numbers, which suggests he has further techniques to eliminate them.
== Sources ==
* : Solutions: $167$. -- The Cancelled Cheque
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/168 - Mental Arithmetic

== by : $168$ ==
;Mental Arithmetic
:Find two whole numbers (each less than $10$)
::such that the sum of their squares, added to their product, will make a square.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $168$. -- Mental Arithmetic
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $233$. Mental Arithmetic
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/168 - Mental Arithmetic/Mistake

== Source Work ==
:
:Arithmetical and Algebraical Problems
::Various Arithmetical and Algebraical Problems
:::$168$ -- Mental Arithmetic
:
:Arithmetical and Algebraical Problems
::Miscellaneous Puzzles
:::$233$ -- Mental Arithmetic
== Mistake ==
:Calling the numbers $a$ and $b$, we have:
::$a^2 + b^2 + a b = \Box = /a - m b/^2 = a^2 = 2 a m b + b^2 m^2$.
::$\therefore b + a = -2 a m + b m^2$,''
::$\therefore b = \dfrac {a \paren {2 m + 1} } {m^2 - 1}$''
:in which $m$ may be any whole number greater than $1$, and $a$ is chosen to make $b$ rational.
== Correction ==
$b$ is already guaranteed to be rational.
What we do is choose $a$ to be an integer.
== Sources ==
* : Solutions: $168$. -- Mental Arithmetic
* : Answers: $233$. Mental Arithmetic
Category:Mistakes/Puzzles and Curious Problems
Category:Mistakes/536 Puzzles & Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/168 - Mental Arithmetic/Solution

== by : $168$ ==
;Mental Arithmetic
== Solution ==
 gives $3$ and $5$:
:$3^2 + 3 \times 5 + 5^2 = 49 = 7^2$
but we also have:
:$7^2 + 7 \times 8 + 8^2 = 169 = 13^2$
== Proof ==
Let $a$ and $b$ be the numbers in question.
We have that:
$m$ can be any integer greater than $1$, and $a$ is chosen to make $b$ an integer.
The general values are:
Setting $m = 2$, we have:
Setting $m = 3$, we have:
and the two solutions are apparent.
== Sources ==
* : Solutions: $168$. -- Mental Arithmetic
* : Answers: $233$. Mental Arithmetic
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/169 - Shooting Blackbirds

== by : $169$ ==
;Shooting Blackbirds
:Twice four and twenty blackbirds
::Were sitting in the rain;
:I shot and killed a seventh part,
::How many did remain?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $169$. -- Shooting Blackbirds
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $234$. Shooting Blackbirds
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/169 - Shooting Blackbirds/Solution

== by : $169$ ==
;Shooting Blackbirds
== Solution ==
:$4$.
== Proof ==
The reader is initially tempted into thinking of:
:Twice four and twenty
as meaning:
:$2 \times 24$
and then is left scratching one's head at the absurdity of finding an integer solution to $x = \dfrac {48} 7$.
In fact:
:Twice four and twenty
is interpreted as:
:$\paren {2 \times 4} + 20$
which gives $28$.
Hence $\dfrac {28} 4 = 4$ were shot, leaving $24$ remaining.
But no! This is a trick puzzle.
The $24$ which were not shot flew away, leaving $4$ (dead) blackbirds remaining.
I hope was pleased with himself.
== Sources ==
* : Solutions: $169$. -- Shooting Blackbirds
* : Answers: $234$. Shooting Blackbirds
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Category:Trick Puzzles
Henry Ernest Dudeney/Puzzles and Curious Problems/16 - The Thrifty Grocer

== by : $16$ ==
;The Thrifty Grocer
:A grocer in a small way of business had managed to put aside (apart from his legitimate profits) a little sum in $\pounds 1$ notes, $10 \shillings$ notes, and crowns,
::which he kept in eight bags,
::there being the same number of crowns and of each kind of note in each bag.
:One night he decided to put the money into only seven bags, again with the same number of each kind of currency in every bag.
:And the following night he further reduced the number of bags to six, again putting the same number of each kind of note and of crowns in every bag.
:The next night the poor demented miser tried to do the same with five bags, but after hours of trial he utterly failed, had a fit, and died, greatly respected by his neighbours.
:What is the smallest possible amount of money he had put aside?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Money Puzzles: $16$. -- The Thrifty Grocer
* : Arithmetical and Algebraical Problems: Money Puzzles: $22$. The Thrifty Grocer
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/16 - The Thrifty Grocer/Solution

== by : $16$ ==
;The Thrifty Grocer
== Solution ==
:$\pounds 294$.
== Proof ==
The number of monetary units of each type is $\lcm {8, 7, 6} = 168$.
Hence the answer is $168 \times \paren {\pounds 1 + {10 \shillings} + {5 \shillings} }$.
Hence the result.
== Sources ==
* : Solutions: $16$. -- The Thrifty Grocer
* : Answers: $22$. The Thrifty Grocer
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/170 - The Six Noughts

== by : $170$ ==
;The Six Noughts
 A B C
 111 111 100
 333 333 000
 555 500 005
 777 077 007
 999 090 999
 ---- ---- ----
 2775 1111 1111
 ---- ---- ----
:Write down the little addition sum $A$, which adds up to $2775$.
:Now substitute $6$ noughts for $6$ of the figures, so that the total sum shall be $1111$.
:It will be seen that in the case $B$ five noughts have been susbtituted, and in the case $C$ nine noughts.
:But the puzzle is to do it with six noughts.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $170$. -- The Six Noughts
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $235$. The Six Zeros
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/170 - The Six Noughts/Solution

== by : $170$ ==
;The Six Noughts
== Solution ==
 100 
 330 
 505 
 077 
 099 
 ---- 
 1111 
 ---- 
== Sources ==
* : Solutions: $170$. -- The Six Noughts
* : Answers: $235$. The Six Zeros
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/171 - Multiplication Dates

== by : $171$ ==
;Multiplication Dates
:In the year $1928$ there were $4$ dates which, when written in the form dd/mm/yy,
::the day multiplied by the month equal the year.
:These are 28/1/28, 14/2/28, 7/4/28 and 4/7/28.
:How many times in the $20$th century -- inclusive -- does this so happen?
:Or, you can try to find out which year in the century gives the largest number of dates that comply with the conditions.
:There is one year that beats all the others.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $171$. -- Multiplication Dates
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $236$. Multiplication Dates
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/172 - Curious Multiplicand

== by : $172$ ==
;Curious Multiplicand
:Readers who remember the Ribbon Problem No. $83$ in , may be glad to have this slightly easier variation of it:
:What number is it that can be multiplied by $1$, $2$, $3$, $4$, $5$, or $6$ and no new figures appear in the result?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $172$. -- Curious Multiplicand
* : Arithmetical and Algebraical Problems: Digital Puzzles: $142$. Curious Multiplicand
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/172 - Curious Multiplicand/Solution

== by : $172$ ==
;Curious Multiplicand
== Solution ==
:$142 \, 857$
== Proof ==
This is, of course, the recurring part of the decimal expansion of the Reciprocal of 7:
:$\dfrac 1 7 = 0 \cdotp \dot 14285 \dot 7$
Hence:
== Also see ==
* Properties of 142,857
== Sources ==
* : Solutions: $172$. -- Curious Multiplicand
* : Answers: $142$. Curious Multiplicand
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Category:142,857
Henry Ernest Dudeney/Puzzles and Curious Problems/173 - Short Cuts

== by : $173$ ==
;Short Cuts
:We have from time to time given various short cuts in mental arithmetic.
:Here is an example that will interest those who are unfamiliar with the process.
:Can you multiply $993$ by $879$ mentally?
:It is remarkable that any two numbers of two figures each,
::where the tens are the same, and the sum of the units make ten, can always be multiplied thus:
:::$97 \times 93 = 9021$
:Multiply the $7$ by $3$ and set it down,
::then add the $1$ to the $9$ and multiply by the other $9$, $9 \times 10 = 90$.
:This is very useful for squaring any number ending in $5$, as $85^2 = 7225$.
:With two fractions, when we have the whole numbers the same and the sum of the fractions equal unity,
::we get an easy rule for multiplying them.
:Take $7 \tfrac 1 4 \times 7 \tfrac 3 4 = 56 \tfrac 3 {16}$.
:Multiply the fractions $= \tfrac 3 {16}$, add $1$ to one of the $7$'s, and multiply by the other, $7 \times 8 = 56$.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $173$. -- Short Cuts
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $237$. Short Cuts
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/173 - Short Cuts/Solution

== by : $173$ ==
;Short Cuts
== Solution ==
To multiply $993$ by $879$, proceed as follows:
:Transfer $7$ from $879$ to $993$, and we get $872$ and $1000$.
:Multiply these together to make $872 \, 000$.
:Subtract $872$ from $993$ to get $121$.
:Multiply $121$ by the $7$ we mentioned above, and get $847$.
Add the two results together, and you get $872 \, 847$.
== Proof ==
== Sources ==
* : Solutions: $173$. -- Short Cuts
* : Answers: $237$. Short Cuts
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/174 - More Curious Multiplication

== by : $174$ ==
;More Curious Multiplication
:What number is it that, when multiplied by $18$, $27$, $36$, $45$, $54$, $63$, $72$, $81$ or $99$,
::gives a product in which the first and last figures are the same as those in the multiplier,
::but which when multiplied by $90$ gives a product in which the last two figures are the same as those in the multiplier?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $174$. -- More Curious Multiplication
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $238$. More Curious Multiplication
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/174 - More Curious Multiplication
Henry Ernest Dudeney/Puzzles and Curious Problems/174 - More Curious Multiplication/Mistake

== Source Work ==
:
:Arithmetical and Algebraical Problems
::Various Arithmetical and Algebraical Problems
:::$174$. -- More Curious Multiplication
:
:Arithmetical and Algebraical Problems
::Miscellaneous Puzzles
:::$238$. More Curious Multiplication
== Mistake ==
:The number is $987,654,321$, which, when multiplied by $18$, gives $17,777,777,778$, with $1$ and $8$ at the beginning and end.
:And so on with the other multipliers, except $90$, where the product is $88,888,888,890$, with $90$ at the end.
== Correction ==
 completely missed all numbers of the form $\sqbrk {1 0 d_1 d_2 \ldots d_n 0 1}_{10}$.
He also missed:
== Sources ==
* : Solutions: $174$. -- More Curious Multiplication
* : Answers: $238$. More Curious Multiplication
Category:Mistakes/Puzzles and Curious Problems
Category:Mistakes/536 Puzzles & Curious Problems
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/174 - More Curious Multiplication
Henry Ernest Dudeney/Puzzles and Curious Problems/174 - More Curious Multiplication/Solution

== by : $174$ ==
;More Curious Multiplication
== Solution ==
=== Solution 1 ===
=== Solution 2 ===
=== Solution 3 ===
== Sources ==
* : Solutions: $174$. -- More Curious Multiplication
* : Answers: $238$. More Curious Multiplication
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Category:Penholodigital Integers
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/174 - More Curious Multiplication
Henry Ernest Dudeney/Puzzles and Curious Problems/174 - More Curious Multiplication/Solution 1

== by : $174$ ==
;More Curious Multiplication
== Solution ==
:$987 \, 654 \, 321$
== Proof ==
We have:
== Sources ==
* : Solutions: $174$. -- More Curious Multiplication
* : Answers: $238$. More Curious Multiplication
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/174 - More Curious Multiplication
Henry Ernest Dudeney/Puzzles and Curious Problems/174 - More Curious Multiplication/Solution 2

== by : $174$ ==
;More Curious Multiplication
== Solution ==
As points out: all numbers such as:
:$1001$
:$10101$
:$100101$
that is, numbers made up of $1$ and $0$, with $1$ at each end and no two consecutive instances of $1$.
 notes that you can't rule out numbers of the form $[1 0 d_1 d_2 \ldots d_n 0 1]_{10}$.
However, at the left hand end you need to make sure your carry digit does not affect the second digit.
== Sources ==
* : Answers: $238$. More Curious Multiplication
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/174 - More Curious Multiplication
Henry Ernest Dudeney/Puzzles and Curious Problems/174 - More Curious Multiplication/Solution 3

== by : $174$ ==
;More Curious Multiplication
== Solution ==
All $d$-digit numbers $n$ in the form of:
:$n = \sqbrk {9 a_1 a_2 \dots a_{d - 2} 1}$
where $n \ge \dfrac 8 {81} \times 10^{d + 1}$ and $d \ge 3$.
The smallest solution of this form is $991$.
== Proof ==
Let $\sqbrk {a b}$ be a $2$-digit integer that is not $91$.
We will show that $n$ satisfies more restrictions than those stated in the problem.
For the case $b \ne 0$, we will show that $n \sqbrk {a b}$ starts with $a$ and ends with $b$.
First we show that $\dfrac {\sqbrk {a b} } {\sqbrk {a 0} } \ge \dfrac {81}{80}$.
This is equivalent to showing $\dfrac b {10 a} \ge \dfrac 1 {80}$,
which in turn is equivalent to showing $\dfrac b a \ge \dfrac 1 8$.
For $a \le 8$:
:$\dfrac b a \ge \dfrac 1 a \ge \dfrac 1 8$
For $a = 9$, we have $b \ge 2$.
Thus:
:$\dfrac b a \ge \dfrac 2 9 > \dfrac 1 8$
and hence our inequality is proved.
Now we have:
therefore $n \sqbrk {a b}$ starts with $a$.
We also have $n \equiv 1 \pmod {10}$ and $\sqbrk {a b} \equiv b \pmod {10}$.
Thus $n \sqbrk {a b}$ ends with $b$.
For $b = 0$, we will show that $n \sqbrk {a 0}$ ends with $\sqbrk {a 0}$.
Both $\sqbrk {a 0}$ and $n \sqbrk {a 0}$ are multiples of $10$, so they end with $0$.
The second-last digit of $n \sqbrk {a 0}$ is the last digit of $n \sqbrk {a 0} \div 10 = n a$.
Since $n \equiv 1 \pmod {10}$, $n a \equiv a \pmod {10}$.
This implies that $n a$ ends with $a$, thus proving our claim.
Finally we show that $n$ starts with a $9$ and $d \ge 3$.
 $n$ does not start with a $9$.
Then $n < 9 \times 10^{d - 1}$.
Thus:
:$9 \times 10^{d - 1} > \dfrac 8 {81} \times 10^{d + 1}$
:$9 > \dfrac {800} {81} \approx 9.87654 \ldots$
which is a contradiction.
Therefore $n$ must start with a $9$ and contain at least $2$ digits.
If $d = 2$, $n = 91$.
However $91 < \dfrac 8 {81} \times 10^3 \approx 98.7654 \ldots$.
Thus $d \ge 3$.
== Example ==
* Solution 1: $987 \, 654 \, 321$, by noting that:
:$987 \, 654 \, 321 > \dfrac 8 {81} \times 10^{11}$
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/174 - More Curious Multiplication
Henry Ernest Dudeney/Puzzles and Curious Problems/175 - Cross-Number Puzzle

== by : $175$ ==
;Cross-Number Puzzle
:Here is a cross-number puzzle on lines similar to those of the familiar crossword puzzle.
:The puzzle is to place numbers in the spaces across and down, so as to satisfy the following conditions:
:
:Across:
::1. a square number
::4. a square number
::5. a square number
::8. the digits sum to $35$
::11. square root of $39$ across
::13. a square number
::14. a square number
::15. square of $36$ across
::17. square of half $11$ across
::18. three similar figures
::19. product of $4$ across and $33$ across
::21. a square number
::22. five times $5$ across
::23. all digits alike, except the central one
::25. square of $2$ down
::27. see $20$ down
::28. a fourth power
::29. sum of $18$ across and $31$ across
::31. a triangular number
::33. one more than $4$ times $36$ across
::34. digits sum to $18$, and the three middle numbers are $3$
::36. an odd number
::37. all digits even, except one, and their sum is $29$
::39. a fourth power
::40. a cube number
::41. twice a square
:Down:
::1. reads both ways alike
::2. square root of $28$ across
::3. sum of $17$ across and $21$ across
::4. digits sum to $19$
::5. digits sum to $26$
::6. sum of $14$ across and $33$ across
::7. a cube number
::9. a cube number
::10. a square number
::12. digits sum to $30$
::14. all similar figures
::16. sum of digits is $2$ down
::18. all similar digits except the first, which is $1$
::20. sum of $17$ across and $27$ across
::21. a multiple of $19$
::22. a square number
::24. a square number
::26. square of $18$ across
::28. a fourth power of $4$ across
::30. a triangular number
::32. digits sum to $20$ and end with $8$
::34. six times $21$ across
::35. a cube number
::37. a square number
::38. a cube number
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $175$. -- Cross-Number Puzzle
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $239$. Cross-Number Puzzle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/175 - Cross-Number Puzzle/Solution

== by : $175$ ==
;Cross-Number Puzzle
== Solution ==
:
== Proof ==
$4$ across is $1$, $4$ or $9$.
As $28$ down is a $4$-digit number, it follows that $4$ across is $9$ and so $28$ down is $6561$.
Hence $28$ across is constrained to be $625$.
As $36$ across is odd, $33$ across must be $53$ and $36$ across must be $13$.
Hence $14$ down, and hence $18$ across.
This gives $14$ across.
Then $7$ down is a cube ending in $61$, which constrains it to being $9261$.
The last digit of $32$ down is given as $8$.
$39$ across is a fourth power beginning with $1$, constraining it to $1296$.
$37$ down follows.
$6$ times $21$ across is a $3$ digit number.
As $21$ across is square, that means $21$ across is $121$.
$34$ down and $18$ down follow.
$31$ across follows, which alerts us to the fact that $31$ down starts with $0$, which we may not have been expecting.
Hence $32$ down follows.
$23$ across follows, as we have sufficient digits for this.
This of course gives $24$ down as $1$, and its square nature is confirmed.
$21$ down follows, and $27$ across then appears.
$34$ across can be completed.
$35$ down follows.
$41$ across is either $6498$ or $6728$.
As $38$ down is a cube, this means $38$ down is $64$ and $41$ across is $6498$.
$26$ down follows from knowing $18$ across.
$29$ across follows from knowing $18$ across and $31$ across.
$30$ down is constrained, and hence follows $37$ across.
$29$ down follows, which gives us $15$ across (which also follows from $36$ across).
$2$ down is evaluated.
$25$ across is evaluated.
$11$ across is evaluated, which leads to $1$ down.
$12$ down follows.
$3$ down is evaluated.
$1$ across follows.
$17$ across is evaluated.
$20$ down is evaluated.
$6$ down is evaluated.
$9$ down is constrained.
The intersecting digits of $13$ across and $10$ down constrain $13$ across.
$5$ across can be deduced.
$8$ down follows.
The remaining $2$ digits of $8$ across sum to $5$, which constrains $10$ down.
$8$ across then follows.
$4$ down follows.
Finally, $22$ across can be evaluated.
== Sources ==
* : Solutions: $175$. -- Cross-Number Puzzle
* : Answers: $239$. Cross-Number Puzzle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/176 - Counting the Loss

== by : $176$ ==
;Counting the Loss
:An officer explained that the force to which he belonged originally consisted of $1000$ men, but that it lost heavily in an engagement,
::and the survivors surrendered and were marched down to a concentration camp.
:On the first day's march one-sixth of the survivors escaped;
::on the second day one-eighth of the remainder escaped, and one man died;
::on the third day's march one-fourth of the remainder escaped.
:Arrived in camp, the rest were set to work in four equal gangs.
:How many had been killed in the engagement?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $176$. -- Counting the Loss
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $240$. Counting the Loss
Category:Henry Ernest Dudeney/Puzzles and Curious Problems