nyuuzyou/wikis huggingface dataset:
Henry Ernest Dudeney/Puzzles and Curious Problems/87 - Reversing the Digits/Solution

== by : $87$ ==
;Reversing the Digits
== Solution ==
:$989 \, 010 \, 989 \times 123 \, 456 \, 789 = 122 \, 100 \, 120 \, 987 \, 654 \, 321$
== Sources ==
* : Solutions: $87$. -- Reversing the Digits
* : Answers: $124$. Reversing the Digits
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/88 - Digital Progression

== by : $88$ ==
;Digital Progression
:If you arrange the nine digits in three numbers thus, $147$, $258$, $369$,
::they have a common difference of $111$ and are therefore in arithmetic progression.
:Can you find $4$ ways of rearranging the $9$ digits so that in each case the number shall have a common difference,
::and the middle number be in every case the same?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $88$. -- Digital Progression
* : Arithmetical and Algebraical Problems: Digital Puzzles: $125$. Digital Progression
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/88 - Digital Progression/Historical Note

== Historical Note on by : $88$ - Digital Progression ==
 informs that subsequent to 's $4$ examples, reported on $760$ such progressions in his from $1952$.
He also found that in addition to $456$ and its permutations, the middle number may be any of the permutations of $258$, $267$, $348$ and $357$.
== Sources ==
* : Answers: $125$. Digital Progression
Category:Historical Notes
Henry Ernest Dudeney/Puzzles and Curious Problems/88 - Digital Progression/Solution

== by : $88$ ==
;Digital Progression
== Solution ==
:$297$, $564$, $831$
:$291$, $564$, $837$
:$237$, $564$, $891$
:$231$, $564$, $897$
where the common differences are respectively $267$, $273$, $327$ and $333$.
== Historical Note ==
== Sources ==
* : Solutions: $88$. -- Digital Progression
* : Answers: $125$. Digital Progression
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/89 - Forming Whole Numbers

== by : $89$ ==
;Forming Whole Numbers
:Can the reader give the sum of all the whole numbers that can be formed with the four figures $1$, $2$, $3$, $4$?
:That is, the addition of all such numbers as $1234$, $1423$, $4312$, etc.
:You can, of course, write them all out and make the addition,
::but the interest lies in finding a very simple rule for the sum of all the numbers that can be made with $4$ different digits selected in every possible way, but $0$ excluded.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $89$. -- Forming Whole Numbers
* : Arithmetical and Algebraical Problems: Digital Puzzles: $126$. Forming Whole Numbers
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/89 - Forming Whole Numbers/Solution

== by : $89$ ==
;Forming Whole Numbers
:Can the reader give the sum of all the whole numbers that can be formed with the four figures $1$, $2$, $3$, $4$?
:That is, the addition of all such numbers as $1234$, $1423$, $4312$, etc.
:You can, of course, write them all out and make the addition,
::but the interest lies in finding a very simple rule for the sum of all the numbers that can be made with $4$ different digits selected in every possible way, but $0$ excluded.
== Solution ==
:$66 \, 660$
== Proof ==
There are $24$ permutations of $1$, $2$, $3$ and $4$.
Of these permutations, each digit appears in each of the $4$ positions $6$ times each.
Let $S$ be the required sum.
We have:
== Sources ==
* : Solutions: $89$. -- Forming Whole Numbers
* : Answers: $126$. Forming Whole Numbers
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/8 - Equal Values

== by : $8$ ==
;Equal Values
:A lady and her daughter set out on a walk the other day,
::and happened to notice that they both had money of the same value in their purses,
::consisting of three coins each, and all six coins were different.
:During the afternoon they made slight purchases,
::and on returning home found that they again had similar value in their purses made up of three coins each, and all six different.
:How much money did they set out with, and what was the value of their purchases?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Money Puzzles: $8$. -- Equal Values
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/8 - Equal Values/Solution

== by : $8$ ==
;Equal Values
== Solution ==
The two ladies had:
:$10 \shillings$, $1 \shillings$ and $6 \oldpence$ in one purse
and:
:$5 \shillings$, $4 \shillings$ and $2 \shillings 6 \oldpence$ in the other.
That is, they had $11 \shillings 6 \oldpence$ each.
They each spent $4 \shillings$ and returned with:
:$5 \shillings$, $2 \shillings$ and $6 \oldpence$ in one purse
and:
:$4 \shillings$, $2 \shillings 6 \oldpence$ and $1 \shillings$ in the other.
They started with $11 \shillings 6 \oldpence$ each and ended with $7 \shillings 6 \oldpence$ each.
== Sources ==
* : Solutions: $8$. -- Equal Values
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/90 - Summing the Digits

== by : $90$ ==
;Summing the Digits
:What is the sum of all the numbers that can be formed with all $9$ digits ($0$ excluded),
::using each digit once and once only, in every number?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $90$. -- Summing the Digits
* : Arithmetical and Algebraical Problems: Digital Puzzles: $127$. Summing the Digits
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Category:Penholodigital Integers
Henry Ernest Dudeney/Puzzles and Curious Problems/90 - Summing the Digits/Solution

== by : $90$ ==
;Summing the Digits
:What is the sum of all the numbers that can be formed with all $9$ digits ($0$ excluded),
::using each digit once and once only, in every number?
== Solution ==
:$201 \, 599 \, 999 \, 798 \, 400$
== Proof ==
There are $9!$ permutations of the $9$ digits, that is: $362 \, 880$.
Of these permutations, each digit appears in each of the $9$ positions a total of $8!$ times each.
We note that $\ds \sum_{n \mathop = 1}^9 n = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45$.
Let $S$ be the required sum.
We have:
== Sources ==
* : Solutions: $90$. -- Summing the Digits
* : Answers: $127$. Summing the Digits
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/91 - Squaring the Digits

== by : $91$ ==
;Squaring the Digits
:Take $9$ counters numbered $1$ to $9$, and place them in a row: $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$, $9$.
:It is required in as few exchanges of pairs as possible to convert this into a square number.
:As an example in $6$ pairs we give the following:
::$\tuple {7, 8}$ (exchanging $7$ and $8$),
::$\tuple {8, 4}$, $\tuple {4, 6}$, $\tuple {6, 9}$, $\tuple {9, 3}$, $\tuple {3, 2}$, which gives us the number $139 \, 854 \, 276$,
::which is the square of $11 \, 826$.
::But it can be done in much fewer moves.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $91$. -- Squaring the Digits
* : Arithmetical and Algebraical Problems: Digital Puzzles: $128$. Squaring the Digits
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/91 - Squaring the Digits/Solution

== by : $91$ ==
;Squaring the Digits
== Solution ==
A $3$-move solution is:
:$\tuple {1, 5}$, $\tuple {8, 4}$, $\tuple {4, 6}$
which gives us:
:$513 \, 814 \, 769 = 22 \, 887^2$
For interest's sake, we also offer up the $4$-move solution given by :
:$\tuple {7, 3}$, $\tuple {3, 4}$, $\tuple {4, 8}$, $\tuple {2, 5}$
which gives us:
:$157 \, 326 \, 849 = 12 \, 543^2$
== Sources ==
* : Solutions: $91$. -- Squaring the Digits
* : Answers: $128$. Squaring the Digits
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/92 - Digits and Squares

== by : $92$ ==
;Digits and Squares
:$(1) \quad$ What is the smallest square number, and
:$(2) \quad$ what is the largest square number
::that contains all the ten digits ($1$ to $9$ and $0$) once, and once only?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $92$. -- Digits and Squares
* : Arithmetical and Algebraical Problems: Digital Puzzles: $129$. Digits and Squares
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Category:Pandigital Integers
Henry Ernest Dudeney/Puzzles and Curious Problems/92 - Digits and Squares/Solution

== by : $92$ ==
;Digits and Squares
== Solution ==
From Smallest Pandigital Square:
:$1 \, 026 \, 753 \, 849 = 32 \, 043^2$
From Largest Pandigital Square:
:$9 \, 814 \, 072 \, 356 = 99 \, 066^2$
== Sources ==
* : Solutions: $92$. -- Digits and Squares
* : Answers: $129$. Digits and Squares
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/93 - Figures for Letters

== by : $93$ ==
;Figures for Letters
Solve this cryptarithm:
 A B C D
 x E F G H I
 -----------------
 A C G E F H I B D
Each of the letters $A$ to $I$ represents one of the digits from $1$ to $9$ inclusive.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $93$. -- Figures for Letters
* : Arithmetical and Algebraical Problems: Cryptarithm Puzzles: $153$. Figures for Letters
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Category:Cryptarithms
Henry Ernest Dudeney/Puzzles and Curious Problems/93 - Figures for Letters/Solution

== by : $93$ ==
;Figures for Letters
== Solution ==
 6 5 4 3
 x 9 8 2 7 1
 -----------------
 6 4 2 9 8 7 1 5 3
== Proof ==
== Sources ==
* : Solutions: $93$. -- Figures for Letters
* : Answers: $153$. Figures for Letters
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/94 - Simple Multiplication

== by : $94$ ==
;Simple Multiplication
 * * * * * * * * * *
x 2
 -------------------
 * * * * * * * * * *
:Substitute for the $*$ symbol each of the $10$ digits in each row,
::so arranged as to form a correct multiplication operation.
:$0$ is not to appear at the beginning or end of either answer.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $94$. -- Simple Multiplication
* : Arithmetical and Algebraical Problems: Skeleton Puzzles: $146$. Simple Multiplication
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Category:Skeleton Puzzles
Henry Ernest Dudeney/Puzzles and Curious Problems/94 - Simple Multiplication/Solution

== by : $94$ ==
;Simple Multiplication
 * * * * * * * * * *
x 2
 -------------------
 * * * * * * * * * *
:Substitute for the $*$ symbol each of the $10$ digits in each row,
::so arranged as to form a correct multiplication operation.
:$0$ is not to appear at the beginning or end of either answer.
== Solution ==
 1 0 9 8 7 6 5 4 3 2
x 2
 -------------------
 2 1 9 7 5 3 0 8 6 4
There are many answers.
The above can be found in Pandigital Integers remaining Pandigital on Multiplication, from which we can also get:
 2 1 9 7 5 3 0 8 6 4
x 2
 -------------------
 4 3 9 5 0 6 1 7 2 8
 4 3 9 5 0 6 1 7 2 8
x 2
 -------------------
 8 7 9 0 1 2 3 4 5 6
 provides:
 4 5 3 9 2 8 1 7 0 6
x 2
 -------------------
 9 0 7 8 5 6 3 4 1 2
who adds:
:If you divide the first number into pairs -- $45$, $39$ etc. -- these can be arranged in any order so long as the $06$ is not at the beginning or the $45$ at the end.
== Sources ==
* : Solutions: $94$. -- Simple Multiplication
* : Answers: $146$. Simple Multiplication
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/95 - Beeswax

== by : $95$ ==
;Beeswax
:The word BEESWAX represents a number in a criminal's secret code,
::but the police had no clue until they discovered among his papers the following sum:
 E A S E B S B S X
 B P W W K S E T Q
------------------
 K P E P W E K K Q
:The detectives assumed that it was an addition sum, and utterly failed to solve it.
:Then one man hit on the brilliant idea that perhaps it was a case of subtraction.
:This proved to be correct, and by substituting a different figure for each letter, so that it worked out correctly,
::they obtained the secret code.
:What number does BEESWAX represent?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $95$. -- Beeswax
* : Arithmetical and Algebraical Problems: Cryptarithm Puzzles: $155$. Beeswax
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/95 - Beeswax
Henry Ernest Dudeney/Puzzles and Curious Problems/95 - Beeswax/Solution

== by : $95$ ==
;Beeswax
== Solution ==
The key is:
 1 2 3 4 5 6 7 8 9 0
 -------------------
 A T Q B K X S W E P
and the solution to the subtraction is:
 9 1 7 9 4 7 4 7 6
- 4 0 8 8 5 7 9 2 3
-------------------
 5 0 9 0 8 9 5 5 3
So BEESWAX represents the number $4 \, 997 \, 816$.
== Proof ==
From the thousands place we have:
:"$S - S = E$"
If there was no borrow from the hundreds place we will have $E = 0$.
But the first number begins with $E$, so $E \ne 0$.
Thus there must be a borrow from the hundreds place, giving $E = 9$.
Now we look at the hundreds place.
If there was a borrow from the tens place we will have:
:$10 + B - 1 - E = B = K$
which is impossible since all letters represent distinct digits.
Thus there is no borrow from the tens place, and
:$10 + B - E = B + 1 = K$
In the ten-thousands place we have:
:"$W + K = B$"
Since $B < K$ and there is a borrow from the thousands place, we have instead:
:$W + K = 10 + B - 1 = 8 + K$
giving $W = 8$.
For the hundred-thousands place we have a borrow from the ten-thousands place.
Therefore we have:
:$P + W = E - 1$
Since $W = 8$ and $E = 9$, this gives $P = 0$.
Combining the millions place and ten-millions place, we have, with no borrows:
:$E + W = 10 A + S$
so we have $A = 1$ and $S = 7$.
The hundred-millions place gives:
:$B + K = E = 9$
Since we have $B + 1 = K$ from earlier, we can conclude:
:$B = 4, K = 5$
We have so far:
 9 1 7 9 4 7 4 7 X
- 4 0 8 8 5 7 9 T Q
-------------------
 5 0 9 0 8 9 5 5 Q
Since the remaining untaken digits are $2, 3$ and $6$, we must have:
:$T = 2, Q = 3, X = 6$
and thus BEESWAX represents $4 \, 997 \, 816$.
=== Addition Case ===
== Sources ==
* : Solutions: $95$. -- Beeswax
* : Answers: $155$. Beeswax
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/95 - Beeswax
Henry Ernest Dudeney/Puzzles and Curious Problems/95 - Beeswax/Solution/Addition Case

== by : $95$ ==
;Beeswax
== Solution ==
There is indeed no solution for the addition case, as claimed:
:The detectives assumed that it was an addition sum, and utterly failed to solve it.:
If this were an addition, we have immediately from the units place:
:$X = 0$
Now in the ten-millions place, if there is no carry from the millions place, we would have:
:$A + P = P$
which gives a repeated $A = 0$.
Hence there is such a carry, and $A = 9$, giving a carry to the hundred-millions place.
The hundreds place and the hundred-millions place have the identical sum:
:"$E + B = K$"
So there must be a carry from the tens place and no carry to the thousands place.
From the thousands place we have:
:$S + S \equiv E \pmod {10}$
so $E$ is even.
We have from the millions place:
:"$S + W = E$"
so this place must have received a carry from the hundred-thousands place, and $W + 1 = S$.
As we now have:
:$S + W + 1 = 10 + E$
there must be a carry from the thousands place to the ten-thousands place.
So far we have:
 E 9 S E B S B S 0
 B P W W K S E T Q
+ 1 1 1 ? 1 0 1 0 - <-- carries
-------------------
 K P E P W E K K Q
Notice that we have:
:$S + T = 10 + K$
:$B + E + 1 = K$
which gives the inequality:
:$4 = 1 + 2 + 1 \le K \le 7 + 8 - 10 = 5$
However, if $K = 5$, $B + E = 4$.
As $B, E \ne 0$ and $E$ is even, we would end up with:
:$B = E = 2$
which is impossible.
Hence $K = 4, B = 1, E = 2$.
From the thousands place:
:$S + S = 12$
so $S = 6$.
From the ten-thousands place:
:$W = B + K + 1 = 6 = S$
which repeats a digit.
So we have shown that there are no solutions to the addition case without repeating digits.
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $95$. -- Beeswax
* : Arithmetical and Algebraical Problems: Cryptarithm Puzzles: $155$. Beeswax
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/95 - Beeswax
Henry Ernest Dudeney/Puzzles and Curious Problems/96 - Wrong to Right

== by : $96$ ==
;Wrong to Right
:Solve this cryptarithm:
 W R O N G
+ W R O N G
-----------
 R I G H T
:Each letter represents a different digit, and no $0$ is allowed.
:There are several different ways of doing this.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $96$. -- Wrong to Right
* : Arithmetical and Algebraical Problems: Cryptarithm Puzzles: $156$. Wrong to Right
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Category:Cryptarithms
Henry Ernest Dudeney/Puzzles and Curious Problems/96 - Wrong to Right/Solution

== by : $96$ ==
;Wrong to Right
== Solution ==
The solution given by :
 2 5 9 3 8
+ 2 5 9 3 8
-----------
 5 1 8 7 6
but there are six more solutions:
 1 2 7 3 4
+ 1 2 7 3 4
-----------
 2 5 4 6 8
 1 2 8 6 7
+ 1 2 8 6 7
-----------
 2 5 7 3 4
 1 2 9 3 8
+ 1 2 9 3 8
-----------
 2 5 8 7 6
 2 5 7 3 4
+ 2 5 7 3 4
-----------
 5 1 4 6 8
 2 5 8 6 7
+ 2 5 8 6 7
-----------
 5 1 7 3 4
 3 7 8 4 6
+ 3 7 8 4 6
-----------
 7 5 6 9 2
== Proof ==
Immediate first deductions:
:$T$ is even.
:$1 \le W \le 4$
== Sources ==
* : Solutions: $96$. -- Wrong to Right
* : Answers: $156$. Wrong to Right
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/97 - Letter Multiplication

== by : $97$ ==
;Letter Multiplication
:In this little multiplication sum the five letters represent $5$ different digits.
:What are the actual figures?
:There is no $0$.
 S E A M
x T
-----------
 M E A T S
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $97$. -- Letter Multiplication
* : Arithmetical and Algebraical Problems: Cryptarithm Puzzles: $157$. Letter Multiplication
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Category:Cryptarithms
Henry Ernest Dudeney/Puzzles and Curious Problems/97 - Letter Multiplication/Solution

== by : $97$ ==
;Letter Multiplication
== Solution ==
 4 9 7 3
x 8
-----------
 3 9 7 8 4
== Proof ==
Note that:
:$M < T$, because $MEATS = SEAM \times T < 10000 \times T = T0000$.
:$M < S$, because $MEATS = SEAM \times T < SEAM \times 10 = SEAM0$, and $M \ne S$.
Also $S \ne T$.
By observing the lower triangular part of the modulo $10$ multiplication table below:
$\begin{array}{c|cccccccc}
T \backslash M & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
\hline
2 & 2 \\
3 & 3 & \color {red} 6 \\
4 & 4 & \color {red} 8 & 2 \\
5 & 5 & 0 & 5 & 0 \\
6 & 6 & 2 & \color {red} 8 & 4 & 0 \\
7 & 7 & \color {red} 4 & 1 & \color {red} 8 & 5 & 2 \\
8 & 8 & \color {red} 6 & \color {red} 4 & 2 & 0 & 8 & 6 \\
9 & 9 & \color {red} 8 & \color {red} 7 & \color {red} 6 & 5 & 4 & 3 & 2 & \\
\end{array}$
we see that only the numbers satisfy the conditions for $S$.
Now we consider an additional condition:
:$M \ge \floor {\dfrac {S \times T} {10}}$
i.e. the tens digit of $S \times T$ cannot exceed $M$.
Otherwise $\floor {\dfrac {S \times T} {10}} \ge M + 1$,
and $SEAM \times T > S000 \times T = 10000 \times \dfrac {S \times T} {10} \ge 10000 \paren {M + 1} > MEATS$.
Hence only the numbers in:
$\begin{array}{c|ccc}
T \backslash M & 2 & 3 & 4\\
\hline
3 & \color {red} 6 \\
4 & 8 & 2 \\
5 & 0 & 5 & 0 \\
6 & 2 & 8 & 4 \\
7 & \color {red} 4 & 1 & 8 \\
8 & 6 & \color {red} 4 & 2 \\
9 & 8 & 7 & 6 \\
\end{array}$
remain.
Using only the values of $M, T, S$, we can determine the value of $SEAM$ by the formula:
:$SEAM = \dfrac {S00M0 - M00TS} {10 - T}$
since:
Now we inspect the $3$ remaining cases using the formula.
$M = 2, T = 3, S = 6$ gives:
:$6EA2 = \dfrac {60020 - 20036} {10 - 3} = 5712$
$M = 2, T = 7, S = 4$ gives:
:$4EA2 = \dfrac {40020 - 20074} {10 - 7} = 6648 \tfrac 2 3$
$M = 3, T = 8, S = 4$ gives:
:$4EA3 = \dfrac {40030 - 30084} {10 - 8} = 4973$
Only $4973$ matches the required form of $SEAM$.
Hence the solution given is the unique solution.
== Sources ==
* : Solutions: $97$. -- Letter Multiplication
* : Answers: $157$. Letter Multiplication
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/98 - Digital Money

== by : $98$ ==
;Digital Money
:Every letter in the following multiplication represents one of the digits, $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$, $9$, all different.
:What is the value obtained if $K = 8$?
 A B C
x K
-----------
 DE FG H
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $98$. -- Digital Money
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Category:Cryptarithms
Henry Ernest Dudeney/Puzzles and Curious Problems/98 - Digital Money/Solution

== by : $98$ ==
;Digital Money
== Solution ==
 3 7 5
x 8
------------
£ 26 19 4
== Proof ==
Given that $K = 8$, recalling that $1$ shilling is $12$ pence:
The possible values of $H$ are $0, 8$ or $4$.
Since both $0, 8$ are not available, we have $H = 4$.
This is only possible if $C = 2, 5$ or $8$.
Again, $8$ is not available, so $C = 2$ or $5$.
This provides a carry of $1$ or $3$, respectively, to the shillings place.
Without the carry, with $K = 8$, recalling that $1$ pound is $20$ shillings, the possible values of $FG$ are:
:$8, 16, 4, 12, 0$
With the carries, the possible values of $FG$ are:
:$9, 17, 5, 13, 1, 11, 19, 7, 15, 3$
Since $FG$ is a $2$-digit number, the only possible values of $FG$ are:
:$17, 13, 19, 15$
which results from:
:$\set {12, 16} + \set {1, 3}$
so the possible values of $B$ are:
:$2, 4, 7, 9$
Since $4$ is taken, we check all cases where:
:$C = 2$ and $B = 7$ or $9$
:$C = 5$ and $B = 2, 7, 9$
== Sources ==
* : Solutions: $98$. -- Digital Money
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/99 - The Conspirators' Code

== by : $99$ ==
;The Conspirators' Code
:Two conspirators had a secret code.
:Their letters sometimes contained little arithmetical sums related to some quite plausible discussion,
::and having an entirely innocent appearance.
:But in their code each of the ten digits represented a different letter of the alphabet.
:Thus, on one occasion, there was a little sum in simple addition which, when the letters were substituted for the figures, read as follows:
 F L Y
 F O R
+ Y O U R
-----------
 L I F E
:It will be found an interesting puzzle to reconstruct the addition sum with the help of the clue that $I$ and $O$ stand for the figures $1$ and $0$ respectively.
=== Click here for solution ===
== Historical Note ==
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $99$. -- The Conspirators' Code
* : Arithmetical and Algebraical Problems: Cryptarithm Puzzles: $158$. The Conspirators' Code
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Category:Cryptarithms
Henry Ernest Dudeney/Puzzles and Curious Problems/99 - The Conspirators' Code/Historical Note

== Historical Note on by : $99$ - The Conspirators' Code ==
 attributes this to a correspondent identified as G. P., whose identity has not been ascertained.
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $99$. -- The Conspirators' Code
* : Arithmetical and Algebraical Problems: Cryptarithm Puzzles: $158$. The Conspirators' Code
Category:Historical Notes