Henry Ernest Dudeney/Puzzles and Curious Problems/Arithmetical and Algebraical Problems/Digital Puzzles

Henry Ernest Dudeney: Puzzles and Curious Problems: Arithmetical and Algebraical Problems

$83$ - Three Different Digits

Find all $3$-digit numbers with distinct digits that are divisible by the square of the sum of those digits.

$84$ - Find the Cube

A number increased by its cube is $592 \, 788$.

What is that number?

$85$ - 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?

$86$ - 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.

$87$ - Reversing the Digits

What $9$-digit number, when multiplied by $123\, 456 \, 789$, gives a product ending in $987 \, 654 \, 321$?

$88$ - 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?

$89$ - 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.

$90$ - 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?

$91$ - 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.

$92$ - 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?

$93$ - 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.

$94$ - 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.

$95$ - 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?

$96$ - 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.

$97$ - 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

$98$ - 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

$99$ - 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.

$100$ - Digital Squares

Find a number which, together with its square, shall contain all the $9$ digits once, and once only, the $0$ disallowed.

Thus, if the square of $378$ happened to be $152 \, 694$, it would be a perfect solution.

But unfortunately the actual square is $142 \, 884$, which gives us repeated $4$s and $8$s, and omits the $6$, $5$, and $9$.

$101$ - Finding a Square

Here are six numbers:

$4 \, 784 \, 887$, $2 \, 494 \, 651$, $8 \, 595 \, 087$, $1 \, 385 \, 287$, $9 \, 042 \, 451$, $9 \, 406 \, 087$

It is known that three of these numbers added together will form a square.

Which are they?

$102$ - Juggling with Digits

Arrange the ten digits in three arithmetical sums,

employing three of the four operations of addition, subtraction, multiplication and division,

and using no signs except the ordinary ones implying those operations.

Here is an example to make it quite clear:

$3 + 4 = 7$; $9 - 8 = 1$; $30 \div 6 = 5$.

But this is not correct, because $2$ is omitted, and $3$ is repeated.

$103$ - Expressing Twenty-Four

In a book published in America was the following:

"Write $24$ with three equal digits, none of which is $8$.

(There are two solutions to this problem.)"

Of course, the answers given are $22 + 2 = 24$, and $3^3 - 3 = 24$.

Readers who are familiar with the old "Four Fours" puzzle, and others of the same class,

will ask why there are supposed to be only these solutions.

With which of the remaining digits is a solution equally possible?

$104$ - Letter-Figure Puzzle

$\text {(0)}: \quad$

$\ds A \times B$

$=$

$\ds B$

$\text {(1)}: \quad$

$\ds B \times C$

$=$

$\ds A C$

$\text {(2)}: \quad$

$\ds C \times D$

$=$

$\ds B C$

$\text {(3)}: \quad$

$\ds D \times E$

$=$

$\ds C H$

$\text {(4)}: \quad$

$\ds E \times F$

$=$

$\ds D K$

$\text {(5)}: \quad$

$\ds F \times H$

$=$

$\ds C J$

$\text {(6)}: \quad$

$\ds H \times J$

$=$

$\ds K J$

$\text {(7)}: \quad$

$\ds J \times K$

$=$

$\ds E$

$\text {(8)}: \quad$

$\ds K \times L$

$=$

$\ds L$

$\text {(9)}: \quad$

$\ds A \times L$

$=$

$\ds L$

Every letter represents a different digit, and, of course, $A C$, $B C$ etc., are two-figure numbers.

Can you find the values in figures of all the letters?

$105$ - Equal Fractions

Can you construct three ordinary vulgar fractions

(say, $\tfrac 1 2$, $\tfrac 1 3$, or $\tfrac 1 4$, or anything up to $\tfrac 1 9$ inclusive)

all of the same value, using in every group all the nine digits once, and once only?

The fractions may be formed in one of the following ways:

$\dfrac a b = \dfrac c d = \dfrac {e f} {g h j}$, or $\dfrac a b = \dfrac c {d e} = \dfrac {f g} {h j}$.

We have only found five cases, but the fifth contains a simple little trick that may escape the reader.

$106$ - Digits and Primes

Using the $9$ digits once, and once only,

can you find prime numbers that will add up to the smallest total possible?

$\qquad \begin{array}{|c|c|c|} \hline 2 & 1 & 8 \\ \hline 4 & 3 & 9 \\ \hline 6 & 5 & 7 \\ \hline \end{array} \qquad \begin{array}{|c|c|c|} \hline 2 & 7 & 3 \\ \hline 5 & 4 & 6 \\ \hline 8 & 1 & 9 \\ \hline \end{array} \qquad \begin{array}{|c|c|c|} \hline 3 & 2 & 7 \\ \hline 6 & 5 & 4 \\ \hline 9 & 8 & 1 \\ \hline \end{array}$

The $9$ digits may be arranged in a square in many ways,

so that the numbers formed in the first row and second row will sum to the third row.

We give $3$ examples, and it will be found that the difference between the first total, $657$, and the second total, $819$,

is the same as the difference between the second, $819$, and the third, $981$ --

that is, $162$.

Now, can you form $8$ such squares, every one containing the $9$ digits,

so that the common difference between the $8$ totals is throughout the same?

$108$ - The Nine Digits

It will be found that $32 \, 547 \, 891$ multiplied by $6$ (thus using all the $9$ digits once, and once only)

gives the product $195 \, 287 \, 346$ (also containing all the $9$ digits once, and once only).

Can you find another number to be multiplied by $6$ under the same conditions?

$109$ - Perfect Squares

Find $4$ numbers such that the sum of every two and the sum of all four may be perfect squares.

$110$ - An Absolute Skeleton

Here is a good skeleton puzzle.

The only conditions are:

$(1)$ No digit appears twice in any row of figures except the dividend.

$(2)$ If $2$ be added to the last figure in the quotient it equals the last but one,

and if $2$ be added to the third figure from the end it gives the last figure but $3$ in the quotient.

That is to say, the quotient might end in, say, $9742$, or in $3186$.

        ********
    ------------
 ***)***********
      ***
       ---
       ***
       ***
       ----
       ****
       ****
       -----
         ***
         ***
         ----
         ****
         ****
         -----
          ****
          ****
          -----
           ****
           ****
           -----
            ****
            ****
            ----

$111$ - Odds and Evens

Every asterisk and letter represents a figure,

and "$\mathtt O$" stands for an odd figure ($1$, $3$, $5$, $7$, or $9$)

while "$\mathtt E$" represents an even figure ($2$, $4$, $6$, $8$, or $0$).

Can you construct an arrangement complying with these conditions?

There are $6$ solutions.

Can you find one, or all of them?

       *******
    ----------
 ***)*********
     OE*
     -----
      ****
      OO**
      -----
        ***
        EE*
        ----
         ***
         EO*
         ----
         ****
         EE**
         -----
           ***
           OO*
           ---

$112$ - Simple Division

There is a simple division sum.

Can you restore it by substituting a figure for every asterisk, without altering or removing the sevens?

If you start out with the assumption that all the sevens are given, and that you must not use another,

you will attempt an impossibility, though the proof is difficult;

but when you are told that though no additional sevens may be used in divisor, dividend or quotient,

any number of extra sevens may be used in the working.

It is comparatively easy.

             **7**
       -----------
 ****7*)**7*******
        ******
        -------
        *****7*
        *******
        -------
          *7****
          *7****
          -------
          *******
           ***7**
          --------
            ******
            ******
            ------

$113$ - A Complete Skeleton

It will be remembered that a skeleton puzzle, where the figures are represented by stars,

has not been constructed without at least one figure, or some added condition, being used.

Perhaps the following comes a little nearer to the ideal,

though there are two division sums and not one,

and they are related by the fact that the six-figure quotient of the first happens to be the dividend of the second.

There appears to be only one solution.

        ******
    ----------
 ***)*********
      ***
     ----
        ****
         ***
        -----
          ***
          ***
          ----
          ****
          ****
          ----

     *****
   -------
 **)******
    **
    ----
     ***
      **
     ----
      ***
      ***
      ----
       ***
       ***
       ---

Henry Ernest Dudeney/Puzzles and Curious Problems/Arithmetical and Algebraical Problems/Digital Puzzles

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