Henry Ernest Dudeney/Modern Puzzles/Arithmetical and Algebraical Problems/Digital Puzzles
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Henry Ernest Dudeney: Modern Puzzles: Arithmetical and Algebraical Problems
$51$ - An Exceptional Number
- A number is formed of $5$ successive digits (not necessarily in regular order)
- so that the number formed by the first $2$ multiplied by the central digit will produce the number expressed by the last $2$.
$52$ - The Five Cards
- I have $5$ cards bearing the figures $1$, $3$, $5$, $7$ and $9$.
- How can I arrange them in a row so that the number formed by the $1$st pair multipied by the number formed with the last pair,
- with the central number subtracted,
- will produce a number composed of repetitions of one figure?
$53$ - Squares and Digits
- What is the smallest square number that terminates with the greatest possible number of similar digits?
- Thus the greatest possible number might be $5$ and the smallest square number with $5$ similar digits at the end might be $24677777$.
- But this is certainly not a square number.
- Of course, $0$ is not to be regarded as a digit.
$54$ - The Two Additions
- Can you arrange the following figures into two groups of $4$ figures each so that each group shall add to the same sum?
- $1 \ 2 \ 3 \ 4 \ 5 \ 7 \ 8 \ 9$
$55$ - The Repeated Quartette
- If we multiply $64253$ by $365$ we get the product $23452345$, where the first $4$ figures are repeated.
- What is the largest number that we can multiply by $365$ in order to produce a similar product of $8$ figures repeated in the same order?
- There is no objection to a repetition of figures -- that is, the $4$ that are repeated need not be all different, as in the case shown.
$56$ - Easy Division
- To divide the number $8 \, 101 \, 265 \, 822 \, 784$ by $8$, all we need to do is transfer the $8$ from the beginning to the end!
- Can you find a number beginning with $7$ that can be divided by $7$ in the same simple manner?
$57$ - A Misunderstanding
- An American correspondent asks me to find a number composed of any number of digits that may be correctly divided by $2$
- by simply transferring the last figure to the beginning.
- He has apparently come across our last puzzle with the conditions wrongly stated.
- If you are to transfer the first figure to the end it is solved by $315 \, 789 \, 473 \, 684 \, 210 \, 526$,
- and a solution may easily be found from this with any given figure at the beginning.
- But if the figure is to be moved from the end to the beginning, there is no possible solution for the divisor $2$.
- But there is a solution for the divisor $3$.
- Can you find it?
$58$ - The Two Fours
- The point [of the Four Fours puzzle] is to express all possible whole numbers with four fours (no more and no fewer), using the various arithmetical signs.
- Thus:
- $17 = 4 \times 4 + \dfrac 4 4$
- and:
- $50 = 44 + 4 + \sqrt 4$
- All numbers up to $112$ inclusive may be solved, using only the signs for addition, subtraction, multiplication, division, square root, decimal points, and the factorial sign $4!$ which means $1 \times 2 \times 3 \times 4$, or $24$, but $113$ is impossible.
- It is necessary to discover which numbers can be formed with one four, with two fours, and with three fours, and to record these for combination as required.
- It is the failure to find some of these that leads to so much difficulty.
- For example, I think very few discover that $64$ can be expressed with only two fours.
- Can the reader do it?
$59$ - The Two Digits
- Write down any $2$-figure number (different figures and no $0$)
- and then express that number by writing the same figures in reverse order,
- with or without arithmetical signs.
$60$ - Digital Coincidences
- If I multiply, and also add, $9$ and $9$, I get $81$ and $18$, which contain the same figures.
- If I multiply and add $2$ and $47$ I get $94$ and $49$ -- the same figures.
- If I multiply and add $3$ and $24$ I get the same figures -- $72$ and $27$.
- Can you find two numbers that, when multiplied and added will, in this simple manner, produce the same three figures?
$61$ - Palindromic Square Numbers
- This is a curious subject for investigation -- the search for square numbers the figures of which read backwards and forwards alike.
- Some of them are very easily found.
- For example, the squares of $1$, $11$, $111$ and $1111$ are respectively $1$, $121$, $12321$, and $1234321$, all palindromes,
- and the rule applies for any number of $1$'s provided the number does not contain more than nine.
- But there are other cases that we may call irregular, such as the square of $264 = 69696$ and the square of $2285 = 5221225$.
- Now, all the examples I have given contain an odd number of digits.
- Can the reader find a case where the square palindrome contains an even number of figures?
$62$ - Factorizing
- What are the factors (the numbers that will divide it without any remainder) of this number -- $1000000000001$?
- This is easily done if you happen to know something about numbers of this peculiar form.
- In fact, it is just as easy for me to give two factors if you insert, say $101$ noughts, instead of $11$, between the two ones.
- There is a curious, easy, and beautiful rule for these cases.
- Can you find it?
$63$ - Find the Factors
- Find $2$ whole numbers with the smallest possible difference between them
- which, when multiplied together, will produce $1234567890$.
$64$ - Dividing by Eleven
- If the $9$ digits are written at haphazard in any order,
- for example $4 \ 1 \ 2 \ 5 \ 3 \ 9 \ 7 \ 6 \ 8$, what are the chances that the number that happens to be produced will be divisible by $11$ without remainder?
- The number I have written at random is not, I see, so divisible, but if I had happened to make the $1$ and the $8$ change places it would be.
$65$ - Dividing by $37$
- I want to know whether the number $49,129,308,213$ is exactly divisible by $37$,
- or if not, what is the remainder when so divided.
- How may I do this quite easily without any process of actual division whatever?
$66$ - Another $37$ Division
- If the $9$ digits are written at haphazard in any order, for example $412539768$,
- what are the chances that the number that happens to be produced will be divisible by $37$ without remainder?
$67$ - A Digital Difficulty
- Arrange the $10$ digits, $1 \ 2 \ 3 \ 4 \ 5 \ 6 \ 7 \ 8 \ 9 \ 0$, in such order that they shall form a number
- that may be divided by every number from $2$ to $18$ without in any case a remainder.
$68$ - Threes and Sevens
- What is the smallest number composed only of the digits $3$ and $7$ that may be divided by $3$ and $7$,
- and also the sum of its digits by $3$ and $7$, without any remainder.
$69$ - Root Extraction
- In a conversation I had with Professor Simon Greathead, the eminent mathematician, ...
- the extraction of the cube root.
- "Ah," said the professor, "it is astounding what ignorance prevails ...
- Thus, ignoring the obvious case of the number $1$, if we want the cube root of $512$, add the digits -- $8$, and there you are!"
- I suggested that that was a special case.
- "Not at all," he replied. "Take another number at random -- $4913$ -- and the digits add to $17$, the cube of which is $4913$."
- I did not presume to argue the point with the learned man,
$70$ - The Solitary Seven
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$71$ - A Complete Skeleton
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$72$ - Alphabetical Sums
RSR
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PR)MTVVR
MVR
---
KKV
KMD
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MVR
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$73$ - Alphabetical Arithmetic
F G
Less A B multiplied by C = D E
---
Leaving H I
---
$74$ - Queer Division
- Find the smallest number which when divided successively by $45$, $454$, $4545$, and $45454$
- leaves the remainders $4$, $45$, $454$, and $4545$ respectively.
$75$ - A Teasing Legacy
- Professor Rackbrain left his typist what he called a trifle of a legacy if she was able to claim it.
- The legacy was the largest amount that she could find in an addition sum,
- Every digit must be used once, a single $0$ may or may not appear, as in the examples below, and the dash may be employed in the manner shown.
£ s. d. £ s. d. - 3 7 4 2 5 - 4 8 6 7 3 - 5 9 -------- 1 6 - £10 9 8 -------- £2 - -
- The young lady was cleverer than he thought.
- What was the largest amount that she could claim?
$76$ - The Nine Volumes
- In a small bookcase were arranged $9$ volumes of some big work,
- numbered from $1$ to $9$ inclusive,
- on $3$ shelves exactly as shown:
26 5 9
7
184 3
- The $9$ digits express money value.
- You will see that they are so arranged that $\pounds 26 \ 5 \shillings \ 9 \oldpence$ multiplied by $7$ will produce $\pounds 184 \ 0 \shillings \ 3 \oldpence$
- Every digit represented once, and yet they form a correct sum in the multiplication of money.
- But in the blank shillings space in the bottom row is a slight defect, and I want to correct it.
- The puzzle is to use the multiplier $3$, instead of $7$, and get a correct result with the $9$ volumes, without any blank space;
$77$ - The Ten Volumes
- As an extension of the last puzzle, let us introduce a $10$th volume marked $0$.
- If we arrange the $10$ volumes as follows, we get a sum of money correctly multiplied by $2$.
54 3 9
2
108 7 6
- Can you do the same thing with the multiplier $4$ so that the $9$ digits and $0$ are all represented,
- once and once only?