Henry Ernest Dudeney/Puzzles and Curious Problems/118 - Six Simple Questions/Solution

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Puzzles and Curious Problems by Henry Ernest Dudeney: $118$

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.


Solution

The answers given by Dudeney 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

\(\text {(1)}: \quad\) \(\ds 12000 + 1200 + 12 - \paren {4000 + 1100 + \tfrac 1 2}\) \(=\) \(\ds 13212 - 5100 \tfrac 1 2\)
\(\ds \) \(=\) \(\ds 8111 \tfrac 1 2\)
$(2): \quad$ 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", Dudeney 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.

$\blacksquare$


Sources

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