[vc_row][vc_column][vc_column_text]As our house is getting rather full of drumming drummers and maids-a-milking, my true love decided on a different gifting strategy this year. I will receive a number of gifts based on the formula $$g=\left[ \sqrt{2n}+1/2 \right]$$ where $g$ is the number of gifts, $n$ is the day of Christmas (e.g. $n=2$ for the second day of Christmas) and $[x]$ denotes the greatest integer less than or equal to $x$. Our regular gift stockist only has the usual Christmas fayre for sale so my true love has decided that [latexpage]
when $g=1$ I will receive $1$ partridge complete with pear tree
when $g=2$ I will receive $2$ turtle doves
when $g=3$ I will receive $3$ French hens
and so on (following items from previous years).
How many “calling birds” $(g=4)$ will I receive?
How many “swimming swans” $(g=7)$ will we have to accommodate?
What will I receive on the $12$th day of Christmas?
How many gifts will I receive overall this year?[/vc_column_text][lvca_accordion][lvca_panel panel_title="Reveal Solution"]Calculating a few terms gives a good feel for things
$n=1$ $g=\left[\sqrt{2}+1/2 \right]$ since $1<\sqrt{2}<1.5$ this gives $1.5<\sqrt{2}+1/2<1.5$ so $g=1.$
$n=2$ $g=\left[ \sqrt{4}+1/2 \right]$ so $g=2.$
$n=3$ $g=\left[ \sqrt{6}+1/2 \right]$ since $2<\sqrt{6}<2.5$ this gives $2.5<\sqrt{6}+1/2<3$ so $g=2.$
$n=4$ $g=\left[ \sqrt{8}+1/2 \right]$ since $2.5<\sqrt{8}<3$ this gives $3<\sqrt{8}+1/2<3.5$ so $g=3.$
The next change will take place when $\sqrt{2n}>3.5$ i.e. $2n>12.25$ so $n>6.125$ so, on the $7$th day $g=4$ and on days $5$ and $6$ $g=3$.
To go to $g=5$, $\sqrt{2n}>4.5$ i.e. $2n>20.25$ so $n>10.125$ . This is day 11.
The sequence would look like this: $1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5.$
There are $4$ days on which I would receive $4$ calling birds so I would have $16$ calling birds in total.
Fortunately, no swimming swans need to be accommodated since $g<7$ for all $12$ days.
On the 12th day of Christmas I will receive $5$ gold rings.
So this year I would receive $1+4+9+16+10=40$ gifts:
$1$ partridge complete with pear tree
$4$ turtle doves – only a small dovecote needed for these
$9$ French hens – or “poules françaises” as we like to call them
$16$ calling birds – a tad noisy but never mind
$10$ gold rings
Not a bad haul and much more manageable than previous years![/lvca_panel][/lvca_accordion][vc_column_text]Problem Page Coordinator: Claire Baldwin – Mathematics in Education and Industry
Acknowledgement: The IMA are indebted to MEI for sourcing and supplying Mathematics Today with these well-known puzzles.
First published in Mathematics Today (December 2017)
