
For the last couple of years, I have written New Year postcards based on an annual number puzzle and thought I would pose it once again this year:
Write down the numbers from 10 down to 0 and the number of the year – leaving gaps between each one:
10 9 8 7 6 5 4 3 2 1 0 xxxx
Now, fill the gaps with one of the arithmetical operations, (+, -, x, ÷), brackets, ( that is ( and ) ), and an equals sign (=) so that you end up with a sum with the first ten numbers on the left hand side of the equation which generate the number of the year.
As it happens, there is usually more than one solution … and it is even easier if one was allowed to use the more advanced functions such as the square root sign (√), and factorial (!) … but it is not allowed is to change the order of the eleven numbers on the left hand side of the equation … of concatenate them (for example, make 109 from 10 and 9, or 98 from 9 and 8 … and so on).
So, for 2013, one solution might be:
(10 + 9 – 8) x ((7 – 6) + (5 x 4 x 3)) x (2 + 1 + 0) =
11 x ((7 – 6) + (5 x 4 x 3)) x (2 + 1 + 0)=
11 x (1 + (5 x 4 x 3) x (2 + 1 + 0) =
11 x (1 + 60) + (2 + 1 + 0) =
11 x 61 x (2 + 1 + 0) =
11 x 61 x 3 =
671 x 3 =
2013
and for 2014:
(10 + 9) x ((8 x 7) + (6 – 5 – 4)) x (3 – 2 + 1 + 0) =
19 x ((8 x 7) + (6 – 5 – 4)) x (3 – 2 + 1 + 0) =
19 x (56 + (6 – 5 – 4)) x (3 – 2 + 1 + 0) =
19 x (56 + (-3)) x (3 – 2 + 1 + 0) =
19 x 53 x (3 – 2 + 1 + 0) =
19 x 53 x 2 =
1007 x 2 =
2014
and for 2015:
10 9 8 7 6 5 4 3 2 1 0 2015
Back in 2015, I said that it was relatively easy to solve, and here’s the solution I offered then:
10 9 8 7 6 5 4 3 2 1 0 2015
(10 + 9) x ((8 x 7) + (6 x 5) + (4 x (3 + 2)) + 1 + 0 =
19 x ((8 x 7) + (6 x 5) + (4 x (3 + 2)) + 1 + 0 =
19 x (56 + (6 x 5) + (4 x (3 + 2)) + 1 + 0 =
19 z (56 + 30 + 4 x (3 + 2)) + 1 + 0 =
19 x (56 + 30 +(4 x 5)) + 1 + 0 =
19 x (56 + 30 + 20) + 1 + 0 =
19 x 106 + 1 + 0 =
2014 + 1 + 0 =
2015
Last year, 2016, was even easier … but, I am not going to give you that solution right now, because … well, the zero at the end plays a significant role.
So, what about this year, 2017?
10 9 8 7 6 5 4 3 2 1 0 2017
If I thought that, last year, the solution was relatively easy … then this one is ‘a piece of cake’ for anyone who solved it …
For the record: here is a solution – but there are others, especially if you decide to use square roots and/or factorials:
10 9 8 7 6 5 4 3 2 1 0 2017
(10 – 9) + (8 x 7 x 6 x (5 – 4) x 3 x 2) – 1 + 0 =
1 + (8 x 7 x 6 x (5 – 4) x 3 x 2) – 1 + 0 =
1 + (8 x 7 x 6 x 1 x 3 x 2) – 1 + 0 =
1 + (56 x 6 x 1 x 3 x 2) – 1 + 0 =
1 + (336 x 1 x 3 x 2) – 1 + 0 =
1 + (336 x 3 x 2) – 1 + 0 =
1 + (1008 x 2) – 1 + 0 =
1 + 2016 – 1 + 0 =
2017 + 1 x 0 =
2017 + 0 =
2017
Yes, it is there! I have written it in white … so you should be able to reveal it if you highlight it … or if you print the page.