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THE MIRROR
The number of the current year has an amazing property: if we add its four digits to the first power and multiply by the square of the sum of the second powers, we get exactly the year 2023 again.
(21 + 01 + 21 + 31)1 * (22 + 02 + 22 + 32)2
= 2023
You can do the math: 7 times 172 is actually 2023.
Is this a coincidence?
Is there a kind of mathematical-divine coincidence behind it?
Or does this happen often?
Find all years in this millennium for which the relationship described above applies!
The only other solution in this millennium is
2400
.
The solution described here (at least in theory) does not require a pocket calculator because only a few simple multiplications have to be carried out.
The trick is to first narrow down the range of digits in question.
The larger the digits in the year, the larger the product of their powers.
However, this product must not be larger than 3000, because we are looking for solutions in this millennium.
If only a single 5 appears as a digit, we end up with a product of more than 5000. Because the following applies:
(21 + 51 + …)*(22 + 52 + …)2 >= 7*292 = 5887
So the largest digit that appears can only be a 4.
We see if we stay under 3000 with 4:
(21 + 41 + …)*(22 + 42 + …)2 >= 6*202 = 2400
We found a solution right away!
We just need to set the missing two digits to zero.
(21 + 41 + 01 + 01)*(22 + 42 + 02 + 02)2 = 2400
However, there is no other solution in which the number 4 appears.
Why?
If only one of the two missing digits is not zero, i.e. at least one 1, the product will be greater than 3000.
(21 + 41 + 11 + …)*(22 + 42 + 12 + …)2 >= 7*212 = 3087
Other solutions are only possible if all digits of the year are less than 4.
The number of variants is thus manageable.
We assume that the first digit is a 2.
(The year 3000 still belongs to the current millennium, but it is not a solution, as one can easily calculate.)
If there are two threes among the remaining three digits, there is no solution because the product is greater than 3000:
(21 + 31 + 31 + …)*(22 + 32 + 32 + …)2 >= 8*222 = 3872
If there is only a 3, the remaining two digits can only be 0, 1 or 2.
We can quickly calculate these few variants.
The product is partly smaller than 2000, partly larger than 3000. But everything only fits with the already known 2023 solution.
If there is not a single 3 among the four digits, we only get a product greater than 2000 for the variant 2222, but no solution.
With this, we have examined all possible cases and found only one further solution.
Many thanks to the mathematician Dierk Schleicher who suggested this puzzle.
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