Problems & Puzzles: Puzzles

 Puzzle 817. 2016  and prime numbers I have found some curio expressions for 2016 in the web. Three very interesting have been published by Claudio Meller in one of his pages. Here are them: 2016 = 2^5 + 2^6 + 2^7 + 2^8 +2^9 + 2^10 2016 = 666 + 666 + 666 + (6 + 6 + 6)                 (aaaa-aa)*aa                  __________ - a - a                        a+a 2016 =  _________________     valid if a is any of {0,1,2, ...,9}                     a + a + a By my own I have found two more: 2016 = T(63) = 63*64/2 2016 = 3^3 + 4^3 + 5^3 + 6^3 + 7^3 + 8^3 + 9^3 But I have not find any interesting expression using only prime numbers, except the very-ugly/non-interesting following expression: 2016 = 71 + 73 + ... +151 +157 (18 consecutive prime numbers) Q Send your best interesting expressions using only prime numbers?

Contributions came from Seiji Tomita

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Seiji wrote:

2016=a^3+b^3+c^3+d^3
a,b,c,d are primes.
{a,b,c,d}<5000
There are five solutions.

2016 = 349^3+ 277^3+ 167^3 -409^3
= 463^3+ 419^3 -557^3 -13^3
= 1123^3+ 571^3+ 149^3 -1171^3
= 2153^3+ 1061^3+ 271^3 -2237^3
= 3709^3+ 2099^3 -3877^3 -1259^3

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T. D. Noe wrote on Jan 24, 2016

There are more solutions to this puzzle in the form of one positive cube
and three negative ones.  For example, in the range up to 5000, there are

2016 = 1637^3 - 1399^3 - 1181^3 - 113^3

2016 = 1877^3 - 1867^3 - 443^3 - 263^3

2016 = 3457^3 - 3359^3 - 1423^3 - 811^3

There appear to be an infinite number of solutions.  I computed 72
solutions less than 200000. They are at
http://IntegerSequences.org/s000832.html

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Pierandrea Formusa wrote on Jan 22, 2019:

We can express 2016 like:

a) p1+p2^2, with p1 and p2 primes and different, in 4 ways:
 p1 p2 p1 p2^2 sum 1847 13 1847 169 2016 1487 23 1487 529 2016 647 37 647 1369 2016 167 43 167 1849 2016

b) p1+p2^2+p3^3, with p1, p2 and p3 primes and different, in 2 ways:
 p1 p2 p3 p1 p2^2 p3^3 sum 1999 3 2 1999 9 8 2016 1669 2 7 1669 4 343 2016
and 2 is mandatory, so if pi is different from 2 for each i, it is impossible to get 2016 with this kind of sum

c) p1+p2^2+p3^3+p4^4, with p1, p2, p3 and p4 primes and different, in 11 ways:
 p1 p2 p3 p4 p1 p2^2 p3^3 p4^4 sum 1567 5 7 3 1567 25 343 81 2016 1471 11 7 3 1471 121 343 81 2016 1423 13 7 3 1423 169 343 81 2016 1303 17 7 3 1303 289 343 81 2016 1231 19 7 3 1231 361 343 81 2016 1063 23 7 3 1063 529 343 81 2016 751 29 7 3 751 841 343 81 2016 631 31 7 3 631 961 343 81 2016 223 37 7 3 223 1369 343 81 2016 1039 3 7 5 1039 9 343 625 2016 523 29 3 5 523 841 27 625 2016
and 3 is mandatory, so if pi is different from 3 for each i, it is impossible to get 2016 with this kind of sum

d) p1+p2^2+p3^3+p4^4+p5^5, with p1, p2, p3, p4 and p5 primes and different, in 35 ways, whose smallest is:

 p1 p2 p3 p4 p5 p1 p2^2 p3^3 p4^4 p5^5 sum 19 3 11 5 2 19 9 1331 625 32 2016
and 2 is mandatory, so if pi is different from 2 for each i, it is impossible to get 2016 with this kind of sum

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