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?

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Contributions came from Seiji Tomita

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Seiji wrote:

About Puzzle 817.

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
three additional solutions:

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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