Problems & Puzzles: Puzzles

Puzzle 620. 2012

On Jan 1 , 2012 Emmanuel Vantieghem sent me an email with the following line:

First of all, a happy 2012 = (1+2-3+4)*(5-6+7*8*9)

In return, my answer to him was:

Same happy wishes to you my friend!, by the way 2012 is also equal to (-139 + 149 + 151 - 157 ) * (163 + 167 +173), using seven consecutive primes only.
His last exchange in this issue was:

2012 is also equal to (163 + 167 +173) * (181 -179) * (193 - 191, using again seven consecutive primes only.

Q1. Perhaps you would like to send some other expressions related to 2012 & consecutive primes.

Good luck with this puzzle & all the year 2012!


Contributions came from Giovanni Resta, Jan van Delden, Farideh Firoozbakht & Jahangeer John Kholdi , Michael Reid, Hakan Summakoğlu, Emmanuel Vantieghem, J. K. Andersen & Carlos Rivera.

***

Giovanni wrote:

Thanks to http://www.trnicely.net/
we can write, in consecutive primes,
2012 = 803351036411255563084726695271453-
803351036411255563084726695269441

***

Jan wrote:

2^11-(13-7)^(5-3)

***

Farideh Firoozbakht & Jahangeer John Kholdi wrote:

2012 = 2*3*5*7*11-13*17+19-23+29-31+37-41+43-47+53-59-61-67+71

2012 = 2*3*5*7*11-13*17-19-23-29+31-37
2012 = 2^(3+5)+7+11+13+1719-23+29

***

Michael Reid wrote:

This year, I sent out a new year's e-mail, the answers to which
were  11 * 5^3 + 13 * 7^2  (I gave the form  a * b^c + d * e^f ,
and riddled to substitute the first six primes in some order)
and  2^11 - 3 * (5 + 7)  (I didn't give the form, just said there
was an expression using the first five primes).

I looked some more at consecutive primes and found the "obvious"
solution using two consecutive primes:
803351036411255563084726695271453 - 803351036411255563084726695269441
(Well, the actual prime values I found at
  http://www.trnicely.net/gaps/g2k.html
surely you know this site.)

I did not find any solution with three consecutive primes.
With four consecutive primes, there is
997 + 1009 - 1013 + 1019
I made a small effort to look at other forms  (p q +/- r) / s
but did not find other solutions.

With five consecutive primes, there is the solution I already sent
using the smallest five primes.

Using the first six primes, there is the nice expression I found
(David Joyner put the puzzling form of my question on his blog:
http://wdjoyner.wordpress.com/2012/01/04/michael-reids-happy-new-year-puzzle/

There are other solutions using the first six primes, e.g.
3^7 - 5 * (2 * 11 + 13)
3^7 - 13^2 - 11 + 5
5^2 * 3^(11 - 7) - 13
but I don't think they're as "cute" as the  a * b^c + d * e^f  one.

It might be nicer to use the primes "in order".
With the first seven primes in order, I found
2 + ((3 + 5) * 7 + 11) * (13 + 17)
2 * (3 + (5 * 7 + 11 + 13) * 17)
2 * (3 - (5 - 7 * 11 + 13) * 17)

A few more solutions:
Using the first six primes:
3^7 - 2^5 - 11 * 13
((3 + 5) * 7 - 11)^2 - 13
(5 * 11 - 3 - 7)^2 - 13
2^11 - (13 - 7)^(5 - 3)
(not as efficient as the first five prime expression with 2^11,
but the difference of powers is nice)

Using consecutive primes "in order":
2 consecutive primes is possible, of the form  -a + b .
4 consecutive primes in order is possible (my previous expression
used them in order)

I believe 5 consecutive primes in order should be possible,
for example, in the form  a * (b - c) + d + e ,
where  b  and  c  are large twins.  But I did not find a solution.
Other similar forms are possibilities using large primes.
Is there a solution using 5 consecutive "small" primes in order?

Using 6 consecutive primes in order is possible:
487 + 491 - 499 + 503 + 509 + 521
The smallest set of 6 consecutive primes in order I found is:
(5 + 7 * 11 * 13) * (-17 + 19)
Is there a solution using  3, ... , 17  or  2, ... , 13  in order?

Another solution using the first 7 primes in order:
-2 * 3 - (5 - 7)^11 - 13 - 17

***

Hakan Summakoğlu wrote:

My expressions equal to 2012 by using smallest n consecutive primes, 
For n=2 to n=30:

n=2:  I couldn't find a solution.
n=3:  I couldn't find a solution.
n=4:  997+1009-1013+1019 
n=5:  (1993-1997)/(1999-2003)+2011
n=6:  (5+7*11*13)*(-17+19)
n=7:  2+((3+5)*7+11)*(13+17)
n=8:  ((2-3)*5-(7*11*13))*(17-19)
n=9:  (2-3-5)+ 7*(11-13+17)*19+23
n=10:  (2-3-5-7)* (11+(13-17)*(19+23))-29
n=11:  (-2+3+5+7+11)+(-13+17)*(19*23+29+31)
n=12:  (-2+3+5+7+11)-(13-17)*(-19+23)+29*(31+37)
n=13:  ((2-3-5-7-11-13-17+19)*23)+29+(31+37)*41
n=14:  ((2-3-5-7-11-13-17)-(19-23-29)*31+37)*(-41+43)
n=15:  (-2+3+5+7+11+13+17+19+23-29-31)/(37-41)+(43*47)  
n=16:  ((-2+3+5+7+11+13+17+19+23-29)*31)+37+41-43-47-53
n=17:  (-2+3+5+7+11+13+17+19+23+29-31-37)*41-43+47*(53-59)
n=18:  ((2-3-5-7-11-13-17-19-23-29-31-37+41)/(43-47))*53)+59-61
n=19:  (-2+3+5+7+11+13+17+19+23+29+31+37)+41*((43+47) / (53-59)+61)-67
n=20:  (2-3-5-7-11-13-17-19-23-29-31-37)-(41-43-47)*(53-(59-61)*(67-71)) 
n=21:  (2-3-5-7-11-13-17-19-23-29-31-37-41-43-47)+(53*(59-61)+67+71)*73 
n=22:  (2-3-5-7-11-13-17-19-23-29-31-37-41-43+47)*(53-59)-(61-67+71-73)*79
n=23:  (2-3-5-7-11-13-17-19-23-29-31-37-41-43-47-53)-(59*61)+67+71+(73*79)+83
n=24:  (-2+3+5+7+11+13+17+19+23+29+31+37+41+43+47+53)*(59-61)+(67+71)*73+79-(83*89)
n=25:  (2-3-5-7-11-13-17-19-23-29-31-37-41-43-47-53-59+61)*(67-71)-73-79-83*(89-97)
n=26:  (-2+3+5+7+11+13+17+19+23+29+31+37+41+43+47+53+59+61+67-71)*73+79-(83+89)*(97+101)
n=27:  (2-3-5-7-11-13-17-19-23-29-31-37-41-43-47-53-59-61-67)*(71+73)-79+83-(89-97)*101*103
n=28:  (-2+3+5+7+11+13+17+19+23+29+31+37+41+43+47+53+59+61+67+71)* ((73-79)*83+89-(97-101)*103)+107
n=29:  (2-3-5-7-11-13-17-19-23-29-31-37-41-43-47-53-59-61-67-71-73+79-83-89-97)*(101-103)+107+109
n=30:  ((2-3-5-7-11-13-17-19-23-29-31-37-41-43-47-53-59-61-67-71-73-79-83)+89*(97-101+103)+107)/(-109+113)

***

Emmanuel Vantieghem wrote:

This is my contribution to Puzzle 620 :
 

It is easy to write 2012 as an alternating sum of consecutive primes.  The one with the smallest initial prime is perhaps : 2012 = -7+11-13+17-19+ ...  -4003+4007-4013+4019 (552 primes) and it is probably the longest alternating sum.  I found a lot of other alternating sums, whence it became a challenge to construct such sums with as few terms as possible.  My shortest solution writes 2012 as a sum of 14 hundred-digit primes :

2012 =-1234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234589141

                +1234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234589519-1234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234589623

                +1234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234589831-1234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234590263

                +1234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234590943-1234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234591067

                +1234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234591207-1234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234591301

                +1234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234591441-1234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234591669

                +1234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234592027-1234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234592039

                +1234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234592147

(all proved prime by Mathematica).

Perhaps it is even possible two find two consecutive primes  p  and  q  with  2012 = -p+q, but it will be a little bit hard to find them.

Other representations are (the +/- signs are not alternating here) :

2012 = 1013+1019+1021+1031-1033-1039
= 983-991+997-1009+1013+1019 = 997+1009-1013+1019
= (5+7*11*13)*(-17+19) 

If we allow concatenations, we have :

2012 = 1923+29+31+37-43+47-53 = -17*19+2329-31+37
= (2*3+5-7)(1113+1719-2329)

and if we impose ourselves to start with  2, we have :
2*(-3+5)*(711-13*17+19+23-29)

***

J. K. Andersen wrote:

All expressions list consecutive primes in increasing order.

2+3+5+(7*11*13)*(-17+19) = 2012
2^(3#+5)-(-7+11)#*(-13+17)# = 2012

The first known occurrence of a prime gap of 2012, found by Helmut Spielauer:
-803351036411255563084726695269441 + 803351036411255563084726695271453 = 2012

The first gap of 2012 for 2012-digit primes, found with PrimeForm/GW and
proved by Primo:  -(10^2011+78740989) + (10^2011+78743001) = 2012

(6661 + ... + 42131)/42139 = 2012, with 3546 primes in the sum.

Let p = 291705862783962687847115915117632823381473,
q = p+1006, r = q+42, s = r+2. Then (p-q)*(r-s) = 2012.

Let p = 6810464776609315912434039063546670736887834732306822\
984289996995993637312616310589589784059838626663524463929705\
788430056256904116496430171290954364620514929256366692069037,
q = p+4024, r = q+66, s = r+2. Then (p-q)/(r-s) = 2012.

***

Carlos Rivera

Using only consecutive primes as needed:

2012 = 503*(503-499)
2012 = 2011 + integer(2027/2017)

***

 

 

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