Jan wrote:

Anton wrote:

J. Tramu wrote:

Farideh wrote:
 

It's obvious that n must be even.

For n = 6 we have :

I. 7*8*9*10*11*12 = 665280 = 110849 + 110863 + 110879 + 110881 + 110899 + 110909

II. 119*120*121*122*123*124 = 3215142342720 = 535857057017+ 535857057049 +
535857057061 + 535857057101 + 535857057233 + 535857057259

III. 197*198*199*200*201*202 = 63032120157600 = 10505353359541 + 10505353359587
+ 10505353359601 + 10505353359607 + 10505353359617 + 10505353359647

Also for n = 8 we have :

I. 9! = 2*3*4*5*6*7*8*9 = 362880 = 45329 + 45337 + 45341 + 45343 + 45361 + 45377 +
45389 + 45403

II. 11*12*13*14*15*16*17*18 = 1764322560 = 220540241 + 220540247 + 220540259 +
220540291 + 220540339 + 220540381 + 220540393 + 220540409

III. 59* 60*61*62*63*64*65*66 = 231580827878400 = 28947603484699 + 28947603484733
+ 28947603484783 + 28947603484817+ 28947603484823 + 28947603484829 +
28947603484837 + 28947603484879

G. Resta wrote:

For Puzzle 477 (Sum of k primes = Product of k integers)
I performed this systematic search: for each even k,
I searched the smallest starting point m such that
m*(m+1)*...*(m+k-1) is the sum of k primes.
For example, for k=6 I obtained m=7, since
7*8*9*10*11*12 is equal to the sum of the 6
primes from 110849 to 110909.
In the following table I listed, for k from 2 to 44,
the smallest starting point m and the first prime of
the k-uple of primes.
(If the prime has more than 70 digits I only report
the number of digits for brevity, but I can supply
the full data if needed).
For K=46,48,50,52,54,56 I was unable to find
the value of m since my search had m<10000 as limit, but
I found the starting point m=8277 for k=58. Here
the starting prime has 226 digits:
36285760306496980971773409816885420514436690853552
62993228760719577673855684755118736033687462345091
54803233304199245511928171267886352177573700255189
70481916901135751983129102494949514706429636517434
04006949519359999999987881

K M P
------------------------------------
2 3 5
4 2 23
6 7 110849
8 2 45329
10 706 3277972714563452726941760377
12 191 275654436711058095634943693
14 66 7747818632800435373183423
16 203 925223589276817217778191355819263479
18 836 2650780340273990096617820296346007921968118128639141
20 362 125054647109174045901310383877940049329352499199177
22 3326 (77 digits)
24 4027 (86 digits)
26 3800 (92 digits)
28 2457 (94 digits)
30 3153 (104 digits)
32 5245 (118 digits)
34 2669 (116 digits)
36 3746 (128 digits)
38 933 (112 digits)
40 2149 (132 digits)
42 1319 (130 digits)
44 2082 (145 digits)

Gennady Gusev wrote:

I found the solutions for the puzzle 477 for even k from 6 to 46 (and the smaller one for k = 4).
They are in attached file "puzzle 477.txt".

For example, the most solution is for k = 46.

In brief form the solution is:

Denote N = 11250 * .. * 11295 (product of 46 consecutive integers).

Let M = N / 46,
then the primes (all of 185 digits) are M-10877, M-10529, M-10093, M-9833, M-9631, M-9397, M-9341, M-8693, M-8009, M-7349, M-6571, M-5167, M-4799, M-4297, M-4283, M-4261, M-3259, M-2749, M-2221,
M-1049, M-1033, M-863, M-743, M-647, M+113, M+307, M+1321, M+2549, M+2689, M+3229, M+3469, M+3659, M+6133, M+6701, M+6899, M+7547, M+7603, M+8009, M+8171, M+8989, M+9167, M+9587, M+9721, M+9739, M+9803, M+10289

The sum of them is N.

J. K Andersen wrote:

n must be even because the prime 2 is not part of any solution,
and the sum of an odd number of odd primes is odd.
In the below list, the n consecutive primes starting at p add up to the
product of the n consecutive numbers starting at b. All primes are proven.

n b p
---------------------
2 3 5
4 2 23
6 7 110849
8 2 45329
10 706 715!/705!/10 - 263
12 191 202!/190!/12 - 307
14 66 79!/65!/14 - 577
16 203 218!/202!/16 - 521
18 836 853!/835!/18 - 859
20 362 381!/361!/20 - 823
22 3326 3347!/3325!/22 - 2887
24 4027 4050!/4026!/24 - 2131
26 3800 3825!/3799!/26 - 2777
28 2457 2484!/2456!/28 - 2351
30 3153 3182!/3152!/30 - 2719
32 5245 5276!/5244!/32 - 3299
34 2669 2702!/2668!/34 - 3467
36 3746 3781!/3745!/36 - 5711
38 933 970!/932!/38 - 3607
40 2149 2188!/2148!/40 - 4967
42 1319 1360!/1318!/42 - 4889
44 2082 2125!/2081!/44 - 6577
46 11250 11295!/11249!/46 - 10877
48 25232 25279!/25231!/48 - 10267

The listed b and p are proven to be the smallest for n <= 20.
For larger n they are probably the smallest. Smaller candidates were
discarded by finding consecutive probable primes with the wrong sum.

J. C. Rosa wrote:

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