Problems & Puzzles: Puzzles

Puzzle 513. Four consecutive primes JM Bergot poses the following nice puzzle: 47*53 + 59*61 = 6090 = 2*3*5*7*29 There are four consecutive primes on each side of the equality. Can this ever happen again?

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Contributions came from J.K. Andersen, Jan van Delden, Jacques Tramu, Farid Lian, Luca Poletti, Giovanni Resta & Farideh Firoozbakht.

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Andersen wrote:

Let n = 1376810447*997#/2. (n-32, n+4, n+8, n+16) are four consecutive
425-digit primes, proved by Marcel Martin's Primo.

(n-32)*(n+4) + (n+8)*(n+16) = 997# * c, where 997# is the product of the 168 consecutive primes below 1000, c = 3011*6719*24749*28859*204913*249442205324565355225951*p, p is a 389-digit prime proved by Primo.

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Jan wrote:

Since the left side is always even, I have been looking for two different types of solution for the right side of the given equation:
[I imposed that all primes should have multiplicity 1]

Type I) (4 consecutive primes starting with prime q=2) * r: They are rather abundant
Type II) 2* (4 consecutive primes, starting with prime q>2) * [r, preferably absent] Far less abundant
Type III) 2*3*(4 consecutive primes, starting with prime q>3) Not found yet

Some results, with starting prime p on the left side of equation:

p<= #Type I #Type II
10^2 1 0
10^3 2 0
10^4 3 0
10^5 21 1
10^6 82 2
10^7 544 16
10^8 3958 85

Some peculiarities:

Type I) The largest solution sofar p=99969239, r=95179532814881.
Type II) Most sets of consecutive primes on the right side start with q=5 or q=7.
Exceptions to this "rule" are:
p=22022773, q=47, r=54098983.
p=46132769, q=17, r=9878509969.
p=53730137, q=13, r=29892579823
p=66337429, q=11, r=95275072507
p=72042389, q=11, r=112366818937
I didn't find any p for which the right side consists of exactly 5 primes, i.e. r is always present.

Changing the order of the primes on the left side of the equation [i.e. a*d+b*c or a*c+b*d] also generates a lot of solutions.

I also found the following solution to a similar type of question:

3096047*3096059*3096061+3096073*3096083*3096103=2*3*5*7*11*13*1976548147556641 [Type I]

Extending this in either of the two following directions, add a multiple of more terms or adding more terms of just one multiplication, will increase the difficulty of finding a solution.

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Tramu wrote:

... there are a LOT (infinite number ? ) of solutions :
for example :

54441379*54441383 + 54441407*54441419 = 5927731414623690 = 2*3*5*7*28227292450589
54479563*54479587 + 54479591*54479599 = 5936050363544490 = 2*3*5*7*28266906493069
54487129*54487157 + 54487187*54487211 = 5937703607167710 = 2*3*5*7*28274779081751
54496511*54496549 + 54496567*54496573 = 5939747923805430 = 2*3*5*7*28284513922883
54533639*54533651 + 54533653*54533657 = 5947837964645010 = 2*3*5*7*28323037926881
54533803*54533849 + 54533861*54533903 = 5947882464187230 = 2*3*5*7*28323249829463
54554897*54554911 + 54554939*54554957 = 5952479901731790 = 2*3*5*7*28345142389199
54618197*54618209 + 54618227*54618271 = 5966301222774690 = 2*3*5*7*28410958203689
54629371*54629387 + 54629411*54629423 = 5968742251685430 = 2*3*5*7*28422582150883
54654857*54654863 + 54654869*54654871 = 5974308536336490 = 2*3*5*7*28449088268269
54702161*54702173 + 54702191*54702247 = 5984659838019030 = 2*3*5*7*28498380181043
and more, more ..........

10257629*10257631 + 10257649*10257659 = 210438438800590 = 2*5*7*11*13*59*701*508301
and more, more ..........

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Farid wrote:

Solution of 513

If, A = a*b + c*d, where: a, b, c and d are all primes, and B = e*f*g*h, where too e, f, g and h are all primes.

Part 1: a = 2 and e = 2

A = always will be odd. B = always will be even, so A / B = odd / even <> Integer, never will be integer No has solution

Part 2: a > 2 and e = 2

A = always will be even. B = always will be even, so even A / B = fac =

 - prime

  - odd

  - composite

Part 3: a > 2 and e > 2

A = always will be even. B = always will be odd, so A / B = even / odd = Integer only even

Examples

Part 2: odd prime Prime

a b c d e f g h A B A / B

47 53 59 61 2 3 5 7 6090 210 29
149 151 157 163 2 3 5 7 48090 210 229
3989 4001 4003 4007 2 3 5 7 32000010 210 152381
11489 11491 11497 11503 2 3 5 7 264270090 210 1258429
12539 12541 12547 12553 2 3 5 7 314754090 210 1498829
16217 16223 16229 16231 2 3 5 7 526501290 210 2507149
18899 18911 18913 18917 2 3 5 7 715176210 210 3405601
19013 19031 19037 19051 2 3 5 7 724510290 210 3450049
19469 19471 19477 19483 2 3 5 7 758551290 210 3612149
19813 19819 19841 19843 2 3 5 7 786378810 210 3744661
20143 20147 20149 20161 2 3 5 7 812045010 210 3866881
20753 20759 20771 20773 2 3 5 7 862287510 210 4106131
22259 22271 22273 22277 2 3 5 7 991905810 210 4723361
24697 24709 24733 24749 2 3 5 7 1222355190 210 5820739
32027 32029 32051 32057 2 3 5 7 2053251690 210 9777389
36523 36527 36529 36541 2 3 5 7 2668881810 210 12708961 

Part 2: even and odd composite composite/even

a b c d e f g h A B A / B

193 197 199 211 2 3 5 7 80010 210 381
199 211 223 227 2 3 5 7 92610 210 441
257 263 269 271 2 3 5 7 140490 210 669
613 617 619 631 2 3 5 7 768810 210 3661
1123 1129 1151 1153 2 3 5 7 2594970 210 12357
1259 1277 1279 1283 2 3 5 7 3248700 210 15470
2063 2069 2081 2083 2 3 5 7 8603070 210 40967
3167 3169 3181 3187 2 3 5 7 20174070 210 96067
3299 3301 3307 3313 2 3 5 7 21846090 210 104029
3583 3593 3607 3613 2 3 5 7 25905810 210 123361
4561 4567 4583 4591 2 3 5 7 41870640 210 199384
6397 6421 6427 6449 2 3 5 7 82522860 210 392966
7591 7603 7607 7621 2 3 5 7 115687320 210 550892
8431 8443 8447 8461 2 3 5 7 142653000 210 679300
8627 8629 8641 8647 2 3 5 7 149161110 210 710291
9539 9547 9551 9587 2 3 5 7 182634270 210 869687 

Part 3: integer even even

a b c d e f g h A B A / B

18859 18869 18899 18911 3 5 7 11 713249460 1155 617532
20483 20507 20509 20521 5 7 11 13 840910070 5005 168014
212203 212207 212209 212227 7 11 13 17 90067441464 17017 5292792
109159 109169 109171 109199 11 13 17 19 23838142900 46189 516100
435221 435223 435247 435257 13 17 19 23 378862492762 96577 3922906

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Luca wrote:

There are lots of that tipe solutions, for example those are the first twenty:

a*b+c*d=x=p*q*r*s*n

47*53+59*61=6090=2*3*5*7*n
149*151+157*163=48090=2*3*5*7*n
193*197+199*211=80010=2*3*5*7*n
199*211+223*227=92610=2*3*5*7*n
257*263+269*271=140490=2*3*5*7*n
613*617+619*631=768810=2*3*5*7*n
1123*1129+1151*1153=2594970=2*3*5*7*n
1259*1277+1279*1283=3248700=2*3*5*7*n
2063*2069+2081*2083=8603070=2*3*5*7*n
3167*3169+3181*3187=20174070=2*3*5*7*n
3299*3301+3307*3313=21846090=2*3*5*7*n
3583*3593+3607*3613=25905810=2*3*5*7*n
3989*4001+4003*4007=32000010=2*3*5*7*n
4561*4567+4583*4591=41870640=2*3*5*7*n
6397*6421+6427*6449=82522860=2*3*5*7*n
7591*7603+7607*7621=115687320=2*3*5*7*n
8431*8443+8447*8461=142653000=2*3*5*7*n
8627*8629+8641*8647=149161110=2*3*5*7*n
9539*9547+9551*9587=182634270=2*3*5*7*n
9859*9871+9883*9887=195031410=2*3*5*7*n

Later he added:

...from this puzzle we could find another interesting puzzle:

a,b,c,d consecutive primes, p,q,r,s consecutive primes, n,z natural numbers, such that:

a*b + c*d = x = p*q*r*s * n

We can see that some x may have multiple quadruplets {p,q,r,s}

example:
18859*18869+18899*18911=713249460=2*3*5*7*n
18859*18869+18899*18911=713249460=3*5*7*11*n

say k the number of different quadruplets, can you find {a,b,c,d} such that k>4?

the least value for k:

k __|_______a________|______________x____________|
2 __|_____18859_____|________713249460_________|
3 __|____133051_____|________35415249870_______|
4 __|____553181_____|________612056054610______|

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Giovanni wrote:

In this period I've been busy enough with alphametics. I have a brand new website
where I implemented an on-line tool that, given two words X and Y, shows the words Z (in English, Spanish and Italian) such that the alphametics X+Y=Z, X-Y=Z, X*Y=Z or X/Y=Z have unique solution. Not prime-related, but if you are curious the address is
http://www.iread.it/cryptarithms.php

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You asked in puzzle 513 if a configuration like
47*53 + 59*61 = 6090 = 2*3*5*7*29
with 4 consecutive prime numbers on both sides
can happen again.

It turns out that such configurations are quite common, the next one being
149*151+157*163=48090=2*3*5*7*229.

So I searched for some simple extensions and generalizations.
With 5 consecutive primes on both sides we can have different patterns, namely:
v=p+q+r+s+t, p=135101 v=3^3*5^3*7*11*13
v=p*q+r+s+t, p=6529 v=2*3^3*5*7*11^3*17
v=p+q*r+s+t, p=2971 v=2^3*3^2*5^3*7*11*13
v=p+q+r*s+t, p=2459 v=2*3^3*5^2*7*11*59
v=p+q+r+s*t, p=7069 v=2^3*3^4*5*7^2*11*29
v=p*q+r*s+t, p=2590771 v=3*5^2*7*11*13*263*679909
v=p*q+r+s*t, p=24043 v=3^2*5*7*11*13*25703
v=p+q*r+s*t, p=3727 v=3*7*11*13*17*19*29
v=p*q*r+s+t, p=427781 v= 3*5*7^2*11*13*1567*475320247
v=p+q*r*s+t, p=818897 v=3*5*7*11*13*23*83*191*3469*28921
v=p*q*r+s*t, p=2069 v=2*3^3*5*7*11*431597
v=p*q+r*s*t, p=159739 v=3*5^2*7*11*13*83*654549013
v=p*q*r*s+t, p=5449 v=2^4*3*5*7*11*48408942683
v=p+q*r*s*t, p=919 v=2*3*5*7*11*335801729

for 6 primes I just checked some patterns:
v=a*b+c*d+e*f, a=14371793 v= 3*5*7*11*13*17*2677*906823
v=a*b*c+d*e*f, a=8999 v=2^2*3^2*5*7*11*13*503*16141
v=a+b*c*d*e+f, a=1310611 v=7*11*13*17*19*23*1373*15719527*18384617

for 8 primes I just checked this:
v=a*b+c*d+e*f+g*h, a=45224057
v=2^3*3^2*5*7*11*13*17*19*70284689

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Farideh Firoozbakht wrote:

It seems that there exist infinitely many such set of four consecutive primes.

149*151 + 157*163 = 2*3*5*7*229
3989*4001 + 4003*4007 = 2*3*5*7*152381
11489*11491 + 11497*11503 = 2*3*5*7*1258429
12539*12541 + 12547*12553 = 2*3*5*7*1498829
...

Between the first 10^7 primes there exist 6582 primes p1, such the
p1*p2 + p3*p4 = 2*3*5*7*q where p1, p2, p3 & p4 are four consecutive
primes (p1 < p2 < p3 < p4 ) and q is prime.

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I found only one set of four consecutive primes p1, p2, p3 & p4 (p1 < p2 < p3 < p4)
where p1 + p2 + p3 + p4 = 2*(q1*q2*q3*q4) such that q1, q2, q3 & q4 are four
consecutive primes (q1 < q2 < q3 < q4), 101+103+107+109 = 2*(2*3*5*7).

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