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?

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

...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

...

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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