Problems & Puzzles: Puzzles

 Puzzle 786. Looking for certain set of integers. Giovanni Resta sent the following nice puzzle. We are looking for 3 set of numbers, which may be composites or primes. Set A, of those numbers which divide the concatenation of the two previous primes. For example, 7 is preceded by the primes 3 and 5, and 7 divides their concatenation 35. Set B, of those n which divide the concatenation of prime preceding n with the prime following n. For example, 9, which is sandwiched between the primes 7 and 11, divides their concatenation 711. Set C, of those numbers which divide the concatenation of the two next primes. For example, the two primes following 801 are 809 and 811, and 801 divides 809811. In practice, if  np(n) is nextprime of n and pp(n) = previous prime of n and & is concatenation, we are looking for (A) x which divides  pp(pp(x)) & pp(x) (B) x which divides pp(x) & np(x) (C) x which divides np(x) & np(np(x)) Find: Q1) the 10 smallest members of the sets A, B and C. Q2) for each set A, B, and C, an element with exactly 30 digits. Q3) for each set A, B, and C, an element with at least 100 digits. Q4) for each set A, B, and C, the largest prime element you can.

Contributions came from Giovanni Resta, Jeff Heleen, Emmanuel Vantieghem.

***

Giovanni wrote:

There are at least 2 approaches, apart pure brute force,
that can be used with these problems.

The first method is systematic and allows us to find all the
solutions up to a certain limit.
The second method is more like a shot in the dark, but allows us
to find larger solutions.

Let's start with the systematic approach applied to problem (C), i.e., I'm looking numbers n which divide the concatenation of the two primes
following n. The other problems can be solved in a similar way.

Let's call p = nextprime(n), and q = nextprime(p).
So we are looking for a number n that divides the concatenation of p and q, denoted with p&q.

Let's suppose we are searching solutions with exactly k digits,
and that n, p and q all have k digits. The remaining cases occur
when a power of 10 is between n and p or between p and q, but these
cases are very few and can be checked directly.

If p and q  have k digits then their concatenation p&q is equal to

p*10^k + q.

Between n and p, and between n and q there are two gaps,
of size g1 and g2 and we can write

p = n + g1 and
q = n + g2.

So, we can rewrite the concatenation of p and q in terms
of n, g1 and g2 as

p&q = (n+g1)*10^k + (n+g2),

and we want this expression to be divisible by n.

Expanding and collecting n we obtain

p&q = n * (10^k+1) + g1*10^k+g2,

and clearly this number can be a multiple of n only if
g1*10^k+g2 is a multiple of n.

Before proceeding, let us talk about the size of the gaps
g1 and g2.
According to an empirical study on the
gaps between primes (you can check Mathworld or wikipedia
or the site http://sweet.ua.pt/tos/gaps.html
about prime gaps), we know that for 4 < n < 10^18,
nextprime(n)-n < (log n)^2,
and I will use this bound as a conjecture also
for larger n.

If we write for brevity L(k) = log(10^k)^2, and we are looking
for numbers n of k digits, i.e., 10^(k-1) <= n < 10^k, then
we can assume that g1 < L(k) and g2 < 2*L(k) or more
precisely, g2 < g1+L(k).

For example, L(16) is about 1357, a number very small
compared to 10^16.

So, what we can do, given k, is to check for all  the odd divisors n
of the number g1*10^k+g2,  as g1 ranges from 2 to L(k) in steps of 2
and g2 ranges from 4 to g1+L(k).

Indeed, we know that the gaps between primes greater than 2
are always even numbers, and since q is odd and p&q is odd also the
divisor n must be odd.

Once we have found a divisor n of g1*10^k+g2, we must check
that n has k digits, and that p=n+g1 and q=n+g2 are prime numbers and
moreover precisely the next prime numbers, i.e., there are no
other primes between n and p=n+g1 or between p=n+g1 and q=n+g2.

For example, let's take k = 12 and assume we are considering
the possibility g1 = 4 and g2 = 18. This means that
we are looking for an odd divisor n of 4*10^12+18 = 4000000000018
of 12 digits. One such divisor is n = 181818181819.

We now check n+g1 = 181818181819 and n+g2 = 181818181837.
They are indeed the two primes following 181818181819,
and thus we have found a solution for (C), since
by construction 181818181819 divides 181818181819181818181837.

When k is large, finding the divisors of g1*10^k+g2 can be
very time consuming, however there is a shorcut.

We want to find a number n of k digits which divides g1*10^k+g2,
so it must exist another number m such that
g1*10^k+g2 = n*m .
It is very easy to see that, for n to have k digits, m
must satisfy  g1 <= m <= 10*g1.

So, instead of factoring g1*10^k+g2, which can be a huge number,
we can simply try to divide it for all the numbers m between
g1 and 10*g1, a much more affordable task.

Moreover, remembering that we are looking for an odd n,
we can step between values of m which are multiple of
appropriate powers of 2.
For example, if g1*10^k+g2 is a multiple of 8, then also
m must be a multiple of 8, otherwise either (g1*10^k+g2)/m
is not an integer or it is not odd.

So, putting together all the pieces,
for numbers of k digits the pseudocode is:

for g1 from 2 to L(k) do
for g2 from 4 to g1+L(k) do
compute z = g1*10^k+g2 and assume that z = 2^u*r
divide z for all the numbers m = h*2^u such that g1<=m<=10*g1
if z/m = n is integer, then
check if n+g1 and n+g2 are the primes following n.

From a practical point of view, the bound L(k) = log(10^k)^2
is quite high, i.e., we expect most of the gaps to be
much smaller.

So, if we are performing a systematic search we can use
the bound L(k), which is conjecturally OK, but if we are
simply looking for larger solutions we can use much smaller bounds
in the search.

For example, if k = 192 a (prime!) solution of (C)
is the 192-digits number n =
7046979865771812080536912751677852348993288590604026845637583892
6174496644295302013422818791946308724832214765100671140939597315
4362416107382550335570469798657718120805369127516778523489932887,

or a prime 115-digits solution of (A)
5684931506849315068493150684931506849315068493150684931506849315
068493150684931506849315068493150684931506849315069,

or a prime 118-digits solution of (B)
23529411764705882352941176470588235294117647058823529411764705882
35294117647058823529411764705882352941176470588235291.

The bound L(192) is about 195450, but the two primes following n
are n+210 and n+326, so using a bound of 1000 instead of 195450
we can still find this n in reasonable time.

Another way, simpler but with no guarantees, to find larger
solutions is to guess them!

For example, as we look at the first solutions for (C),
1, 3, 7, 61, 167, 801, 1143, 2001, 6001, 8001, 125001,...
we notice some solutions of the form t*10^s+1, like
8001 = 8*10^3 + 1.
This suggests us to look for other, larger solutions of this
form, and indeed they exists.
For example, we obtain
n = 8*10^111+1 or 25*10^133+1 or 92*10^159+1 for problem (C).
Other patterns works for (C) and some (different) patterns
work for (A) and (B). Looking at the lists below you can guess
which pattern to try.

Summarizing, subject to the conjecture that nextprime(p)-p < log(p)^2,
(which is verified up to 10^18), the solutions up to 36 digits are

(A) 7, 43, 4167, 857143, 909091, 1443299, 4166667, 92857143, 2205882353, 2792792793, 1046511627907, 5737704918033, 19083969465649, 53947368421053, 55882352941177, 772727272727273, 2962962962962962963, 5806451612903225807, 263888888888888888889, 1826923076923076923077, 5081967213114754098361, 15086782376502002670227, 35042735042735042735043, 48648648648648648648649,
53658536585365853658537, 312182741116751269035533,
28342245989304812834224599, 160220994475138121546961325967,
457142857142857142857142857143, 3541666666666666666666666666666667,
19181946403385049365303244005641749, 132369299221357063403781979977753059,
422131147540983606557377049180327869.

(B) 7, 9, 49, 111, 1090909, 28571427, 111111111, 3333333327, 25641025641, 10576923076923, 59090909090909, 2631578947368421, 4827586206896549, 8947368421052631, 18644067796610169, 111111111111111111, 812499999999999999, 1889250814332247557, 9189189189189189189, 39999999999999999999, 76829268292682926829, 645161290322580645161, 2457002457002457002457,
4242424242424242424241, 1090909090909090909090909, 2272727272727272727272727, 3064516129032258064516129, 1583011583011583011583011583, 10752688172043010752688172043, 32258064516129032258064516129, 276923076923076923076923076923, 537037037037037037037037037037, 5999999999999999999999999999999, 20833333333333333333333333333333, 57142857142857142857142857142851, 333333333333333333333333333333333, 2227272727272727272727272727272727, 4366197183098591549295774647887323, 6999999999999999999999999999999997, 7142857142857142857142857142857139, 16413373860182370820668693009118541, 352941176470588235294117647058823529.

(C) 1, 3, 7, 61, 167, 801, 1143, 2001, 6001, 8001, 125001, 25000001,
181818181819, 2500000000001, 16666666666667, 45000000000001, 640000000000001, 1142857142857143, 4000000000000001, 37500000000000001, 153846153846153847, 937500000000000001, 2881355932203389831, 270270270270270270271, 44827586206896551724137933, 240174672489082969432314411, 461538461538461538461538463, 785714285714285714285714289, 7294117647058823529411764707, 196428571428571428571428571429, 1618497109826589595375722543353, 1715328467153284671532846715329, 2450980392156862745098039215687, 2758620689655172413793103448277, 3220338983050847457627118644069, 5955056179775280898876404494383, 255319148936170212765957446808511, 6250000000000000000000000000000003, 50000000000000000000000000000000001, 62264150943396226415094339622641511.

***

Jeff wrote:

For puzzle 786 I have found these solutions for primes less than 2^32:

 P1 P2 A(x) 3 5 7 37 41 43 4157 4159 4167 857107 857137 857143 909071 909089 909091 1443271 1443293 1443299 4166647 4166651 4166667 92857117 92857139 92857143 2205882293 2205882337 2205882353 2792792731 2792792747 2792792793 P1 P2 B(x) 7 11 9 47 53 49 109 113 111 28571423 28571449 28571427 111111109 111111113 111111111 3333333323 3333333403 3333333327 P1 P2 C(x) 5 7 3 11 13 7 67 71 61 173 179 167 809 811 801 1151 1153 1143 2003 2011 2001 6007 6011 6001 8009 8011 8001 125003 125017 125001 25000009 25000033 2500000

***

Emmanuel wrote:

Q1.
Smallest element in  A : 7 ( A = { 7, 43, 4167, 857143, 909091, 1443299, 4166667, 92857143, ... } ).
Smallest element in  B : 7  ( B = { 7, 9, 49, 111, 1090909, 28571427, ... } ).
Smallest element in  C : 3  ( C = { 3, 7, 61, 167, 801, 1143, 2001, 6001, 8001, 125001, 2500000, ... } ).

Q2.
A : 132325141776937618147448015123 (divides 132325141776937618147448014703&132325141776937618147448014721).
B : 276923076923076923076923076923 (divides 276923076923076923076923076851&276923076923076923076923076943).
C : 196428571428571428571428571429  (divides 196428571428571428571428571451&196428571428571428571428571477).

Q3.
To save place, let's  call the element of the set  m.  If  p  and  q  are the concatenating primes that are divisible by  m, I print them in the form  m+/-u, p+/-v..
A : m  = 24265842349304482225656877897990726429675425038639876352395672333848531684698608
9644513137557959814528593508501 (111  digits) ;
p = m - 942, q = m - 882 ; m  is prime and is my answer to Q4.
B : m =  23529411764705882352941176470588235294117647058823529411764705882352941176470588
23529411764705882352941176470588235291 (118 digits) ;
p = m - 8, q = m + 106 ; m  is prime and is my answer to  Q4.
C : m = 13720930232558139534883720930232558139534883720930232558139534883720930232558139534883
720930232558139534883720930232558139534883720930232558139534883720930233 (158  digits) ;
p = m + 118, q = m + 380.

Q4.
See  Q3  for the answers for  A  and  B.
The biggest prime I know in  C  is  167 (despite great efforts to find a bigger one).

***

On Jun1, 2015, Emmanuel wrote:

During last week I was not able to use my PC nor had I access to th Internet.
My first job when I returned at home was to doublecheck my contribution to puzzle 786.  I discovered an error in the program that I used to compute set A (I simply had to change  >=  into  <= ).
So, my answer to Q2,A  is wrong (my corrected program gives the two numbers that are in Giovanni's list).
The answer to  Q3,A  should be :
m  = 5684931506849315068493150684931506849315068493150684931506849
315068493150684931506849315068493150684931506849315069 (115  digits) ;
p = m - 166, q = m - 148 ; m  is prime and is my answer to Q4.

***

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