Problems & Puzzles: Puzzles

 Puzzle 447. Zeros on p.q J. M. Bergot sent the following puzzle: Take 89443^2 = 8,000,050,249.  You notice that 89443 has NO zeros.  Can you find two primes with no or few zeros to give a product with many zeros? Q. In order to be a little more precise, let's ask for p & q zero-free primes such that p.q has no less that 50% of zero digits, as in the example given above by Bergot. Swift addendum suggested by a reader: Maybe it would be more challenging with restrictions on p and q, for example p=q as in the example, or p and q having the same length

Several nice & smart contributions came this time too. Digit patterns are enjoyable candies anythime. As usual the found patterns could be used to go finding larger records than the already sent.

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Seiji Tomita wrote:

Search condition. 10^n+r=p*q, 20<=n<=30, r<10000, p and q have same length

10^21+1357= 1000000000000000001357= 28614892763* 34946837239
10^21+2373= 1000000000000000002373= 64255394153* 15562895741
10^21+5397= 1000000000000000005397= 59778265957* 16728487921
10^21+8617= 1000000000000000008617= 34566364237* 28929857741
10^23+1917= 100000000000000000001917= 471441453971* 212115415727
10^25+3811= 10000000000000000000003811= 6782289615587* 1474428337153
10^25+8203= 10000000000000000000008203= 2162199953531* 4624919163313
10^27+4339= 1000000000000000000000004339= 11391522377393* 87784579345123
10^27+5157= 1000000000000000000000005157= 64725645618113* 15449826578789

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Juan José Alba González wrote:

p=
3432783127434647596485212138318347689318561278444491951481537169418131543233137
46491354298159397864587626678833643921524258157258673892383

p^2 =
11784000000000000000000000000000000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000000000459485013793291905190617
59612016745901763527338632975303614771631434238434875073120117674727807884898261
232613252752722368018862571865418689

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Anton Vrba wrote:

A solution with p and q having equal length:
1131161597818429811 * 6188328894386331569 =
7000000000000000000000000000000003459

and a fun solution to the original question:
156421 * 63942822894879875468127725821981 =
10002000300040005000600007000800090001

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Farideh Firoozbakht wrote:

If n>2 & p=q=99*10^n-1 then number of zero digits of
p*q=1/2*length(p*q).
In this case we have p*q=9800.9(n-3).802.0(n-1).1 where dot
between digits means concatenation.

For example if p=q=99*10^5-1 then p*q=9800.9(5-3).802.0(5-1).1=
98009980200001.

Numbers n such that 99*10^n-1 is prime are :

3, 4, 5, 9, 19, 25, 34, 70, 82, 718, 2035, ...

Also if n>2, p=99*10^n-1 & q=9*10^(n-1)-1 then number of zero digits
of p*q=1/2*length(p*q).
In this case we have p*q=890.9(n-3).001.0(n-2).1

For example if p=99*10^4-1=989999 & q=9*10(4-1)-1=8999
then p*q=890.9(4-3).001.0(4-2).1=8909001001.

Numbers n such that q=9*10^(n-1)-1 is prime are:

2, 4, 8, 20, 30, 38, 94, 936, ...

So 4 is the only known number n such that both numbers p=99*10^n-1
& q=9*10^(n-1)-1 are primes.

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JC Rosa wrote:

About the puzzle 447, I have found many solutions where
the percentage of zeros digits in the product p.q is >=50.
Examples :
P=2 Q=5 PQ=10 %=50
P=3 Q=26669 PQ=80007 %=60
P=17 Q=5882353 PQ=100000001 %=77.77...(a well-known number ! see Puzzle 180)

With P=3 and Q=2*(10^k-1)/3+1 we have the following pattern:
P.Q=2*10^k+1 and %=100*(k-1)/(k+1)
Example : k=6 P=3 Q=2*(10^6-1)/3+1=666667 P.Q=2000001 and %=71.4...

My best result with this pattern is for k=66:
P=3 Q=2*(10^66-1)/3+1 (Q is prime verified with ecm ) P.Q=2*10^66+1 and %=97.01...

Now if we add the following condition: length of P=length of Q my best result is:
8279539*8454577=70000000000003 and %=85.71

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J. K. Andersen wrote:

Let p = 3 and q = (2*10^n+1)/3, for n = 6, 8, 9, 11, 20, 23, 41, 63, 66, 119, 122, 149, 252, 284, 305, 592, 746, 875, 1204, 1364, 2240, 2403,
5106, 5776, 5813, 12456, 14235, found in A096507.
Then p*q = 2*10^n+1 has 2 non-zero digits and n-1 zeroes.
Maybe it would be more challenging with restrictions on p and q,
for example p=q as in the example, or p and q having the same length.

...

The following submission is for the addendum with equal length of p and q.

If p and q each end in n nines then p*q ends in n-1 zeros and a one, because:
(a*10^n-1) * (b*10^n-1) = (a*b*10^n-a-b)*10^n + 1.
This can be used to construct large solutions if a*b*10^n-a-b has enough zeros.

The largest known near-repdigit prime at http://primes.utm.edu/top20/page.php?id=15 is p = 99*10^139670-1, found in 2006 by Larry Soule with NewPGen, LLR and PrimeForm/GW. p^2 has 2*139672 digits and exactly half of them are zeros. The decimal form of p^2 is 98009(139667)8020(139669)1, where d(r) means the digit d repeated r times.

The only solution with a single non-zero digit is 2*5 = 10.

The first 5 (and possibly only) solutions with 2 non-zero digits:
17 * 59 = 1003
19 * 53 = 1007
31 * 97 = 3007
15299 * 52291 = 800000009
8279539 * 8454577 = 70000000000003
There are no others up to 70 digits.

For 3 non-zero digits, the largest found solution has 58 digits:
66457199844545535453798119489 * 75236392921998416666319329863
= 5000000000000000000000000000000000000000000000000080000007
It is the only solution with 34 to 61 digits.
There are 176 smaller solutions, ending with 33 digits:
18187462949997379 * 27491464937943533 = 500000000000000000500000000000007

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