Problems & Puzzles: Puzzles

 Puzzle 1117 2023 At first sight 2023 is a notorious inconspicuous integer, but... A. On my request Rodolfo Kurchan sent the following puzzle about it... 2023 is a composite integer such that abs(20-23)=3 and 20+23=43, both primes! Under the same scheme the first prime integer is 41 [abs(4-1)=3, 4+1=5] Q1. Find the smallest pandigital prime as with the same scheme of 2023, having a) the following decimal digits, 1 to 9 and b) having the 0 to 9, decimal digits. B. From the Giovanni Resta's page we know that: "The prime factors of 2023, concatenated, give a palindrome: 71717." Q2. What is the next year that produces a) another palindrome b) a palprime? Q3. Can you provide another interesting prime property and puzzle related to 2023?

During the week from Jan 1-7, 2023, contributions came from Emmanuel Vantieghem, J-M Rebert, Jim Howell, Oscar Volpatti, Gennady Gusev

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Emmanuel wrote:

Q1
Here I assumed that pandigitallity means "having at least all digits 1, ..., 9 or 0, ..., 9"
a) p = 1125647893
11256 + 47893 = 59149, prime
Abs[11256 - 47893] = 36637, prime
b) p = 100126478593
100126 + 478593 = 578719, prime
Abs[100126 - 478593] = 378467, prime.
(here I assumed the number of digits of  p  must be even)

Q2. Next palindrome year : 2048 = 2*2*2*2*2*2*2*2*2*2*2
Next palprime year : 3039 = 3*1013

Q3. a) Find the smallest prime whose sum of digits is  2023 (easy)
b) Find the smallest prime whose square has sum of digits  2023 (hard).

______________________________

Sincere greetings and a nice (9 + 8)*(6*5*4 - 3 +2*1) !

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Jean-Marc wrote:

Q1.a.The smallest pandigital prime, I found is 1125647893.

1125647893, is a prime such that abs(11256 - 47893) = 36637 and 11256 + 47893 = 59149, both primes !

Q1.b.The smallest pandigital prime, I found is 100126478593.

100126478593, is a prime such that abs(100126 - 478593) = 378467 and 100126 + 478593 = 578719, both primes !

Q2.a.The next year that produces another palindrome (22222222222) is 2048 = 2^11.

Q2.b.The next year that produces another palprime (31013) is 3039 = 3*1013.

Q3. Units-digit 7 occurs in the following results :

From 2023 there are 77 = 7*11 years to 2100.

2 + 0 + 2 + 3 = 7 is prime.

precprime(2023) = 2017

nextprime(2023) = 2027

2027 - 2017 = 17-7

and each prime factor of 2023 is ending with 7. (2023 = 7*17^2)

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Jim wrote:

For Question 2:

The next years where factorization is a palindrome are:

2048=2*2*2*2*2*2*2*2*2*2*2
2097=3*3*233
2187=3*3*3*3*3*3*3
2319=3*773
2321=11*211
2359=7*337
2401=7*7*7*7
2649=3*883
2701=37*73
3125=5*5*5*5*5
3421 = 11*311 and 11311 is prime

So we need to wait almost 1400 years for the first palprime.

Another palprime (probably not the second smallest) is:

10301 = 10301 = prime

Question 3.

I am not sure if this qualifies as “interesting”, but if we concatenate the puzzle number with 2023, we get 11172023 which is a prime number.

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Oscar wrote:

Q1
Both 2023 and 41 have an even number of digits, which are split into two blocks of equal length.
Under this "equal-length"  constraint, the smallest pandigital prime solution is:
1125647893 for the "zeroless" case a;
100126478593 for the "full" case b.

abs(11256-47893) = 36637, prime;
(11256+47893) = 59149, prime;
abs(100126-478593) = 378467, prime;
(100126+478593) = 578719, prime.

Without the "equal-length" restriction, there are smaller solutions; in some cases, digits can be split in k different ways.
"Zeroless" pandigital prime case a.
k = 1, p = 1123754869;
abs(112-3754869) = 3754757, prime;
(112+3754869) = 3754981, prime;
k = 2, p = 1128657349;
abs(112-8657349) = 8657237, prime;
(112+8657349) = 8657461, prime;
abs(1128-657349) = 656221, prime;
(1128+657349) = 658477, prime;
k = 3, p = 1325487629;
abs(132-5487629) = 5487497, prime;
(132+5487629) = 5487761, prime;
abs(132548-7629) = 124919, prime;
(132548+7629) = 140177, prime;
abs(132548762-9) = 132548753, prime;
(132548762+9) = 132548771, prime.

"Full" pandigital prime case b.
k = 1, p = 10123548679;
abs(1012-3548679) = 3547667, prime;
(1012+3548679) = 3549691, prime;
k = 2, p = 10125367849;
abs(10-125367849) = 125367839, prime;
(10+125367849) = 125367859, prime;
abs(101253678-49) = 101253629, prime;
(101253678+49) = 101253727, prime;
k = 3, p = 10274863549;
abs(10-274863549) = 274863539, prime;
(10+274863549) = 274863559, prime;
abs(102-74863549) = 74863447, prime;
(102+74863549) = 74863651, prime;
abs(102748-63549) = 39199, prime;
(102748+63549) = 166297, prime;
k = 4, p = 15272406389;
abs(152-72406389) = 72406237, prime;
(152+72406389) = 72406541, prime;
abs(15272-406389) = 391117, prime;
(15272+406389) = 421661, prime;
abs(1527240-6389) = 1520851, prime;
(1527240+6389) = 1533629, prime;
abs(1527240638-9) = 1527240629, prime;
(1527240638+9) = 1527240647, prime;.

Q2
The prime factors of 2023 are concatenated with repetition and in ascending order, obtaining a composite palindrome:
2023 = 7*17^2,
7.17.17 = 71717 = 29*2473.
Under this "ascending-order" constraint, next few years producing a composite palindrome (case a) are:
2048 = 2^11,
2.2.2.2.2.2.2.2.2.2.2 = 22222222222, repdigit;
2097 = 3^2*233,
3.3.233 = 33233 = 167*199;
2187 = 3^7,
3.3.3.3.3.3.3 = 3333333, repdigit;
Next few years producing a palprime (case b) are more sparse:
3039 = 3*1013,
3.1013 = 31013;
3421 = 11*311,
11.311 = 11311;
4303 = 13*331,

13.331 = 13331.

Without the "ascending-order" constraint, there are smaller solutions.
Case a: 2025 = 3^4*5^2,
3.3.5.5.3.3 = 335533 = 11^2*47*59,
3.5.3.3.5.3 = 353353 = 7*11*13*353,
5.3.3.3.3.5 = 533335 = 5*11*9697.
Case b: 2205 = 3^2*5*7^2,
3.7.5.7.3 = 37573, palprime.

7.3.5.3.7 = 73537 = 151*487, composite.

Q3
The number 2023 can be represented as the difference of two 5-smooth integers:
2023 = 3^4*5^2 - 2,
2023 = 2^11 - 5^2.
Can you find one more 5-smooth representation of 2023, or some proof that no more representations exist?
What about 7-smooth representations of 2023?

A p-smooth number is an integer whose prime factors are all less than or equal to p.

For example
2025 = 3^4*5^2 is 5-smooth,
2048 = 2^11 is 2-smooth, so it is 5-smooth too,
2023 = 7*17^2 is not 5-smooth but it is 17-smooth

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Q1.
In order for the number to be divided into 2 parts equal in number of digits,
then the number of digits in the number must be even.

a) if both parts have an equal number of digits:
1125647893: abs(11256-47893)=36637 & 11256+47893=59149 - all 3 numbers are primes;
if not:
1123754869:  abs(112-3754869)=3754757 & 112+3754869=3754981;

b) if both parts have an equal number of digits:
100126478593: abs(100126-478593)=378467 & 100126+478593=578719;
if not:
10123548679: abs(1012-3548679)=3547667 & 1012+3548679=3549691;

Q2.
a) 2048, 22222222222

b) 3039, 31013

Q3.
Express consecutive primes using only the digits 2, 0, 2, 3 and exactly in this order using the operation signs: +, -, *, /, ^, ! ( factorial and multifactorial) and concatenation of digits before the first pass.
Beginning (for example):
2 = 2 + 0 * 2 * 3
3 = -20 + 23
5 = 2 * 0 + 2 + 3
7 = 2 + 0 + 2 + 3
11 = 2 + 0! + 2^3 and so on.

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