Define d(x) as sum of the digits of x, where x is a hexadecimal positive integer.
d(d(x)) denotes the sum of digits of d(x).
For example, when x=(ABC)16
Then, d(x) = (A)16+(B)16+(C)16 = (21)16
and, d(d(x)) = 2 + 1 = 3
Consider the first 1011 (base ten) values of a hexadecimal prime number N.
Devise an algorithm such that:
• Each of d(N) and d(d(N)) is divisible by a prime power.
Note: A prime power is a number of the form pn, where p is a prime number, and n is an integer greater than 1.
clearvars,clc
ct=0;
for n0=1:1011
n=nthprime(n0);
hex=[];
while n>0
q=floor(n/16);
r=n-16*q;
hex=[r hex];
n=q;
end
n=hex;
d=sum(n);
f=factortable(d);
q=f(:,1);
q=q(q>1);
if length(q)>0
n=d;
hex=[];
while n>0
q=floor(n/16);
r=n-16*q;
hex=[r hex];
n=q;
end
n=hex;
dd=sum(n);
f=factortable(dd);
q=f(:,1);
q=q(q>1);
if length(q)>0
ct=ct+1;
fprintf('%4d %7d %5s %3d %3d\n',n0,nthprime(n0),dec2base(nthprime(n0),16),d,dd)
end
end
end
ct
finds 22 such primes in the range specified
ordinal prime
prime prime in hex d dd
8 19 13 4 4
9 23 17 8 8
16 53 35 8 8
23 83 53 8 8
30 113 71 8 8
56 263 107 8 8
62 293 125 8 8
71 353 161 8 8
103 563 233 8 8
108 593 251 8 8
136 769 301 4 4
137 773 305 8 8
208 1283 503 8 8
245 1553 611 8 8
565 4099 1003 4 4
568 4129 1021 4 4
569 4133 1025 8 8
597 4373 1115 8 8
627 4643 1223 8 8
632 4673 1241 8 8
687 5153 1421 8 8
711 5393 1511 8 8
ct =
22
That's all there was found through the 1011th prime.
To get some real hex digits you need to go higher:
1029 8209 2011 4 4
1035 8243 2033 8 8
1038 8273 2051 8 8
1061 8513 2141 8 8
1092 8753 2231 8 8
1114 8963 2303 8 8
1174 9473 2501 8 8
1470 12289 3001 4 4
1472 12323 3023 8 8
1581 13313 3401 8 8
1905 16433 4031 8 8
1929 16673 4121 8 8
2312 20479 4FFF 49 4
2313 20483 5003 8 8
2338 20753 5111 8 8
2726 24593 6011 8 8
3124 28669 6FFD 49 4
3485 32479 7EDF 49 4
3511 32719 7FCF 49 4
3512 32749 7FED 49 4
3808 35839 8BFF 49 4
3855 36319 8DDF 49 4
3876 36559 8ECF 49 4
4172 39679 9AFF 49 4
4222 40189 9CFD 49 4
4239 40429 9DED 49 4
4260 40639 9EBF 49 4
4263 40699 9EFB 49 4
4281 40879 9FAF 49 4
4287 40939 9FEB 49 4
4557 43759 AAEF 49 4
4583 44029 ABFD 49 4
4608 44269 ACED 49 4
4671 44959 AF9F 49 4
4907 47599 B9EF 49 4
4934 47869 BAFD 49 4
4952 48079 BBCF 49 4
4954 48109 BBED 49 4
4998 48589 BDCD 49 4
(The d and dd are shown in decimal, and 49 decimal is 31 in hex, whose digits add to 4.)
The first one exhibiting "real" hex digits is shown by
2312 20479 4FFF 49 4
where the hex representation of the 2312th prime is 4FFF, so the digits add to 4+3*15 = 49, which is 7^2, and is represented in hex as 31, which adds up to 4, the square of 2.
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Posted by Charlie
on 2023-05-15 13:57:09 |
computer solution
