Problems & Puzzles: Puzzles

 Puzzle 525. D= d1*P1 + d2*P2 +...dn*Pn JM Bergot sent the following puzzle: Given a prime D with digits d1, d2, d3 .... dn, find a set of primes P1, P2, ...Pn such that:  D = d1*P1 + d2*P2 +...dn*Pn A small example is D=47 since  47 = 4*3 + 7*5 = 12 + 35  Q. Can you find a larger example than this?

Contributions came from Jim Howell, Torbjörn Alm, Faridh Lian, Flavio Torasso, JC Colin, Seiji Tomita & Farideh Firoozbakht

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Jim Howell wrote:

113 = 1*7 + 1*13 + 3*31

127 = 1*2 + 2*3 + 7*17

6700417 = 6*191 + 7*3 + 0*7 + 0*11 + 4*5 + 1*13 + 7*957031

Perhaps the problem meant to ask for CONSECUTIVE primes for P1...Pn. Here is a solution with P1, P2,...Pn being consecutive primes:

197 = 1*7 + 9*11 + 7*13

Torbjörn Alm wrote:

Solutions with contigous primes.
All digits in prime >0
Lowest solution for length 3-6:

D = 197
197 = 1*7 + 9*11 + 7*13

D = 1217
1217 = 1*103 + 2*107 + 1*109 + 7*113

D = 12161
12161 = 1*1093 + 2*1097 + 1*1103 + 6*1109 + 1*1117

D = 114661
114661 = 1*6007 + 1*6011 + 4*6029 + 6*6037 + 6*6043 + 1*6047

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Farid Lian wrote:

a(i)n, P1, P2, P3, P4, P5, P6, P7, P8, P9
217,2,3
3997,73,47,7
41019,97,67,2,79
511159,1069,97,97,97,859
6100057,9973,997,997,997,997,25261
71461001,352249,9973,9973,9973,9973,9973,999049
810006121,9916351,9973,9973,9973,9973,9973,9973,13
9100002011,99962119,9973,9973,9973,9973,9973,9973,7307,9973
i =prime size
n =main prime
Pi =primes factors

Flavio Torasso wrote:

I found the following solutions, with p(x) the starting prime of the set

3-digits solutions
D x p(x)
197 4 7 197 =1*7 + 9*11+ 7*13
263 8 19 263 =2*19+ 6*23+ 3*29
353 10 29 353 =3*29+ 5*31+ 3*37

4-digits solutions
D x D x D x
1439 21 4483 50 7451 84
1217 27 2311 66 7541 85
1051 34 3701 67 8641 87
1231 39 6263 72 7213 101
4649 44 8951 76

5-digits solutions
D x D x D x
12161 183 34039 275 66179 337
14303 210 34123 379 67211 547
16273 146 35993 201 70019 565
16963 121 36011 461 71287 412
23561 219 38567 214 81031 810
23741 220 39937 207 89501 538
25121 338 40487 271 90473 544
26171 241 43541 373 90619 505
28621 238 44119 342 91283 547
29501 269 48397 244 92639 450
30293 274 49019 319 94321 663
30853 255 53047 404 94811 566
32213 421 59029 348 98213 585
33931 275 60337 447

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JC Colin wrote:

I found D with set of consecutive increasing Pi up to 10 and for decreasing to 7.

increasing set

2567821 = 2 * 82799 + 5 * 82811 + 6 * 82813 + 7 * 82837 + 8 * 82847 + 2 * 82883 + 1 * 82889
2418257 = 2 * 83341 + 4 * 83357 + 1 * 83383 + 8 * 83389 + 2 * 83399 + 5 * 83401 + 7 * 83407
2585641 = 2 * 83389 + 5 * 83399 + 8 * 83401 + 5 * 83407 + 6 * 83417 + 4 * 83423 + 1 * 83431

22614299 = 2 * 646039 + 2 * 646067 + 6 * 646073 + 1 * 646099 + 4 * 646103 + 2 * 646147 + 9 * 646157 + 9 * 646159
26491847 = 2 * 646073 + 6 * 646099 + 4 * 646103 + 9 * 646147 + 1 * 646157 + 8 * 646159 + 4 * 646169 + 7 * 646181

225578741 = 2 * 5501827 + 2 * 5501843 + 5 * 5501879 + 5 * 5501921 + 7 * 5501933 + 8 * 5501939 + 7 * 5501941 + 4 * 5501953 + 1 * 5501959
...
2313781891 = 2 * 53808827 + 3 * 53808829 + 1 * 53808857 + 3 * 53808863 + 7 * 53808869 + 8 * 53808871 + 1 * 53808889 + 8 * 53808899 + 9 * 53808917 + 1 * 53808919

decreasing set

2865923 = 2 * 81919 + 8 * 81901 + 6 * 81899 + 5 * 81883 + 9 * 81869 + 2 * 81853 + 3 * 81847
2385191 = 2 * 82279 + 3 * 82267 + 8 * 82261 + 5 * 82241 + 1 * 82237 + 9 * 82231 + 1 * 82223
2575129 = 2 * 83093 + 5 * 83089 + 7 * 83077 + 5 * 83071 + 1 * 83063 + 2 * 83059 + 9 * 83047
2581357 = 2 * 83311 + 5 * 83299 + 8 * 83273 + 1 * 83269 + 3 * 83267 + 5 * 83257 + 7 * 83243
2914487 = 2 * 83311 + 9 * 83299 + 1 * 83273 + 4 * 83269 + 4 * 83267 + 8 * 83257 + 7 * 83243
see the two last D which have the same set of Pi

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Seiji Tomita wrote:

We can do it as follows only to get a larger example.

Let D = d1*10^n+d2 = d1*p1+d2*p2.
Then, p1 = 10^n-d2, p2 = d1+1.

Search condition: n<30,D>47

D p1,p2

67 [3, 7]
103 [97, 2]
1009 [991, 2]
2003 [997, 3]
4003 [997, 5]
200009 [99991, 3]
400009 [99991, 5]
100000000000000003 [99999999999999997, 2]
200000000000000003 [99999999999999997, 3]

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Farideh wrote:

A large example :

p = 10^4056 + 1111 = 1.0(4052).1111
= 1*(p - 4860) + 1*11 + 1*19 + 1*101 + 1*4729

Since both numbers p & p - 4860 are prime (probable prime) if q, r, s & t be four
primes such that q+r+s+t = 4860 then p = 1*(p - 4860 ) + 1*q + 1*r + 1*s + 1*t.

It seems that all primes with at least 3 nonzero digits are of the mentioned form.
Also all primes greater than 100 with only two nonzero digits where one of these
two digits is even, are of such form.

So primes which aren't of this form are more interesting

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