20060807, 17:13  #1 
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}×3^{3}×19 Posts 
Number 100
We are all familiar with the Name Edouard Lucas. Like many mathematicians he spent his leisure moments solving or compiling puzzles. The following is one attributed to him. He found seven solutions and expressed doubts for any more ways. Actually today we know 11 ways and no more. The Problem: To write 100 using all 9 digits (1  9) in the form of a mixed number; Eg: 100 = 91 + 5742/638 or 91(5742/638) a mixed number and no repitiion of digits but just the nine. If you can give more than seven you beat Lucas at his own game. Mally 
20060807, 17:41  #2 
"Robert Gerbicz"
Oct 2005
Hungary
23×71 Posts 
Are parentheses allowed?

20060807, 23:10  #3  
Aug 2005
Brazil
2·181 Posts 
Quote:
On a side note, I bet Lucas didn't have a computer! 94+1578/263 = 100 91+7524/836 = 100 3+69258/714 = 100 91+5823/647 = 100 96+1428/357 = 100 96+2148/537 = 100 82+3546/197 = 100 81+5643/297 = 100 81+7524/396 = 100 96+1752/438 = 100 91+5742/638 = 100 Last fiddled with by fetofs on 20060807 at 23:34 

20060808, 04:38  #4  
Jun 2005
2×191 Posts 
Quote:


20060808, 07:37  #5  
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}·3^{3}·19 Posts 
Computers!
Quote:
Fetofs has certainly got it straight off the bat! Well rather than start a new thread and a new problem I'm presenting this one, a simple one, to which no computer is needed at all. Maybe a calculator at the most, if you go about it the right way. If you use a computer, the compilers from Cambridge Univ. recommend C++. Actually there is no need of this and I got it straight away. I am nil in it anyway! Puzzle: Given the expression 1/p + 1/q + 1/r < 1 Find values of p, q, r, such that its the largest fraction but less than 1 p, q, r, are whole numbers and no other symbols, parenthesis etc. In other words no funny stunts! Having said that, the use of a comp would be cheating and please try to prove the answer in some way or the other. Kindly spoilerise your answers. Mally 

20060808, 09:33  #6  
"Robert Gerbicz"
Oct 2005
Hungary
23·71 Posts 
Quote:
The answer is 1/2+1/3+1/7=41/42 Proof Let s=1/p+1/q+1/r and indirectly let 1>s>41/42. We can suppose that 1<=p<=q<=r If p=1 then s>=1, contradiction. If p>3 then s<=1/4+1/4+1/4=3/4<41/42, contradiction. If p=3 then case1: r=3, then s=1, contradiction. case2: r>3, then s<=1/3+1/3+1/4=11/12<41/42, contradiction. So we can suppose that p=2 ( There is no other case ). If q=2, then s>=1, contradiction. If q>4, then s<=1/2+2/5=9/10<41/42. If q=4, then case1: r=4, then s=1, contradiction. case2: r>4, then s<=1/2+1/4+1/5=19/20<41/42, contradiction. So we can suppose that q=3. ( There is no other case). If q<7, then s>=1/2+1/3+1/6=1, contradiction. If q>7, then s<=1/2+1/3+1/8=23/24<41/42, contradiction. So q=7, and for that s=1/p+1/q+1/r=41/42. 

20060808, 11:11  #7 
Dec 2005
2^{2}×7^{2} Posts 
Continuing this train of thought the values for s,t,u etc will be x[1]=2 x[n]=1/(1PROD(1/x[i]))+1 So s=43, t=1807 The obvious lurking question: which numbers in this sequence are prime ? 
20060808, 15:25  #8  
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}·3^{3}·19 Posts 
Number 100
Quote:
Well Kees thats an interesting train of thought and u should be in the region of 9 digits long at least. However, t breaks the prime run and I doubt very much if there will be any more primes at all. after s Still I would like it worked out for a few more digits v,w,x.... to make a conjecture. Mally 

20060808, 18:50  #9 
Aug 2002
Buenos Aires, Argentina
2·761 Posts 
Using my factorization applet in batch mode I found:
2 = 2 3 = 3 7 = 7 43 = 43 1807 = 13 * 139 3263443 = 3263443 10650056950807 = 547 * 607 * 1033 * 31051 113423713055421844361000443 = 29881 * 67003 * 9119521 * 6212157481 12864938683278671740537145998360961546653259485195807 = 5295435634831 * 31401519357481261 * 77366930214021991992277 165506647324519964198468195444439180017513152706377497841851388766535868639572406808911988131737645185443 = 181 * 1987 * 112374829138729 * 114152531605972711 * 35874380272246624152764569191134894955972560447869169859142453622851 Since the next member in the sequence is about the square of the previous one (like the Fermat number sequence), I think that there are no other primes (or maybe only one more). 
20060808, 19:11  #10 
"Robert Gerbicz"
Oct 2005
Hungary
23×71 Posts 
This is a well known sequence, see http://en.wikipedia.org/wiki/Sylvester's_sequence
including some more factorizations also. 
20060808, 21:41  #11 
Aug 2002
Buenos Aires, Argentina
5F2_{16} Posts 
I started my program to find small factors before trying to find the sequence in other Web site (anyway this is OK in the Puzzles forum).
These are the small prime factors I found with a program written in C: k=0: 2 k=1: 3 k=2: 7 k=3: 43 k=4: 13, 139 k=5: 3263443 k=6: 547, 607, 1033, 31051 k=7: 29881, 67003, 9119521 k=9: 181, 1987 k=10: 2287, 2271427 k=11: 73 k=13: 52387, 5020387 k=14: 13999, 74203, 9638659, 57218683 k=15: 17881 k=16: 128551 k=17: 635263, 1286773, 21269959 k=20: 352867 k=21: 387347773, 1620516511 k=23: 74587 k=26: 27061 k=27: 164299, 3229081 k=28: 20929 k=29: 1171, 298483, 97562299 k=31: 1679143 k=34: 120823, 447841 k=35: 2408563 k=36: 38218903, 818864029 k=37: 333457, 117602053 k=38: 30241 k=39: 4219, 1372512301 k=40: 1085443, 156507607, 171818923 k=41: 7603 k=42: 1861, 84819199 k=44: 23773, 290791 k=45: 51769 k=46: 1285540933 k=47: 429547, 196973983 k=51: 179585227 k=53: 9829 k=57: 763176709 k=59: 2029, 38329, 320329, 3567469 k=61: 50023, 1445860807 k=64: 1459, 11234941 k=65: 72091, 609421 k=67: 245563, 3346633, 343870981 k=68: 6367063 k=69: 12763 k=70: 384061 k=71: 3799489, 8401963 k=73: 11844787 k=74: 35869 k=75: 20161, 42428887 k=76: 123427, 160200259 k=79: 51973, 195502537 k=80: 2437 k=82: 6230971 k=86: 9436201 k=88: 151477 k=89: 41077, 2171593 k=92: 15373, 21661, 1386013 k=93: 429446209 k=94: 12593197 k=95: 535879 k=96: 1407223, 1143074323 k=98: 20479 The value k is the index of the sequence. The upper bound is 2*10^{9}. I also found that the prime factors greater than 3 must be congruent to 1 (mod 6): Let p be a prime factor of a member of Sylvester's sequence greater than 3 (all equal signs below are congruences mod p). u^{2}  u + 1 = 0 Let u = t+b. (t+b)^{2}  (t+b) + 1 = 0 t^{2} + 2bt + b^{2}  t  b + 1 = 0 t^{2} + (2b  1)t + b^{2}  b + 1 = 0 Let 2b  1 = 0, so b = 1/2 (valid because gcd(p,2)=1). t^{2} + 1/4  1/2 + 1 = 0 t^{2} + 3/4 = 0 4t^{2} + 3 = 0 Let w = 2t: w^{2} + 3 = 0 w^{2} = 3 3 is a quadratic residue mod p. Since 3 is congruent to 1 (mod 4), from the law of quadratic reciprocity we get that p is a quadratic residue mod 3 (or 3). This implies that p = 1 (mod 3). So all prime factors of Sylvester's sequence greater than 3 must be congruent to 1 (mod 6). 
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