[math-fun] RE: Unsolved? problem
Dear Richard: It is indeed a good idea to collect Murray Klamkin problems. I am particularly fond of one his problems-but it did not appear in Math journals, but in the book by L.A.Graham : Ingenious Mathematical problems and methods, Dover. Murray's problem,which appears as #72 in the book,reads: find the smallest number such that if the last digit is removed and placed at the beginning to become the first digit, this new number is nine times the original one. The solution given in the book itself is the astronomical fijure 10,112,359,550,561,797,752,808,988,764,044,943,820,224,719. What would be the solution if we do similar operation with the last two digits and ask '99 times the original number", or replace the last three digits similarly to get 999 times the original number etc. Subbarao
===== Original Message From Richard Guy <rkg@cpsc.ucalgary.ca> ===== I'm collecting Murray Klamkin problems and solutions and am currently going thru Math Mag.
I came across Problem 886, Math Mag 48(1975) 57--58 [nothing to do with Murray] which isn't properly stated but should read as in OEIS A003508 :
a(n) = a(n-1) + 1 + sum of distinct prime factors of a(n-1) that are < a(n-1).
This leads to 1,2,3,4,7,8,11,12,18,24,30,41,42,55,...
The original problem asked that if you start elsewhere, e.g.,
5,6,12, ... or 9,13,14,24, ... or 10,18, ... or 15,24, ...
do you always merge with the original sequence? Evidently
91,112,122,186,... takes a little while.
Has anyone ... Can anyone prove Charles Trigg's guess ? R.
Dear Subbarao, Good to hear from you and to know that the old brain is still ticking over. I think that Murray's problem goes back quite a bit earlier. Singmaster's Sources doesn't mention this particular problem, but has references to similar ones back to 1854. On writing it in the form 9 * (10x + y) = y * 10^k + x we have 89x = y(10^k - 9) and 89 is a `semi-long' prime with the period for 1/89 having length (89 - 1)/2 = 44, the number of digits in the answer. If you want even more spectacular examples, look at (10^k - 1) * 10^k - 1 where Murray's problem has k = 1. k = 6 and k = 9 give primes, but I don't know what their order is, mod 10. Best, R. On Thu, 28 Apr 2005, msubbara wrote:
Dear Richard: It is indeed a good idea to collect Murray Klamkin problems. I am particularly fond of one his problems-but it did not appear in Math journals, but in the book by L.A.Graham : Ingenious Mathematical problems and methods, Dover. Murray's problem,which appears as #72 in the book,reads: find the smallest number such that if the last digit is removed and placed at the beginning to become the first digit, this new number is nine times the original one. The solution given in the book itself is the astronomical fijure
10,112,359,550,561,797,752,808,988,764,044,943,820,224,719. What would be the solution if we do similar operation with the last two digits and ask '99 times the original number", or replace the last three digits similarly to get 999 times the original number etc. Subbarao
===== Original Message From Richard Guy <rkg@cpsc.ucalgary.ca> ===== I'm collecting Murray Klamkin problems and solutions and am currently going thru Math Mag.
I came across Problem 886, Math Mag 48(1975) 57--58 [nothing to do with Murray] which isn't properly stated but should read as in OEIS A003508 :
a(n) = a(n-1) + 1 + sum of distinct prime factors of a(n-1) that are < a(n-1).
This leads to 1,2,3,4,7,8,11,12,18,24,30,41,42,55,...
The original problem asked that if you start elsewhere, e.g.,
5,6,12, ... or 9,13,14,24, ... or 10,18, ... or 15,24, ...
do you always merge with the original sequence? Evidently
91,112,122,186,... takes a little while.
Has anyone ... Can anyone prove Charles Trigg's guess ? R.
PS. It's well-known to those who well know it that 1/999998999999 = 0.000000 000001 000001 000002 000003 000005 000008 000013 000021 000034 000055 000089 000144 000233 000377 000610 000987 001597 002584 004181 006765 010946 017711 028657 046368 075025 121393 196418 317811 514229 8320413462711783125245837028962.... R. On Thu, 28 Apr 2005, msubbara wrote:
Dear Richard: It is indeed a good idea to collect Murray Klamkin problems. I am particularly fond of one his problems-but it did not appear in Math journals, but in the book by L.A.Graham : Ingenious Mathematical problems and methods, Dover. Murray's problem,which appears as #72 in the book,reads: find the smallest number such that if the last digit is removed and placed at the beginning to become the first digit, this new number is nine times the original one. The solution given in the book itself is the astronomical fijure
10,112,359,550,561,797,752,808,988,764,044,943,820,224,719. What would be the solution if we do similar operation with the last two digits and ask '99 times the original number", or replace the last three digits similarly to get 999 times the original number etc. Subbarao
===== Original Message From Richard Guy <rkg@cpsc.ucalgary.ca> ===== I'm collecting Murray Klamkin problems and solutions and am currently going thru Math Mag.
I came across Problem 886, Math Mag 48(1975) 57--58 [nothing to do with Murray] which isn't properly stated but should read as in OEIS A003508 :
a(n) = a(n-1) + 1 + sum of distinct prime factors of a(n-1) that are < a(n-1).
This leads to 1,2,3,4,7,8,11,12,18,24,30,41,42,55,...
The original problem asked that if you start elsewhere, e.g.,
5,6,12, ... or 9,13,14,24, ... or 10,18, ... or 15,24, ...
do you always merge with the original sequence? Evidently
91,112,122,186,... takes a little while.
Has anyone ... Can anyone prove Charles Trigg's guess ? R.
What would be the solution if we do similar operation with the last two digits and ask '99 times the original number", or replace the last three digits similarly to get 999 times the original number etc. Subbarao
If you take n digits from the right of a number and prepend them back, the result has the same number of digits as the original number, and must be less than 10 times the original number. It cannot be 99, 999, etc, times the original number. We could stay with 9 times the original number, but I'm too lazy to solve it. ----- Original Message ----- From: "msubbara" <msubbara@ualberta.ca> To: <ham>; "Math Fun" <math-fun@mailman.xmission.com>; "Richard Guy" <rkg@cpsc.ucalgary.ca>; "seqfan" <seqfan@ext.jussieu.fr> Sent: Thursday, April 28, 2005 7:35 PM Subject: [math-fun] RE: Unsolved? problem
Dear Richard: It is indeed a good idea to collect Murray Klamkin problems. I am particularly fond of one his problems-but it did not appear in Math journals, but in the book by L.A.Graham : Ingenious Mathematical problems and methods, Dover. Murray's problem,which appears as #72 in the book,reads: find the smallest number such that if the last digit is removed and placed at the beginning to become the first digit, this new number is nine times the original one. The solution given in the book itself is the astronomical fijure
10,112,359,550,561,797,752,808,988,764,044,943,820,224,719. What would be the solution if we do similar operation with the last two digits and ask '99 times the original number", or replace the last three digits similarly to get 999 times the original number etc. Subbarao
Oh yes, BTW, if you allow zero prefixes on your numbers, there are solutions with multiplier 99, 999, etc. For example, the smallest solution for 99 is this monster: 99*(10^486-1)/9899 = 01000101020305081321345590463683200323264976260228305889483786241034 44792403273057884634811597131023335690473785230831397110819274674209 51611273866047075462167895747045156076371350641478937266390544499444 38832205273259925244974239822204263056874431760783917567431053641781 99818163450853621577937165370239418123042731589049398929184766137993 73674108495807657339125164157995757147186584503485200525305586422870 997070411152641680977876553187190625315688453379129204970199 but we need that initial 0 to make it work. Also note initial elements of the Fibonacci sequence hiding near the beginning of the number. This is a feature of this sort of number. ----- Original Message ----- From: "msubbara" <msubbara@ualberta.ca> To: <ham>; "Math Fun" <math-fun@mailman.xmission.com>; "Richard Guy" <rkg@cpsc.ucalgary.ca>; "seqfan" <seqfan@ext.jussieu.fr> Sent: Thursday, April 28, 2005 7:35 PM Subject: [math-fun] RE: Unsolved? problem
What would be the solution if we do similar operation with the last two digits and ask '99 times the original number", or replace the last three digits similarly to get 999 times the original number etc. Subbarao
msubbara wrote:
...find the smallest number such that if the last digit is removed and placed at the beginning to become the first digit, this new number is nine times the original one. The solution given in the book itself is the astronomical fijure
10,112,359,550,561,797,752,808,988,764,044,943,820,224,719. What would be the solution if we do similar operation with the last two digits and ask '99 times the original number", or replace the last three digits similarly to get 999 times the original number etc.
For 99, the answer is 10001010203050813213455904636832003232649762602283 05889483786241034447924032730578846348115971310233 35690473785230831397110819274674209516112738660470 75462167895747045156076371350641478937266390544499 44438832205273259925244974239822204263056874431760 78391756743105364178199818163450853621577937165370 23941812304273158904939892918476613799373674108495 80765733912516415799575714718658450348520052530558 64228709970704111526416809778765531871906253156884 53379129204970199 For 999, the answer has 146011 digits. Here are the beginning and end. 10009019028047075122197319516836353189542732275007 ... 17596013609623232856088945033979012992004997001999 -- Don Reble djr@nk.ca
participants (4)
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David Wilson -
Don Reble -
msubbara -
Richard Guy