- Dear Calculators,

For the 8 May weekend I organised a MC weekend. My esteemed guests were Dr.Dr. Gert Mittring, Jan van Koningsveld and Dr. Andrew Robertshaw, quality street so to say.

We did a lot of exercizes, and inspired by the George Lane file I proposed a big division with a remainder. The challenge was to find the remainder without first finding the quotient.

The dividend I choosed randomly, the divisor could be factorised. I had the silent hope that modulo calculation could help us, unknowing how.

121068 = 2²×3³×19×59.

139875219643 ÷ 121068.

It was Andy who ran the gauntlet. He immediately calculated that the dividend = 3(2²), 13(3³), 8(19) and 37(59). After that he explained us the chinese remainder theorem, eventually to be found at internet.

First step: to find a number which meets 3(2²) and 13(3³) . 3= 3(3³) so for 13 (27) we miss 10(27). Taking 2² × 4 we have the missing 10 but we are not 3(2²). Taking 13½×4 we are (0)27 and 2(2²) so then we have reached our first goal: 67 which meets the requirements.

Now from starting point 67 we have to find a number 67(108) and 8(19).

67=10(19), 108 = 13(19) and number ?=8(19) . So how many times we have to add 13(19) from 10(19) to get 8(19). In fact we miss 17(19). Adding 13-26-39 etc we arrive at 169 (13×13) then we have 17(19).

So by adding 67 + (13×108) we arrive at 1471 which - indeed- meets the third requirement.

Now the 37(59).

We muliply 2²×3³×19 = 2052= 46(59) and have to add this so many times till we have a number which is 37(59).

1471= 55(59).

So 55(59) + ??×46(59) = 37(59). So we miss 41(59).

Lady luck gave us 9×46= 1(59). So now we miss 9×41=15(59).

Finally: when we take 15×2052 and add this at 1471 we have reached our final goal: 32251 which, a check learns this to us indeed meets all the requirements.

I think this question is too big for a tournament, a smaller one would to my opinion be an enrichment in a tournament.

What I fear - being no mathematician but a simple arithmetician - is if this type of questions could be answerd with a prime number as a divider.

Nevertheless; it has put me on fire and I'll train every day 1 type of this question. The answer is always a smaller number than the product of the numbers, in this case 121068.

Kindest regards,

Willem Bouman

[Non-text portions of this message have been removed] - Respected Mr Bouman

I did not understand what is this chinese remainder theorm but I am very much interested to learn the same. Can you give 1 or 2 simple examples like the one you have shown .It helps us a lot to know whether there will be a remainder in the answer when we are deviding by big numbers .Pl mail one or two examples describing the steps to be followed.Pl oblige.You can send the mail to gmanchali@... or manchaali@... I also have a method to findout but it is different from what you have shown.

--- On Fri, 5/15/09, A.W.A.P. Bouman <awap.bouman@...> wrote:

From: A.W.A.P. Bouman <awap.bouman@...>

Subject: [Mental Calculation] MC weekend

To: "mental calculation" <MentalCalculation@yahoogroups.com>

Date: Friday, May 15, 2009, 4:10 PM

Dear Calculators,

For the 8 May weekend I organised a MC weekend. My esteemed guests were Dr.Dr. Gert Mittring, Jan van Koningsveld and Dr. Andrew Robertshaw, quality street so to say.

We did a lot of exercizes, and inspired by the George Lane file I proposed a big division with a remainder. The challenge was to find the remainder without first finding the quotient.

The dividend I choosed randomly, the divisor could be factorised. I had the silent hope that modulo calculation could help us, unknowing how.

121068 = 2²×3³×19×59.

139875219643 ÷ 121068.

It was Andy who ran the gauntlet. He immediately calculated that the dividend = 3(2²), 13(3³), 8(19) and 37(59). After that he explained us the chinese remainder theorem, eventually to be found at internet.

First step: to find a number which meets 3(2²) and 13(3³) . 3= 3(3³) so for 13 (27) we miss 10(27). Taking 2² × 4 we have the missing 10 but we are not 3(2²). Taking 13½×4 we are (0)27 and 2(2²) so then we have reached our first goal: 67 which meets the requirements.

Now from starting point 67 we have to find a number 67(108) and 8(19).

67=10(19), 108 = 13(19) and number ?=8(19) . So how many times we have to add 13(19) from 10(19) to get 8(19). In fact we miss 17(19). Adding 13-26-39 etc we arrive at 169 (13×13) then we have 17(19).

So by adding 67 + (13×108) we arrive at 1471 which - indeed- meets the third requirement.

Now the 37(59).

We muliply 2²×3³×19 = 2052= 46(59) and have to add this so many times till we have a number which is 37(59).

1471= 55(59).

So 55(59) + ??×46(59) = 37(59). So we miss 41(59).

Lady luck gave us 9×46= 1(59). So now we miss 9×41=15(59).

Finally: when we take 15×2052 and add this at 1471 we have reached our final goal: 32251 which, a check learns this to us indeed meets all the requirements.

I think this question is too big for a tournament, a smaller one would to my opinion be an enrichment in a tournament.

What I fear - being no mathematician but a simple arithmetician - is if this type of questions could be answerd with a prime number as a divider.

Nevertheless; it has put me on fire and I'll train every day 1 type of this question. The answer is always a smaller number than the product of the numbers, in this case 121068.

Kindest regards,

Willem Bouman

[Non-text portions of this message have been removed]

[Non-text portions of this message have been removed] - --- In MentalCalculation@yahoogroups.com, gururaj manchali <gmanchali@...> wrote:
>

Before asking other people for help, I would recommend to do your homework:

> Respected Mr Bouman

> I did not understand what is this chinese remainder theorm but I am very much interested to learn the same.

Option 1: Lookup "Chinese Remainder Theorem" at Wikipedia

Option 2: Search for "Chinese Remainder Theorem" at Google

Option 3: Search the archive of this group for "Chinese Remainder Theorem" (where you will find my message #1815 with a good link)

Ralf Laue