Find the remainder of the division of by

Could someone please explain how I would do this without a calculator??

Thanks in advance!

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- February 17th 2008, 04:59 AMhunkydory19finding remainder of a division
Find the remainder of the division of by

Could someone please explain how I would do this without a calculator??

Thanks in advance! - February 17th 2008, 06:38 AMTriKri

...

and so on until you have come down to , is one way. Then you will have to work your way up again, but use the modulus rule this time to get the values in the interval .

Another way is to make a table over modulus values for different exponents of 23.

...

and so on. Eventually, you will get the same modulu value as you have had before, then you know that the modulu values will begin to repeat. When you have found the period of repetition (in this case , so the period will be 21), you can calculate which of the exponents in your table that has the same modulu value as . - February 17th 2008, 07:45 AMSoroban
Hello, hunkydory19!

Quote:

Find the remainder of the division of by

*primitive*method for solving these . . .

. . (I'll omit a LOT of the trial-and-error that I used.)

We try: .

. . Then: .

Hence: .

Then: .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

Hence: .

Then: .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

Hence: .

That is: . has a remainder of 1.

Then: .

Now go to work on

. . [Hint: . ]

- February 17th 2008, 07:48 AMgalactus
There are 21 mod 49 residues.

We note that

This tells us that 666342 is 12 more than a multiple of 21.

So,

The remainder is 8. - February 17th 2008, 08:15 AMTriKri
Soroban, why did you choose 7 as an exponent? And galactus, how did you know that there are 21 mod 49 residues? Are there any quick ways?