Reminder Practice Qs – Fermat’s Little Theorem

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Fri, 2008-02-01 14:43

praveen_84

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I have posted a topic on Fermat’s little theorem @ http://www.cat4mba.com/node/6104

Few questions relevant to the above topic

Q1. What is the reminder when 11³⁹ is divided by 19
Q2. Find the reminder when 5⁹¹ is divided by 91
Q3. Find the following reminders
     a. ₇₅75⁷⁵is divided by 37
     b. 2¹⁰⁰ is divided by 101
     c. ₂₀ 51 ⁹⁷ is divided by 17

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Fri, 2008-02-01 16:10

OSO

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Q3 answers

Q3a. 75 = 37 (mod 1)

So ₇₅ 75⁷⁵divided by 37 'll give reminder 1.

Q3b. 2¹⁰⁰ is divided by 101
reminder 1

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Fri, 2008-02-01 16:43

nishit

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Reminder of 2017 divided by

Reminder of 20¹⁷ divided by 17 = 20 = 3

=> Reminder of 20⁵¹ divided by 17 = 27 = 10

Now Reminder of 10¹⁷ divided by 17 = 10

=> Reminder of 10³⁴ divided by 17 = 100 =-2

=> Reminder of 10⁶⁸ divided by 17 = 4

=> Reminder of 10⁸⁵ divided by 17 = 40 = 6

=> Reminder of 10⁹⁷ divided by 17 = 6x10¹²

Again Reminder of 10² divided by 17 = -2

=> Reminder of 10⁴ divided by 17 = 4

=> Reminder of 10⁸ divided by 17 = 16 = -1

=> Reminder of 10¹² divided by 17 = -4

So Reminder of 10⁹⁷ divided by 17 = 6x10¹²

= Reminder of 6 x (-4)divided by 17

= Reminder of -24 divided by 17

= 10 (that’s the answer)

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Sat, 2008-02-02 15:13

jigar_er_civil

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ans for 1st

Q1. What is the reminder when 11³⁹ is divided by 19

11 and 19 is prime to each other so 11^36/19=1 reminder so final reminder=11*11*11/19=1 reminder

Little Star

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Sat, 2008-02-02 15:15

jigar_er_civil

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ans 2nd qus

Q2. Find the reminder when 5⁹¹ is divided by 91

91 and 5 prime to each other so 5^90/91=1 reminder so reminder =5

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Wed, 2008-07-23 20:13

ssss (not verified)

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doubt

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Thu, 2008-08-07 07:58

ashishlukka

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Hi..

Can you please explain the above soluution?
how the step 11^36/19 comes?

thanks

Thu, 2008-08-07 12:10

paridhi ajmera

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hihere is the solutionQ) rem

hi
here is the solution
Q) rem when 11^39 divided by 19
here since 11 and 19 are co prime so according to euler theorem e(19)=18
so acc to euler 11^18*11^18*11^18 (i.e. 11^36) /19 rem=1
so rest is 11^3/19 so rem is 1

Thu, 2008-08-07 20:51

ashishlukka

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solution q) rem

thanks a lot..:-)

i guess i need lot more practise in apt....

Fri, 2008-08-08 17:40

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Solution to

Solution to Q1....

11^39/19

=> (11^3)^13 /19

=> (1331)^13 /19

=> (1330 + 1)^13/19

=> as 1330 is divisible by 19, every factor of the expansion will contain a 1330...
except

(1)^13/19

therefore: remainder is 1.

Hope you get it now.

CAT4MBA

Reminder Practice Qs – Fermat’s Little Theorem

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