Remainder Theorem and related concepts for CAT Preparation  Ravi Handa

Fermat's Theorem
What is the remainder of 57^67^77/17 ?
Rem [57^67^77/ 17] = Rem [6^67^77/17]
Now, by Fermat's theorem, which states Rem [a^(p1)/p] = 1, we know
Rem [6^16/17] = 1The number given to us is 6^67^77
Let us find out Rem[Power / Cyclicity] t0 find out if it 6^(16k+1) or 6^(16k+2). We can just look at it and say that it is not 6^16k
Rem [67^77/16] = Rem [3^77/16] = Rem[(3^76*3^1)/16]
= Rem[((81^19) * 3)/16] = Rem [1 * 3/16] = 3=> The number is of the format 6^(16k + 3)
=> Rem [6^67^77 /17] = Rem [6^(16k + 3)/17] = Rem [6^3/17]
= Rem [216/17] = 12What is the remainder when 2^1000 is divided by 59?
We have to find out Rem [2^1000/59]
As per Fermat's Theorem, [a^(p1)/p] = 1 where p is a prime number and HCF(a,p) = 1
Rem [2^58 / 59] = 1
=> Rem [(2^58)^17 / 59] = 1
=> Rem [2^986 / 59] = 1Rem [2^1000/59]
= Rem [2^986 x 2^14 / 59]
= Rem [1 x 128 x 128 / 59]
= Rem [1 x 10 x 10 / 59]
= Rem [100/59]
= 41What is the remainder when 2^89 is divided by 89?
We can use Fermat's Theorem here which says
Rem [a^(p1)/p] = 1 where p is a prime number
=> Rem [2^88/89] = 1
=> Rem [2^89/89] = Rem [2^88/89] * Rem [2/89] = 1*2 = 2What is the remainder when 17^432 is divided by 109?
As per Fermat's Theorem, [a^(p1)/p] = 1 where p is a prime number and HCF(a,p) = 1
Rem [17^108 / 109] = 1
=> Rem [(17^108)^4 / 109] = 1
=> Rem [17^432 / 109] = 1What is the remainder when 17^325 is divided by 109?
As per Fermat's Theorem, [a^(p1)/p] = 1 where p is a prime number and HCF(a,p) = 1
Rem [17^108 / 109] = 1
=> Rem [(17^108)^3 / 109] = 1
=> Rem [17^324 / 109] = 1
=> Rem [17^324 * 17 / 109] = 1*17
=> Rem [17^325 / 109] = 17What is the remainder when 2^1040 is divided by 131?
As per Fermat's Theorem, [a^(p1)/p] = 1 where p is a prime number and HCF(a,p) = 1
Rem [2^130 / 131] = 1
=> Rem [(2^130)^8 / 131] = 1
=> Rem [2^1040 / 131] = 1Euler's Theorem
What is the remainder of (121) ^(121) divided by 144?
Euler Totient (144) = 144*(11/2)(11/3) = 48
By Euler's Theorem we can say that
Rem[a^48/144] = 1 if a and 144 are coprime to each other
=> Rem [a^240/144] = 1
=> Rem [11^240/144] = 1Now, we need to find out Rem [121^121/144]
= Rem [11^242/144]
= Rem [11^240/144] * Rem [11^2/144]
= 1 * 121
= 121Wilson's Theorem
What is the remainder when 97! is divided by 101?
Wilson's Theorem says For a prime number 'p'
Rem [ (p1)! / p] = p1This can be extended to say,
Rem [ (p2)! / p] = 1Let us use that here. We need to find out Rem [97! / 101] = r
We know from the above theorem,
Rem [99! / 101] = 1
=> Rem [99 * 98 * 97! / 101] = 1
=> Rem [ (2)*(3)*r / 101] = 1
=> Rem [6r / 101] = 1
=> 6r = 101k + 1
We need to think of a value of k, such that 101k + 1 is divisible by 6.
If we put, k = 1, we get 101 + 1 = 102, which is divisible by 6.
=> 6r = 102
=> r = 17What is the remainder when 21! is divided by 361?
Rem [21!/361]
= Rem [(21 * 20 * 19 * 18!)/361]
Using Rem [ka/kb] = k Rem[a/b]
= 19 Rem [(21 * 20 * 18!)/19]
Using Wilson's Theorem says For a prime number 'p' Rem [ (p1)! / p] = p1
= 19 Rem [(21 * 20 * 18)/19]
= 19 Rem [(2 * 1 * (1))/19]
= 19 * (2)
= 38
= 323Pattern Recognition / Cyclicity Method
What is the remainder when 7^99 is divided by 2400?
Let us try doing it by the pattern recognition / cyclicity method. It can be really long if you do not get the pattern quickly. Use other methods like Euler's Totient if you do not get a pattern quickly.
Rem[7^1 / 2400] = Rem [7 / 2400] =7
Rem[7^2 / 2400] = Rem [49 / 2400] =49
Rem[7^3 / 2400] = Rem [343/ 2400] = 343
Rem[7^4 / 2400] = Rem [2401 / 2400] = 1After this the same pattern will keep on repeating because you got a 1.
Once we have obtained the cyclicity (number of terms in the pattern), all we need to do is to find out the Remainder of Power when divided by the Cyclicity. Whatever is this remainder, that particular value in the cycle is our answer.
In this case, power is 99 and cyclicity is 4.
Rem [Power / Cyclicity] = Rem [99/4] = 3
=> Our answer will be the third value in the cycle = 343What is the remainder when 32^32^32 is divided by 7?
Rem [32^32^32 / 7] = Rem [4^32^32 /7]
Now, we need to observe the pattern
4^1 when divided by 7, leaves a remainder of 4
4^2 when divided by 7, leaves a remainder of 2
4^3 when divided by 7, leaves a remainder of 1And then the same cycle of 4, 2, and 1 will continue.
If a number is of the format of 4^(3k+1), it will leave a remainder of 4
If a number is of the format of 4^(3k+2), it will leave a remainder of 2
If a number is of the format of 4^(3k), it will leave a remainder of 1The number given to us is 4^32^32
Let us find out Rem[Power / Cyclicity] t0 find out if it 4^(3k+1) or 4^(3k+2). We can just look at it and say that it is not 4^3k
Rem [32^32/3] = Rem [(1)^32/3] = 1
=> The number is of the format 4^(3k + 1)
=> Rem [4^32^32 /7] = 4What is the remainder when 32^ (32^ (32^...infinite times)) is divided by 9?
Rem [32^32^32... / 9] = Rem [4^32^32... /9]
Now, we need to observe the pattern
4^1 when divided by 9, leaves a remainder of 4
4^2 when divided by 9, leaves a remainder of 7
4^3 when divided by 9, leaves a remainder of 1And then the same cycle of 4, 7, and 1 will continue.
If a number is of the format of 4^(3k+1), it will leave a remainder of 4
If a number is of the format of 4^(3k+2), it will leave a remainder of 7
If a number is of the format of 4^(3k), it will leave a remainder of 1The number given to us is 4^32^32....
Let us find out Rem[Power / Cyclicity] to find out if it 4^(3k+1) or 4^(3k+2). We can just look at it and say that it is not 4^3k
Rem [32^32^32.../3] = Rem [(1)^32^32.../3] = 1
=> The number is of the format 4^(3k + 1)
=> Rem [4^32^32 /9] = 4What is the remainder when 34^31^301 is divided by 9?
Rem [34^31^301 / 9] = Rem [7^31^301 /9]
Now, we need to observe the pattern
7^1 when divided by 9, leaves a remainder of 7
7^2 when divided by 9, leaves a remainder of 4
7^3 when divided by 9, leaves a remainder of 1
And then the same cycle of 7, 4, and 1 will continue.
If a number is of the format of 7^(3k+1), it will leave a remainder of 7
If a number is of the format of 7^(3k+2), it will leave a remainder of 4
If a number is of the format of 7^(3k), it will leave a remainder of 1The number given to us is 7^31^301
Let us find out Rem[Power / Cyclicity] to find out if it 7^(3k+1) or 7^(3k+2). We can just look at it and say that it is not 7^3k
Rem [31^301/3] = Rem [1^301/3] = 1
=> The number is of the format 7^(3k + 1)
=> Rem [7^31^301 /9] = 7What is the remainder of 30^72^87 when divided by 11?
Rem [30^72^87 / 11] = Rem [(3)^72^87 / 11] = Rem [3^72^87 / 11]
Now, we need to observe the pattern
3^1 when divided by 11, leaves a remainder of 3
3^2 when divided by 11, leaves a remainder of 9
3^3 when divided by 11, leaves a remainder of 5
3^4 when divided by 11, leaves a remainder of 4
3^5 when divided by 11, leaves a remainder of 1
And then the same cycle of 3, 9, 5, 4 and 1 will continue.
If a number is of the format of 3^(5k + 1), it will leave a remainder of 3
If a number is of the format of 3^(5k + 2), it will leave a remainder of 9
If a number is of the format of 3^(5k + 3), it will leave a remainder of 5
If a number is of the format of 3^(5k + 4), it will leave a remainder of 4
If a number is of the format of 3^(5k), it will leave a remainder of 1The number given to us is 3^72^87
Let us find out Rem[Power / Cyclicity] t0 find out if it 3^(5k + what?)Rem [72^87 / 5]
= Rem [2^87 / 5]
= Rem [2 * 4^43/5]
= Rem [2 * (1) / 5]
= 2
= 3
=> The number is of the format 3^(5k + 3)
=> Rem [3^72^87 / 11] = 5What is the remainder when 1! +2 * 2! + 3 * 3! + 4 * 4! +... +12 * 12! Is divided by 13?
The trick in these type of questions is often observing the pattern
1! + 2 * 2! = 1 + 4 = 5 = 3!  1
1! + 2 * 2! + 3 * 3! = 1 + 4 + 18 = 23 = 4!  1
1! + 2 * 2! + 3 * 3! + 4 * 4! = 1 + 4 + 18 + 96 = 119 = 5!  1
1! + 2 * 2! + 3 * 3! + 4 * 4! + 5 * 5! = 1 + 4 + 18 + 96 + 600 = 719 = 6!  1
So, we can say
1! + 2 * 2! + 3 * 3! + 4 * 4! +... + 12 * 12! = 13!  1
=> Rem [(1! + 2 * 2! + 3 * 3! + 4 * 4! +... + 12 * 12!) / 13] = 1 = 12What is the remainder of 57^67^77/17 ?
Rem [57^67^77/ 17] = Rem [6^67^77/17]
Now, by Fermat's theorem, which states Rem [a^(p1)/p] = 1, we know
Rem [6^16/17] = 1The number given to us is 6^67^77
Let us find out Rem[Power / Cyclicity] to find out if it 6^(16k+1) or 6^(16k+2). We can just look at it and say that it is not 6^16k
Rem [67^77/16] = Rem [3^77/16] = Rem[(3^76 * 3^1)/16]
= Rem[((81^19) * 3)/16] = Rem [1 * 3/16] = 3=> The number is of the format 6^(16k + 3)
=> Rem [6^67^77 /17] = Rem [6^(16k + 3)/17] = Rem [6^3/17]
= Rem [216/17] = 12How do you find the remainder when 7^26 is divided by 100?
Finding out the remainder from 100, is the same as finding out the last two digits of a number
Last two digits of 7^1 are 07
Last two digits of 7^2 are 49
Last two digits of 7^3 are 43
Last two digits of 7^4 are 01
After this, the same pattern will keep on repeating.
So, 7^(4n+1) will end in 07, 7^(4n+2) will end in 49, 7^(4n + 3) will end in 43, and 7^4n will end in 017^26 = 7^(4n+2) will end in 49
=> Rem [7^26/100] = 49Using Binomial Theorem
What is the remainder when 25^10 is divided by 576?
We need to find out the remainder of 25^10 when divided by 576.
Please note that 576 = 24^2
There are couple of methods of solving this.Using Binomial Theorem
25^10 = (24 + 1)^10
In the expansion, there will be 11 terms where the powers of 24 will vary from 0 to 10.
If the power of 24 is greater than or equal to 2 in a term, that term will be divisible by 576
The terms that will not be divisible by 576 are the terms that have powers of 24 as 0 or 1.
Those terms are
10C1 * 24^1 * 1^9 + 10C0 * 24^0 * 1^10
= 10 * 24 * 1 + 1 * 1 * 1
= 241
So, Rem [25^10/576] = 241