What is the remainder when (18^22)^10 is divided by 7 ?

Fermat's little theorem
If p is a prime number

then for any integer a, the number (a^p –a ) is an integer multiple of p, that is, p〡(a^p –a )
and if p & a mutually prime, then p〡(a^p-1 –1)

I was wondering how could we, by using this theorem, to apply it to this question

the question can be simplified into 4^220 = 2^440, so we have to find the number of the exponent that can be divided by 7 as well as be closet to 440

from above theorem, I intuitively think of
2^441- 2 = 441 * x = 7 * 63 * x
2(2^440- 1) = 7 * 63 * x
also 2&7 mutually prime
2^440- 1 = 7 * y (must be multiple of 7)
2^440= 1 (mod7)…..remainer is 1, totally wrong

I check where my missing point lead to this big mistake
I notice in definition it require that p be “prime number”
but 441 = 7^2 * 9 , not prime
and this is how it lead to my false remainer

if we do the calculation in the definition of Fermat’s theorem, which said p must be prime
now assume a = 2^x , x be the unknown and can be any given integer, in the polynomial of
[(2^x)7 - (2^x)] = 7 * z
(2^x)[ (2^x)6 – 1] = 7 * z

also 2^x & 7 mutually prime
7〡[(2^x)6 – 1]
[ (2^x)6 – 1] = 7 * n ……(1)

Here for 6x, find the one closet to 440 since this question mainly concern about (2^440)/7
440 / 6 = 73 …2
440 = 73 * 6 + 2
2^2 [ (2^6)73 – 1] = 2^2 (7 * n)

Thus 2^440 = 2^2 (mod 7), remainer 4 , correct

Bunuel's approach is the best approach (as usual). If you do not know or understand the Binomial Theorem or Fermat's Little Theorem or Wilson's Theorem (or any other theorem for that matter), there is a bit longer way to proceed.

Concept: Given --- X * Y / 7 ---- and you want to know the Remainder, you can take the Rem of the terms separately and then Multiply the Remainders. In the end you remove the Excess Rem greater than 7 by continually dividing 7 (the Divisor) from the Excess Rem


Step 1: Nested Exponent Rule

(18 ^ 22) ^ 10 = 18^22 * 18^22 * 18^22 ........ * 18^22 ----- 10 Times Multiplied

When you Add the Powers of the Same Base 18, the Result = (18)^220

Step 2: 18 = 2 * 3^2 ----- Substitute this Prime Fact. in for 18

(2 * 3^2) ^ 220

Again applying the Nested Exponent Rule

(2) ^220 * (3) ^ 440

Divide all of this by 7

[ (2)^220 * (3)^440 ] / 7 --- Rem = ?

Using the 1st Concept of Splitting the Terms and Multiplying the Remainders:

( [ 2^220 / 7 ] Remainder of ) * ( [ 3 ^440 / 7 ] Remainder of ) = Remainder of above

Step 3: Find the Cyclicity of the Remainder Pattern when you continually Divide Consecutive Powers of 2 by 7 and Consecutive Powers of 3 by 7


2/7 ------- Rem = 2

4/7 ------ Rem = 4

8/7 ------- Rem = 1 *you can stop here, after a Rem = 1, the Pattern will keep repeating



Remainder pattern when 2^x is Divided by 7 ----- [2 - 4 - 1] --- 3 Remainders keep repeating over and over

Dividing the Exponent (220) by 3 --- 220/3 = Rem of 1


****Thus the Remainder when 2^220 / 7 = Rem of 2


When you do the Same for 3^x / 7, you find the Remainder Pattern ---- [3 - 2 - 6 - 4 - 5 - 1]

Dividing the Exponent of 440 by the 6 Cyclicity of the Rem Pattern ---- 440/6 = Rem of 2

***Thus the Remainder of: (3)^440 / 7 --- Rem = 2



(2^220 / 7) Rem of 2 * (3^440 / 7) Rem of 2 = Total Remainder of 4



Since this Remainder of 4 is Less Than the Divisor this is your Answer.



Really long way to do it, Bunuel's approach in the beginning is much faster. Understanding the Binomial Theorem is helpful. For this approach, knowing the Powers of 2 and Powers of 3 (thanks Karishma for advising that) really helps it move along faster.

Hello, VeritasKarishma mam.Do you have any article regarding this topic?

Use Binomial Theorem.

Hello, VeritasKarishma mam.Thank you so much for providing me with your amazing article.Can I do this math using the following method?

First of all working with 18^22/7
here, 18^22=(14+4)^22
(14+4)^22 leaves a remainder 4 when divided by 7 ,as 4^22=4^(3k+1)
now, 4^10=4^3k+1; so it will also leave 4 as a remainder.

Can you please verify? TIA

Yes, this is correct. I am assuming here that you realise that
4^1 leaves rem 4
4^2 leaves rem 2
4^3 leaves rem 1
and the cycle repeats.

(18^22)^10

Replace the numbers by their respective remainders when divided by 7:

(4^1)^3 = 4^3 = 256

Rof 256 = Rof4 * Rof64 = 4*1 = 4

Bunuel KarishmaB
It did not strike me to expand 18 as (14+4) and thus proceeded to find the units digit of the problem given and I did end up getting the units digit of (18^220) as '6'. However, I could not conclude from the obtained answer what might be the remainder when divided by 7 (as it could be '16'; '46'; '56'; '76' etc). Is there any way forward with my solution?

Units digit doesn't help you find the remainder except in certain specific cases (when the divisor is 2 or 5 or 10).
You will need to use either pattern recognition (which will get very tedious very quickly here) or binomial theorem.