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KarishmaB
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What is the remainder when (18^22)^10 is divided by 7 ? Updated on: 03 Oct 2023, 09:11
Originally posted by KarishmaB on 19 May 2018, 23:31.
Last edited by KarishmaB on 03 Oct 2023, 09:11, edited 1 time in total.
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No, the answer would be the same

\((21 - 3)^{220}\)

The last term will be \((-3)^{220}\) which is same as \(3^{220}\).

\(3* 3^{3*73} = 3 * (27)^{73} = 3 * (28 - 1)^{73}\)

The last term now will be \(3*(-1)^{73}\) which is -3 (a negative remainder)

So the actual remainder will be 7 - 3 = 4

Thank you! I have tried just to find the last digit of \((- 3)^{220}\), since 3 is already less than 7. I thought it's a last stop. Is the last stop the format (x+/-1)/x or we can stop with any number that is less than devisor? Or the power should be odd? I'm confused. Thank you in advance.
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Yes, but 3^220 is much much larger than 7. What you need is to bring it to the form (7a + 1)^n or (7a - 1)^n.
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What is the remainder when (18^22)^10 is divided by 7 ? 11 May 2020, 09:06
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Bunuel
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What is the remainder when (18^22)^10 is divided by 7 ?

А 1
B 2
C 3
D 4
E 5

I think this question is beyond the GMAT scope. It can be solved with Fermat's little theorem, which is not tested on GMAT. Or another way:

\((18^{22})^{10}=18^{220}=(14+4)^{220}\) now if we expand this all terms but the last one will have 14 as multiple and thus will be divisible by 7. The last term will be \(4^{220}\). So we should find the remainder when \(4^{220}\) is divided by 7.

\(4^{220}=2^{440}\).

2^1 divided by 7 yields remainder of 2;
2^2 divided by 7 yields remainder of 4;
2^3 divided by 7 yields remainder of 1;

2^4 divided by 7 yields remainder of 2;
2^5 divided by 7 yields remainder of 4;
2^6 divided by 7 yields remainder of 1;
...

So the remainder repeats the pattern of 3: 2-4-1. So the remainder of \(2^{440}\) divided by 7 would be the same as \(2^2\) divided by 7 (440=146*3+2). \(2^2\) divided by 7 yields remainder of 4.

Answer: D.
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VeritasKarishma

I've understood the format the way Bunuel has explained but still have a doubt in how we arrive at 4^22^10=4^220

I saw a similar ques for 32^32^32, and in that case wen broke the power 32 as 2^5 and got 32^2^5*32=32^2^160

My question is how can we break the power of 22 in a similar way
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What is the remainder when (18^22)^10 is divided by 7 ? 13 May 2020, 22:25
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What is the remainder when (18^22)^10 is divided by 7 ?

А 1
B 2
C 3
D 4
E 5

I think this question is beyond the GMAT scope. It can be solved with Fermat's little theorem, which is not tested on GMAT. Or another way:

\((18^{22})^{10}=18^{220}=(14+4)^{220}\) now if we expand this all terms but the last one will have 14 as multiple and thus will be divisible by 7. The last term will be \(4^{220}\). So we should find the remainder when \(4^{220}\) is divided by 7.

\(4^{220}=2^{440}\).

2^1 divided by 7 yields remainder of 2;
2^2 divided by 7 yields remainder of 4;
2^3 divided by 7 yields remainder of 1;

2^4 divided by 7 yields remainder of 2;
2^5 divided by 7 yields remainder of 4;
2^6 divided by 7 yields remainder of 1;
...

So the remainder repeats the pattern of 3: 2-4-1. So the remainder of \(2^{440}\) divided by 7 would be the same as \(2^2\) divided by 7 (440=146*3+2). \(2^2\) divided by 7 yields remainder of 4.

Answer: D.

VeritasKarishma

I've understood the format the way Bunuel has explained but still have a doubt in how we arrive at 4^22^10=4^220

I saw a similar ques for 32^32^32, and in that case wen broke the power 32 as 2^5 and got 32^2^5*32=32^2^160

My question is how can we break the power of 22 in a similar way
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The two questions are different.

\((18^{22})^{10}\)
Here, 18^{22} is raised to the power 10 so the exponents are multiplied since 18 has two exponents - 22 and 10.

vs

\(32^{32^{32}}\)
Here 32 is raised to the power 32^{32}.
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What is the remainder when (18^22)^10 is divided by 7 ? 15 May 2020, 18:07
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Bunuel
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What is the remainder when (18^22)^10 is divided by 7 ?

А 1
B 2
C 3
D 4
E 5

I think this question is beyond the GMAT scope. It can be solved with Fermat's little theorem, which is not tested on GMAT. Or another way:

\((18^{22})^{10}=18^{220}=(14+4)^{220}\) now if we expand this all terms but the last one will have 14 as multiple and thus will be divisible by 7. The last term will be \(4^{220}\). So we should find the remainder when \(4^{220}\) is divided by 7.

\(4^{220}=2^{440}\).

2^1 divided by 7 yields remainder of 2;
2^2 divided by 7 yields remainder of 4;
2^3 divided by 7 yields remainder of 1;

2^4 divided by 7 yields remainder of 2;
2^5 divided by 7 yields remainder of 4;
2^6 divided by 7 yields remainder of 1;
...

So the remainder repeats the pattern of 3: 2-4-1. So the remainder of \(2^{440}\) divided by 7 would be the same as \(2^2\) divided by 7 (440=146*3+2). \(2^2\) divided by 7 yields remainder of 4.

Answer: D.
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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
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What is the remainder when (18^22)^10 is divided by 7 ? 26 Aug 2020, 17:47
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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.
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What is the remainder when (18^22)^10 is divided by 7 ? 05 May 2021, 04:33
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VeritasKarishma
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Can you point out the mistake here:
R(2^440)/7:

2^440 = 2*2^339 = 2*8^113=2*(7+1)^113=2*(7*I+1^113), here I is an integer.

R(2*(7*I+1^113))/7 = R(2*7*I+2)/7 = R(2/7) = 2

There's your mistake. If you take a 2 separately, you are left with 439 2s and not 339 2s.

\(2^{440} = 4*2^{438} = 4*8^{146} = 4*(7 + 1)^{146}\)

When you divide it by 7, remainder is 4.
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Hello, VeritasKarishma mam.Do you have any article regarding this topic?
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What is the remainder when (18^22)^10 is divided by 7 ? Updated on: 03 Oct 2023, 09:11
Originally posted by KarishmaB on 06 May 2021, 03:07.
Last edited by KarishmaB on 03 Oct 2023, 09:11, edited 1 time in total.
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cumulonimbus

Can you point out the mistake here:
R(2^440)/7:

2^440 = 2*2^339 = 2*8^113=2*(7+1)^113=2*(7*I+1^113), here I is an integer.

R(2*(7*I+1^113))/7 = R(2*7*I+2)/7 = R(2/7) = 2

There's your mistake. If you take a 2 separately, you are left with 439 2s and not 339 2s.

\(2^{440} = 4*2^{438} = 4*8^{146} = 4*(7 + 1)^{146}\)

When you divide it by 7, remainder is 4.
Hello, VeritasKarishma mam.Do you have any article regarding this topic?
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Use Binomial Theorem.
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What is the remainder when (18^22)^10 is divided by 7 ? 06 May 2021, 11:47
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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 :)
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What is the remainder when (18^22)^10 is divided by 7 ? 06 May 2021, 23:05
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Jihyo
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 :)
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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.
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What is the remainder when (18^22)^10 is divided by 7 ? 14 May 2021, 07:33
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(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
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What is the remainder when (18^22)^10 is divided by 7 ? 03 Oct 2023, 05:44
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What is the remainder when (18^22)^10 is divided by 7 ?

А 1
B 2
C 3
D 4
E 5

I think this question is beyond the GMAT scope. It can be solved with Fermat's little theorem, which is not tested on GMAT. Or another way:

\((18^{22})^{10}=18^{220}=(14+4)^{220}\) now if we expand this all terms but the last one will have 14 as multiple and thus will be divisible by 7. The last term will be \(4^{220}\). So we should find the remainder when \(4^{220}\) is divided by 7.

\(4^{220}=2^{440}\).

2^1 divided by 7 yields remainder of 2;
2^2 divided by 7 yields remainder of 4;
2^3 divided by 7 yields remainder of 1;

2^4 divided by 7 yields remainder of 2;
2^5 divided by 7 yields remainder of 4;
2^6 divided by 7 yields remainder of 1;
...

So the remainder repeats the pattern of 3: 2-4-1. So the remainder of \(2^{440}\) divided by 7 would be the same as \(2^2\) divided by 7 (440=146*3+2). \(2^2\) divided by 7 yields remainder of 4.

Answer: D.
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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?
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What is the remainder when (18^22)^10 is divided by 7 ? 03 Oct 2023, 09:18
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What is the remainder when (18^22)^10 is divided by 7 ?

А 1
B 2
C 3
D 4
E 5

I think this question is beyond the GMAT scope. It can be solved with Fermat's little theorem, which is not tested on GMAT. Or another way:

\((18^{22})^{10}=18^{220}=(14+4)^{220}\) now if we expand this all terms but the last one will have 14 as multiple and thus will be divisible by 7. The last term will be \(4^{220}\). So we should find the remainder when \(4^{220}\) is divided by 7.

\(4^{220}=2^{440}\).

2^1 divided by 7 yields remainder of 2;
2^2 divided by 7 yields remainder of 4;
2^3 divided by 7 yields remainder of 1;

2^4 divided by 7 yields remainder of 2;
2^5 divided by 7 yields remainder of 4;
2^6 divided by 7 yields remainder of 1;
...

So the remainder repeats the pattern of 3: 2-4-1. So the remainder of \(2^{440}\) divided by 7 would be the same as \(2^2\) divided by 7 (440=146*3+2). \(2^2\) divided by 7 yields remainder of 4.

Answer: D.

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?
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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.
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What is the remainder when (18^22)^10 is divided by 7 ? 12 Jul 2024, 05:19
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What is the remainder when (18^22)^10 is divided by 7 ?

А. 1
B. 2
C. 3
D. 4
E. 5
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­Solved it the old fashion way...

(18^22)^10 = ((2x3^2)^22)^10 = [(2^220)x(3^440)­]

So, (2^1)/7 -> 0R2; (2^2)/7 -> 0R4; (2^3) ->1R1
(2^220)/7 -> R2 (because the pattern repeats every 3x as shown above)

(3^440)/7 -> R2 (with 3 raised to an exponent the pattern repeats every 6x)

So the final remainder will be R2xR2=4

Answer D.
 ­
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What is the remainder when (18^22)^10 is divided by 7 ? 17 Sep 2024, 07:20
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We need to find the remainder when \((18^{22})^{10}\) is divided by 7?

\((18^{22})^{10}\) = \(18^{22*10}\) = \(18^{220}\)

We solve these problems by using Binomial Theorem, where we split the number into two parts, one part is a multiple of the divisor(7) and a big number, other part is a small number.

=> \(18^{220}\) = \((14+4)^{220}\)

Watch this video to MASTER BINOMIAL Theorem

Now, if we use Binomial Theorem to expand this then all the terms except the last term will be a multiple of 7
=> All terms except the last term will give remainder of 0 when divided by 7
=> Problem is reduced to what is the remainder when the last term (i.e. 220C220 * 14^0 * 4^220) is divided by 7
=> Remainder of 4^220 is divided by 7
=> Remainder of 2^440 is divided by 7

To solve this problem we need to find the cycle of remainder of power of 2 when divided by 7

Remainder of \(2^1\) (=2) by 7 = 2
Remainder of \(2^2\) (=4) by 7 = 4
Remainder of \(2^3\) (=8) by 7 = 1
Remainder of \(2^4\) (=16) by 7 = 2
Remainder of \(2^5\) (=32) by 7 = 4
Remainder of \(2^6\) (=64) by 7 = 1

=> Cycle is 3

Remainder of 440 by 3 = 2
=> Remainder of \(2^{440}\) by 7 = Remainder of \(2^{2}\) by 7 = 4

So, Answer will be D
Hope it Helps!

Watch following video to MASTER Remainders by 2, 3, 5, 9, 10 and Binomial Theorem

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