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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.

When we expand \((7 + 1)^{146}\), we get lots of terms such that \(7^{146}\) is the first term and 1 is the last term. If you multiply \((7 + 1)^{146}\) by 4, you get the same terms except each is multiplied by 4 so the first term is \(4*7^{146}\) and the last one is 4. Every term will still have a 7 in it except the last term. Since the last term is 4, the remainder will be 4.
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Re: What is the remainder when (18^22)^10 is divided by 7 ? [#permalink]

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10 Jun 2013, 03:50

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Financier wrote:

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

А. 1 B. 2 C. 3 D. 4 E. 5

^ 1.(18)^220 = (((18) ^4)^5)^11 2. When 18 is divided by 7 the remainder is 4 3. Now 4^4 is 256. when divided by 7, the remainder is 4 4. Since the remainder is again 4, compute 4^5 = 1024. when divided by 7, the remainder is 2 5. Since 2 is the remainder, now we compute 2^11 = 2048. When divided by 7, the remainder is 4. This is the answer.

Re: What is the remainder when (18^22)^10 is divided by 7 ? [#permalink]

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17 May 2014, 07:23

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Bunuel wrote:

gaurav1418z wrote:

Bunuel

(18^22)^10 = 18^220

I remember you quoted that for (xyz)^n, if we are asked to find the remainder, we can find remainder for z^n

so i found remainder for 8^220, and got answer as 1

Where am i going wrong?

Where did I write that? I think that you mean the following part from Number Theory booksaying that the last digit of \((xyz)^n\) is the same as that of \(z^n\). But the last digit of a number does not determine its remainder upon dividing by 7. For example, 8 divided by 7 gives the remainder of 1 while 18 divided by 7 gives the remainder of 4.

Another approach to solve such kind of questions is using cyclicity. Lemme try and explain this.

We have (18^22)^10 = 18^220

Now we try to see what are the remainders for various powers of 18 when it is divided by 7

18^1 leaves 4 as remainder 18^2 leaves 2 as remainder 18^3 leaves 1 as remainder 18^4 leaves 4 as remainder

Hence, we see after 3 set of powers, the remainder starts repeating. Now, this means if the power is multiple of 3, the remainder is 1, if the power leaves remainder 1 when divided by 3, the actual answer will be 4, and so forth.

We see, when 220 is divided by 3, we get 1 as remainder, showcasing 18^1 case, hence the final remainder is 4.

This is because 18^220 = 18^219 x 18 = (18^3)^73 x 18

Thus, we need to find the remainder when 18 is divided by 4 as 18^219 gives 1 as remainder.

Just remember that all terms in the expression (a+b)^n are divisible by a except for the last term i.e b^n. My solution uses just this much.

I need someone to validate this approach or did I just go absolutely berserk

Yes, that's your binomial theorem concept applied to remainders. In case you are interested in checking out the details of this approach, look at this post: http://www.veritasprep.com/blog/2011/05 ... ek-in-you/ _________________

Bunuel what is wrong with the approach below- 18^220 = 2^220 3^440 this product will have last digit 6... Can't we use cyclisity or some other approach to solve? On the face of it it looks like a GMAT type problem...

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.

Hi Karishma/Bunnel,

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.

Re: What is the remainder when (18^22)^10 is divided by 7 ? [#permalink]

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22 Apr 2014, 18:07

(18^{22})^{10}=18^{220}=(7+11)^{220}. If we expand this equation all terms will be divisible by 7 except the last one.

The last one will be 11^{220}. So we should find the remainder when 11^{220} is divided by 7.

11^{220}=?

11^1 divided by 7 yields remainder of 4; 11^2 divided by 7 yields remainder of 2; 11^3 divided by 7 yields remainder of 1; 11^4 divided by 7 yields remainder of 4; Now we have a pattern 4,2,1,4,2,1

Conclusion: the remainder repeats the pattern of 3: 4-2-1. So the remainder of 11^{220} divided by 7 would be the same as 11*1 (that is because 220 is 73*3+1) Answer: D.

I remember you quoted that for (xyz)^n, if we are asked to find the remainder, we can find remainder for z^n

so i found remainder for 8^220, and got answer as 1

Where am i going wrong?

Where did I write that? I think that you mean the following part from Number Theory book (math-number-theory-88376.html) saying that the last digit of \((xyz)^n\) is the same as that of \(z^n\). But the last digit of a number does not determine its remainder upon dividing by 7. For example, 8 divided by 7 gives the remainder of 1 while 18 divided by 7 gives the remainder of 4.
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How did you get 12 as remainder when -1 is divided by 13, i didnt quite understand how did you substitute quotient and the remainder in the formula below

How did you get 12 as remainder when -1 is divided by 13, i didnt quite understand how did you substitute quotient and the remainder in the formula below

-1 = 13(-1) + 12

Hi there,

Remember that remainder can never be negative so when you get remainder negative as in above case then you add divisor to it

So we have remainder -1+13 =12

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