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

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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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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Using Binomial, we can say that when we divide \((7 + 1)^{146}\) by 7, the remainder will be 1.


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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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.

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

^
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.

The answer is choice D.
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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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Here,
18^22
can be written as 18^(3k+1),k being muliple of 7
Hence (3k+1)^10 div by 7 will yield rem 1 in any case
so,18^1/7
Rem=4

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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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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Yes, that's your binomial theorem concept applied to remainders.
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Responding to a pm:

Yes, correct.


Not correct. Unit's digit cyclicity is applicable only in case the divisor is 2 or 5 or 10 (or a multiple of 10).


You need to handle it using binomial.

\(3^{220} = 3 * 3^{219} = 3 * 27^{73} = 3 * (28 - 1)^{73}\)

Remainder from \((28 - 1)^{73}\) will be -1 so overall remainder will be -3. Since divisor is 7, that is the same as remainder of 4.

If you are not sure about negative remainders, check: https://anaprep.com/number-properties-t ... emainders/
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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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Use Binomial Theorem.
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How can you say that remainder is 4?

(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.

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?

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.

(18^22)^10 = (3*3*2)^22^10 = ((7-1)* 3 )^22^10 = [(7-1)^22 * 3^22]^10

(7-1)^10 divided by 7 will have Remainder = 1 ---> A

Now for 3^22 =9^11= (7+2)^11 , the whole term will be divisible by 7 except 2^11

2^11= 2*2* 8^3
2*2* (7+1)^3 or, Remainder =4 ---->B

From statement A and B ,

18^22^10 = (Remainder 1*Remainder4)^10

4^10 = 4 * 64^3 = 4* (63+1)^3

4* remainder 1

Answer D

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

\(\frac{(18^{22})^{10}}{7}\)

\(18 = 4 (mod 7)\)

\(\frac{(4^{22})^{10}}{7} = \frac{4^{220}}{7} = \frac{2^{440}}{7}\)

\(2^3 = 1 (mod 7)\)

\(\frac{2^{438}*2^2}{7} = \frac{(2^3)^{146}*2^2}{7} = \frac{1*4}{7}\)

Remainder is \(4\).

Hi,

At this point 2^440/7.

2 has a cycle of 2-4-8-6 and 440 is divisible by 4. thus the units digit = 6
6/7 = remainder = 6..

What am I missing here? Please explain