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What is the remainder when the number 3^1989 is divided by 7

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SrinathVangala
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What is the remainder when the number 3^1989 is divided by 7 [#permalink] New post   Updated on: 17 May 2013, 17:19
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What is the remainder when the number 3^1989 is divided by 7?

A. 1
B. 5
C. 6
D. 4
E. 3
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Official Answer
C

Originally posted by SrinathVangala on 17 May 2013, 07:43.
Last edited by Bunuel on 17 May 2013, 17:19, edited 1 time in total.
Edited the question.
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Bunuel
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What is the remainder when the number 3^1989 is divided by 7 [#permalink] New post   16 Apr 2014, 05:51
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What is the remainder when the number 3^1989 is divided by 7?

A. 1
B. 5
C. 6
D. 4
E. 3

\(3^{1989}=3^{3*663}=27^{663}=(21+6)^{663}\).

Now if we expand this, all terms but the last one will have 21 as a multiple and thus will be divisible by 7. The last term will be \(6^{663}\). So we should find the remainder when \(6^{663}\) is divided by 7.

6^1 divided by 7 yields remainder of 6;
6^2 divided by 7 yields remainder of 1;
6^3 divided by 7 yields remainder of 6 again;
...

The remainder repeats in blocks of two: {6-1}{6-1}{6-1}... When the power is odd the remainder is 6 and when the power is even the remainder is 1. So, the remainder when \(6^{663}=6^{odd}\) is divided by 7 is 6.

Answer: C.

Units digits, exponents, remainders problems: https://gmatclub.com/forum/new-units-di ... 68569.html

Hope it helps.
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Re: What is the remainder??? [#permalink] New post   17 May 2013, 07:48
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3 / 7 rem = 3
3^2 = 9 / 7 rem = 2
3 ^ 3 = 27 / 7 rem = 6 or -1 --------(1)

Now, 1989/3 = 663

From (1) above,
3 ^ 1989 = (3^3) ^ 663 ;
rem = (-1) ^ 663 = -1 or 6
Ans: 6

Hope it is clear.
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Re: What is the remainder??? [#permalink] New post   17 May 2013, 10:09
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This is how I solved it:-
\(\frac{3^{1989}}{7}=\frac{(7-4)^{1989}}{7}\)
Every term in the expansion of \((7-4)^{1989}\) would contain the number '7' except \((-4)^{1989}\)
So it ultimately reduces to finding the the remainder when \((-4)^{1989}\) is divided by 7.
\(\frac{(-4)^{1989}}{7}=\frac{(-1).(4)^{1989}}{7}=\frac{(-1).(64)^{663}}{7}\)
Now 64 would leave a remainder of 1 when divided by 7.
Hence the final remainder would be = -1x1=-1.
This is a negative remainder,hence for finding the actual remainder we just have to add this negative remainder to the divisor i.e. 7
Therefore, the final remainder is (-1+7)=6
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Re: What is the remainder??? [#permalink] New post   17 May 2013, 08:19
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coolpintu wrote:
Can someone please explain me in detail how we arrived at the problem? I solved the Q for unit digit of the expression. Is this approach wrong? How to arrive at the solution?


Unit digit is remainder when divided by 10, what we are asked is remainder when we divide the no. by 7, so finding unit unit digit wont help you.

A rule:
If a when divided by b leaves remainder c,
then, a^x, when divided by b will leave the remainder c^x.

So, to approach the problem, we can start from raised to power 1, and go on and stop when we get 1 or -1 as remainder, then it becomes easy to solve it.
Like in this case, 3^3 leaves remainder -1 when divided by 7,
so using the above rule, we can say that 3^1989 = (3^3)^ 663 will leave remainder (-1)^663 or -1, when divided by 7.
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What is the remainder when the number 3^1989 is divided by 7 [#permalink] New post   Updated on: 17 Oct 2022, 03:12
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SrinathVangala wrote:
What is the remainder when the number 3^1989 is divided by 7?

A. 1
B. 5
C. 6
D. 4
E. 3


Use the concepts of Binomial theorem and negative remainders:

\(3^{1989} = 3^{3*663} = 27^{663} = (28 - 1)^{663}\)

When we use binomial to open this, we will get all terms with 28 (which is divisible by 7) except that last term which will be \((-1)^{663} = -1\)

So the remainder will be -1 which is the same as 7 - 1 = 6 (using the concept of negative remainders)

Originally posted by KarishmaB on 22 May 2017, 05:44.
Last edited by KarishmaB on 17 Oct 2022, 03:12, edited 1 time in total.
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Re: What is the remainder when the number 3^1989 is divided by 7 [#permalink] New post   11 Feb 2019, 20:24
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Cyclicity when 3^n divided by 7 is 6, thus divided 1989 by 6 we get reminder as 3 and and when 3^3 is divided by 7 it gives reminder as 6, thus the answer will be 6

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Re: What is the remainder??? [#permalink] New post   17 May 2013, 08:09
Can someone please explain me in detail how we arrived at the problem? I solved the Q for unit digit of the expression. Is this approach wrong? How to arrive at the solution?
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Re: What is the remainder when the number 3^1989 is divided by 7 [#permalink] New post   24 Sep 2015, 13:15
3/7 remainder 3, 9/7 remainder 2, 27/7 remainder 6

so the cyclicity is 3264000
1989/7 gives remainder 1,
so remainder for 3^1989 should be 3, what am i missing?
Please explain?
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Re: What is the remainder when the number 3^1989 is divided by 7 [#permalink] New post   24 Sep 2015, 17:16
SahilKataria wrote:
3/7 remainder 3, 9/7 remainder 2, 27/7 remainder 6

so the cyclicity is 3264000
1989/7 gives remainder 1,
so remainder for 3^1989 should be 3, what am i missing?
Please explain?


My question to you is: how are you getting 'cyclicity" as 3264000? Cyclicity is defined as number of terms after which a particular pattern will repeat itself be it in remainders or unit's digits etc. How is the cyclicity 32640000 and then based on 1989/7, how can you relate the remainder to what the is asking?

For this question, the best approach is Bunuel's at what-is-the-remainder-when-the-number-3-1989-is-divided-by-152951.html#p1356693

One way to solve these questions is to make sure to express the given exponent in some 'relatable' form wrt the denominator which is what is done above.
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Re: What is the remainder when the number 3^1989 is divided by 7 [#permalink] New post   12 Sep 2023, 21:55
Maybe I'm terribly missing a concept here, but why is it not possible in this example to get the units digit using cyclicity and divide by 7? Why is it that in some instances it may be used, but in other instances you're better off using the binomial theorem? Thanks
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Re: What is the remainder when the number 3^1989 is divided by 7 [#permalink] New post   13 Sep 2023, 00:25
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Pereln2 wrote:
Maybe I'm terribly missing a concept here, but why is it not possible in this example to get the units digit using cyclicity and divide by 7? Why is it that in some instances it may be used, but in other instances you're better off using the binomial theorem? Thanks


The point is that the units digit of an integer does not define its remainder upon division by 7. For example, 15 and 25 both have a units digit of 5. However, when divided by 7, 15 gives a remainder of 1, while 25 gives a remainder of 4. Hence, the cyclicity of the units digit won't help with determining the remainder when dividing by 7.
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Re: What is the remainder when the number 3^1989 is divided by 7 [#permalink]
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