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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.

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.

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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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?

Re: What is the remainder when the number 3^1989 is divided by 7 [#permalink]

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01 May 2015, 18:00

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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?

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]

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06 May 2016, 06:05

I have done this using binomial. 3^1989= 3.3^1988

Leave 3 aside for the moment. now, 3^1988= ((3^2))^994. = 9^994 = (7+2)^994 All the terms in the expression will be divisible by 7 except last one which is 2. So we get here 2. Now we get, 3*2/7(i had kept 3 aside in the beginning) Hence, the remainder 6.

Re: What is the remainder when the number 3^1989 is divided by 7 [#permalink]

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06 May 2016, 07:17

tallyho_88 wrote:

I have done this using binomial. 3^1989= 3.3^1988

Leave 3 aside for the moment. now, 3^1988= ((3^2))^994. = 9^994 = (7+2)^994 All the terms in the expression will be divisible by 7 except last one which is 2. So we get here 2. Now we get, 3*2/7(i had kept 3 aside in the beginning) Hence, the remainder 6.

No issues with your way, you might consider this as a possible way of doing the problem as well -

\(3^{1989}\)= \(3^{663}\) = \(3^{3*221}\)

\(\frac{3^3}{7}\)= \(\frac{27}{7}\) =6

So, \(\frac{3^{663}}{7}\) = Remainder 6

Hence answer will be C. 6 _________________

Thanks and Regards

Abhishek....

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