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

Question

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

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

Bunuel · Accepted Answer

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.

mkdureja · Answer

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.

coolpintu · Answer

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?

mkdureja · Answer

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.

GMATtracted · Answer

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

SahilKataria · Answer

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?

ENGRTOMBA2018 · Answer

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.

KarishmaB · Answer

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)

vishumangal · Answer

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

Posted from my mobile device

Pereln2 · Answer

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

Bunuel · Answer

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.

PA0037 · Answer

Hello  Bunuel   KarishmaB 
for this sum I did 
3^1 / 7 = R 3
3^2 / 7 = R 2
3^3 / 7 = R6
3^4 / 7 = R4
3^5 / 7 = R5
3^6 / 7 = R0 
same same continues ,
so 1989/6 gives a remainder of 3 and 3^3 / 7 = R6
Hence , I chose 6
is this the correct way to go about this

Bunuel · Answer

Yes, your approach is correct, but note that the remainder of 3^6/7 is 1, not 0.

KarishmaB · Answer

Your solution is correct, as Bunuel already pointed out. Pattern recognition is a great skill to have for GMAT.

Personally though, for such questions, I favour the binomial approach (discussed above), if possible. It takes less time.

Kinshook · Answer

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

The remainder when the number 3^1989= 3ˆ{3*663} is divided by 7
= The remainder when 27ˆ663 is divided by 7
= The remainder when (28-1)ˆ663 is divided by 7
= The remainder when (-1)ˆ663 is divided by 7
= The remainder when 7-1=6 is divided by 7
=6

IMO C

findingmyself · Answer

Doubt on this question: Expanding 3^1989 as (10-7)^1989.

(10^3978)-(7*10*2)+7^3978
Terms 2 and 3 in above equation is divisible by 7. Leaving with 10^3978

Rewriting 10^3978 as (14-4)^3978

The above expression becomes: (14^7956)-(2*14*4)+(4^7956)

All other terms except 4^7956 is divisible by 7. To know the reminder when 4^7956 is divided by 7..

4 when divided by 7---> reminder 4
4^2 when divided by 7---> reminder 2
4^3 when divided by 7----> reminder 1

The pattern repeats:
7956/3=2652 implies reminder is 1

What is wrong in this approach???

Bunuel · Answer

There are several mistakes.

First, how does (10 - 7)^1989 = (10^3978) - (7 * 10 * 2) + 7^3978? Are you treating this as the  square  of a difference? (10 - 7)^1989 expands into many terms — not just 3 terms.

Same with (14 - 4)^3978 = (14^7956) - (2 * 14 * 4) + 4^7956 — that's just wrong.

williamwordsworth · Answer

Let's solve the problem using **Fermat's Little Theorem** step-by-step to find the remainder when $ 3^{1989} $ is divided by 7.

---

### **Fermat's Little Theorem**
Fermat's Little Theorem states that if $ p $ is a prime number and $ a $ is an integer not divisible by $ p $, then:

\

In this problem:
- $ p = 7 $ (which is prime),
- $ a = 3 $ (which is not divisible by 7).

Thus, applying Fermat's Little Theorem:

\

This tells us that powers of 3 modulo 7 repeat every 6 exponents.

---

### **Step 1: Express the exponent in terms of the cycle length (6)**
We need to compute $ 3^{1989} \mod 7 $. First, divide 1989 by 6 to find how many full cycles fit and the remaining exponent:

\

So, we can write:

\

Thus:

\

---

### **Step 2: Apply Fermat's Little Theorem**
From Fermat's Little Theorem, $ 3^6 \equiv 1 \pmod{7} $. Therefore:

\

So, the equation simplifies to:

\

---

### **Step 3: Compute $ 3^3 \mod 7 $**
Calculate $ 3^3 $:

\

Now, find $ 27 \mod 7 $:

\

Thus:

\

---

### **Final Result**
Substituting back:

\

Therefore, the remainder when $ 3^{1989} $ is divided by 7 is **6**.

---

### **Answer**
The correct choice is **C. 6**.

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

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

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