12 Days of Christmas GMAT Competition - Day 8: If n is the number of

Question

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If n is the number of multiples of 3 between 3^13 and 3^10, not inclusive, what is the remainder when n is divided by 13?

A. 0
B. 1
C. 3
D. 10
E. 12

Bunuel · Accepted Answer

GMAT Club's Official Explanation:

The number of multiples of an integer in a range is given by the formula  \frac{	ext{last multiple in the range - first multiple in the range}}{	ext{multiple}}+1 . Hence:

Since both 3^13 and 3^10 are multiples of 3, the last multiple of 3 in the range will be 3^13 - 3, and the first multiple will be 3^10 + 3. Therefore, the number of multiples of 3 between 3^13 and 3^10, not inclusive, is:

\frac{3^{13} - 3 - (3^{10} + 3)}{3} + 1=

=3^{12} - 3^{9} - 2 + 1=

=3^{12} - 3^{9} -1=

=3^9(3^3 - 1) - 1=

=3^9*26 - 1=

=3^9*2*13 - 1=

The first term, 3^9 * 2 * 13, is divisible by 13, so the remainder of the entire expression comes from -1 divided by 13. The remainder when -1 is divided by 13 is 12.

Answer: E.

P.S. The process for finding the remainder when dividing a negative integer by a positive integer follows the same principles as when dividing a positive integer by a positive integer.

For example:

What is the remainder when dividing 21 by 6? We find the closest multiple of 6 that is  less  than 21, which is 18. Then, we calculate (dividend) - (closest multiple  less  than the dividend) = 21 - 18, yielding a remainder of 3. 
 What is the remainder when dividing -23 by 7? Here, we find the closest multiple of 7 that is  less  than -23, which is -28. Then, we calculate (dividend) - (closest multiple  less  than the dividend) = -23 - (-28), resulting in a remainder of 5.

What about dividing -100 by 30? The closest multiple of 30  less  than -100 is -120. So, the remainder is (dividend) - (closest multiple  less  than the dividend) = -100 - (-120) = 20.

Alternatively, consider this method:

What is the remainder when dividing -23 by 7? Dividing 23 by 7 gives a remainder of 2. To find the remainder for -23 divided by 7, subtract this 2 from the divisor. Thus, the remainder when dividing -23 by 7 is 7 - 2 = 5.

Similarly, what is the remainder when dividing -100 by 30? Dividing 100 by 30 gives a remainder of 10. Therefore, the remainder when dividing -100 by 30 is 30 - 10 = 20.

GraCoder · Answer

By AP formula an = a0 + (n-1)d

3^13 = 3^10 + (n-1)3
n = 26*3^9 + 1
Excluding the first and last term => 26*3^9 - 1

26 is divisible by 13, hence remainder is -1 mod 13 => 12

Kinshook · Answer

If n is the number of multiples of 3 between 3^13 and 3^10, not inclusive, what is the remainder when n is divided by 13?

In arithmetic progression (applicable in case of multiples ) : - 
First term = a
Last term = l
Common difference = d
Number of terms, not inclusive = (l-a)/d - 1

n = Number of multiples of 3 between 3^10 and 3^13, not inclusive = (3^13 - 3^10)/3 - 1 = 3^10*(3^3-1)/3 - 1 = 3^9*26 - 1

The remainder when n is divided by 13 = The remainder when (3^9*26 - 1) is divided by 13 = 13-1 = 12

IMO E

RamaSubramanian · Answer

We are asked to find number of 3's between 3^10 and 3^13 (exclusive) and find remainder when divided by 13

say, lets count number of 3's from 3^1 to 3^3 (exclusive)

which would be {3,6,9,12,15,18,21,24,27} - {3,27}

= (3^3 - 3^1) / 3 + 1 - 2
= 7

similarly extending

we have (3^13 - 3^10) / 3 + 1 - 2

= (3^13 - 3^10) / 3 + 1 - 2

3^10 (3^3 - 1) / 3 - 1

=3^9 * 26 - 1
= when mod 13, we get 12

So , option E

hr1212 · Answer

For an AP, nth term Tn = a + (n - 1)d

Here,  Tn = 3^{13} - 3, a = 3^{10} + 3, d = 3

3^{13} - 3 = 3^{10} + 3 + (n - 1)*3 
 3^{10}*(3^3 - 1) - 6 = (n - 1)*3 
 (3^9)*26 - 2 = n - 1 
n =  (3^9)*26 - 1

Here, first term is divisible by 13, so subtracting 1 from a number that is divisible by a divisor will always give us a remainder of (divisor - 1) => 13 - 1 = 12

Answer: E

missionmba2025 · Answer

We can take a small subset to understand the number of terms that are divisible by 3

Between 3 and 3^2 , we have 1 terms which is divisible by 3

We can find this by

\frac{(3^2 - 3)}{3} - 1   = 1

Similarly, Between 3 and 3^3 , we have 8 terms which is divisible by 3

We can find this by

\frac{(3^3 - 3)}{3} - 1   = 8

In similar fashion between  3^{10}  and  3^{13} , we have  \frac{3^{13} - 3^{10}}{3} - 1  terms which are divisible by 3

n =  \frac{3^{13} - 3^{10}}{3} - 1

n =  \frac{3^{10}( 3^2- 1)}{3} - 1

n =  3^{9}*26 - 1

Remainder when n is divided by 13 = -1

13 - 1 = 12

Option E

nishantswaft · Answer

The stem tells us that;
 
  n is the number of multiples of 3 between 3^13 and 3^10, not inclusive

And we have to select;  
 
  what is the remainder when n is divided by 13?

Let's find the number of multiples of 3 or value of n first

n= 3^{13} - 3^{10} }/3] -1 
  { 3^{12} - 3^{9} } -1 
   3^{9} (  3^{3} -1 ) -1

When n is divided by 13, 
 R( \frac{n}{13} ) = R(  3^{9} (26)/13 ) -R( \frac{1}{13} ) 
  R( 0 ) -R( \frac{1}{13} ) 
  0-1= -1

Since -1 can't be the remainder; 
 actual remainder = 13-1 =12

SafSin28 · Answer

If n is the number of multiples of 3 between 3^13 and 3^10, not inclusive, what is the remainder when n is divided by 13?
Solution:
n is divisible by both 3^11 and 3^12. So Take the last and divide by 13.
3^12=9^6=81^3=(6*13+3)^3.
So 3^3=27.
27/13=2 (1 remains)

B. 1

pintukr · Answer

If n is the number of multiples of 3 between 3^13 and 3^10, not inclusive, what is the remainder when n is divided by 13?

A. 0
B. 1
C. 3
D. 10
E. 12

Number of multiples, n = (3^13 - 3^10) -1 
  = 3^10 (3^3 -1) -1 
  = 3^10 (26) -1 
  = 3^10 (2)(13) -1

thus , n will leave the remainder of (-1), when divided by 13

or, n will leave the remainder of 13 + (-1), when divided by 13 
  or, n will leave the remainder of 12, when divided by 13

(E) is the CORRECT answer

Lizaza · Answer

We're dealing with an arithmetic sequence here, with a difference of 3. The number of items in it is: 
  \frac{a_n-a_1}{3}=n-1

Therefore, as neither the first nor the last member is included, we need to subtract yet another 1 from this value. So: 
  \frac{3^{13}-3^{10}}{3}=\frac{3^{10}(3^3-1)}{3}=3^9(27-1)=3^9*26  
 From that, we're subtracting 1 more, and the number of values is  n=3^9*26-1

As we see,   3^9*26   is definitely divisible by 13. Subtracting 1, we get the remainder of 12. 
 Therefore,    the answer is E.

sequimaxime · Answer

This is a good quality question
So First lets find out number of multiples of 3 
This can be done by using \frac{(Highest Multiple of 3 - Lowest Multiple of 3) 3 and we need to substract 1 since we not including upper and lower bounds
 (3^13 - 3^10)}3 -1 
Which eventually lead us to 
 
(26*3^9 -1) 13 Hence -1/13 the remainder will be +12

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