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

So numbers has to be from 3^10 to 3^13 not inclusive.

So total number of numbers 3^10 + 3 ...............3^13-3.

Number of such multiples is \((3^13 - 3^10)/3+1\)

3^12-1 - 3^9 -1+ 1 => \(3^9(26) -1\)

26k-1 format. So when 13 is dividing this number it will leave 12 as the remainder. (-1mod 13 = 12)

IMO E

Is the answer choice A?

We need to find are the number of multiples of 3 between 3^10 < 3k < 3^13, which simplifies to 3^9 < k < 3^12.

Since it's exclusive, the number of multiples between them will be (3^12 -1) -(3^9 + 1) +1 = 3^12 - 3^9 -1 = 3^9 (3^3 -1) -1 = 3^9(26) -1

Now this last number (3^9(26) -1) divided by 13 leaves a remainder based on

3^9 last digit is, given 3^3 = 27 and 3^9 =(3^3)^3 then it will have a last digit of 3
26/12 leaves a remainder of 0
Then the remainder will be, given the last digits, 3*0 -1 = -1,

Remainder will be 12

The number of multiples of 3 between 3^10 and 3^13 exclusive is based upon a number in the form of 3*A

Thus we need to find values given 3^9 < A < 3^12

As we know, number of total number in a series is given by last number - first + 1, in this case it'd be:

3^12 - 1 is last number, since it's exclusive
3^9 + 1 is first number, since, again, it's exclusibe

3^12 - 3^9 -1 = 3^9(3^3 -1) -1 = (3^9*26) - 1

As this is a remainder questions, we need to focus on last digits, since those will determine the final remainder when we divide by 13

So we know that 3^n follows a pattern every 4th term, given this, 3^9's last digit will be 3
26 is a multiple of 13, leaving 0 remainder

So we'd get 3*0 -1= -1 is the number to divide by 13 and find the remainder, remainder will be 12.

The number of multiples:
n=(Last multiple - First multiple)/3 +1 -2
n=(3^13-3^10)/3 -1 = 3^12-3^9 -1 = 3^9(27-1) -1=26*3^9 -1

26*3^9 -1 = (2*3^9)*13 -1 = (2*3^9 -1+1)*13 -1 = (2*3^9 -1)*13 +13-1= (2*3^9 -1)*13 +12

Remainder is 12

Answer E

Since 3^10 and 3^13 are multiples of 3, the first multiple of 3 is 3^10 + 3 and the last is 3^13 - 3.

So n =


3^13 - 3 -3^10 -3

3^13 - 3^10 - 6/3

3^12 - 3^9 - 2 + 1

n = 3^12 - 3^9 - 1

Now we need to find the remainder when n is divided by 13

3^9 (3^3 - 1)/13 - 1/13

3^9( 26)/13 - 1/13

0 - 1 = -1


Therefore -1 + 13 = 12

Answer E

Here's how I approached it

The smallest multiple of 3 greater than 3^10 is 3^10 + 3, and the largest multiple of 3 less than 3^13 is 3^13 - 3.

3^13 - 3 = (3^10 + 3) + (n-1) * 3
3^13 - 3^10 - 6 = (n-1) * 3
3^12 - 3^9 - 2 = (n-1)
3^12 - 3^9 - 1 = n

So, the total number of multiples of 3 in the range is: 3^12 - 3^9 -1

To find out the remainder of this when divided by 13:

--> (( 3^12 - 3^9) - 1)/13
--> (3^9(3^3 - 1))/13 - 1/13
--> (3^9 * 26)/13 - 1/13

Remainder for (3^9 * 26)/13 will be 0 and for 1/13 it will be 1

Final remainder is 0 - 1 = -1 or basically 12 (To convert a negative remainder to a positive one, simply add the divisor to the negative remainder)

Answer : E

Number of multiples of 3^10 and 3^13
To solve this ,we need to find connection between these two numbers using 3 multiples.
3^10 * 1,3^10 * 2 ,3^10 *3 ,..... 3^10 * 27(3^3) which is 3^13.
so excluding 3^10 and 3^13 we have .. multiples from 2....26.. (25)
So n is 25
and Re{25/13} =12.

3^13 - 3^10 = 3^10(27-1) = 3^10*26 = 3^10*13*2;

(n) Multiples of three = (3^10*13*2) / 3 = 3^9*13*2;

n/13 = (3^9*13*2)/3 = 3^9 * 2; Leaving 0 as remainder.

Since 3^10 & 3^13 are not included, so the first number would be 3^10 + 3 & last number would be 3^13 - 3
Using AP formula we get,
3^13 - 3 = (3^10 + 3) + (n-1)3
3^13 - 3^10 - 6 = (n-1)3
(3^10(3^3 - 1) - 6)/3 = (n - 1)
3^9 * 26 - 2 = n - 1
3^9 * 26 - 1 = n

We know that 26 is divisible by 13, but the number is one less than a multiple of 13 (we can ignore the 3^9)
Hence, the remainder will be 12

IMO E

lets take a few example:-
the number of factors between 3 to the power 1 and 3 to the power 4 that is divisible by 3 is 25
When 25 is divisible by 13 it leaves a remainder of 12
Similarly when we divide the no of factors between 3 to the power 2 and 3 to the power 5 by 13 we get the remainder 12
Following the same pattern we will get remainder of 12

Solve this step by step:

1. Between 3^10 and 3^13 (not including them):
   
   • 3^10 = 59,049
   • 3^13 = 1,594,323
2. Since we want multiples of 3:
   
   • The count will be (1,594,323 - 59,049 - 2)/3
   • We subtract 2 because endpoints are not included
3. Simplifying:
   
   • 1,594,323 - 59,049 = 1,535,274
   • 1,535,274 - 2 = 1,535,272
   • 1,535,272 ÷ 3 = 511,757
4. When 511,757 is divided by 13:
   
   • 511,757 = 39,366 × 13 + 12

Therefore, the remainder is 12.
Answer: E. 12

We are tasked with finding the number of multiples of strictly between and (not inclusive), and then finding the remainder of this count when divided by .


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Step 1: Find the range of multiples of 3

Every integer between 3^10 and 3^13 that is a multiple of can be written as:

3.k where k is an integer

k = 3^{10} + 1, 3^{10} + 2.......... 3^{13} - 1.


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Step 2: Total count of multiples of 3 between 3^10 and 3^13

Since 3^10 and 3^13 are both themselves multiples of , the total number of multiples strictly between them is:

Count of multiples= (3^13 - 3^10)/3 - 1.

On Simplifying:

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

Further simplify:

3^3 - 1 = 27 - 1 = 26, 3^10(26)/3-1
3^9(26)-1
On dividing this value by 13 we have 26 being divisible by 13 so the remainder will be 0-1=-1 or the positive remainder would be 12.

Hence the correct answer is option (E) 12

Hi All,

the no of multiple of 3 we need tofind out is in the range 3 ^12 -3 ^9 -1

3 ^9(3 ^3 -1) /13

3 ^9(26)/13

as 26 is divisible by 12,

3^9/13

(27)^3/13

therefore -1 /13 hence remainder is 12

n=3^13-3^10-1

3^3=1 mod13 , 3^9=1 mod13 , 3^12=1 mod13
3^13=3 mod13, 3^10=3 mod13
therefore by substituting,
n mod13=(3-3-1) mod13 =-1 mod13= 12

Please find the attached solution
my answer is E

My approach :

For n,
3^13=3^10+(n-1)3

Since both the values are not inclusive,

[3^10(3^3-1)]/3=n-1

Dividing Both sides by 13 gives remainder 0.

My ans is A