For a positive integer n, what is the remainder when (13^4n+3)(9^2n)

Question

For a positive integer n, what is the remainder when  (13^{4n+3})(9^{2n})  is divided by 10?

A. 1
B. 3
C. 5
D. 7
E. 9

GMATinsight · Answer

CONCEPT: Remainder when N is divided by 10 is the same as the Unit digit of N Now, unit digit of (13^4n+3)(9^2n) = (3^3)*(9^2n) = 7*1 = 7 CONCEPT: Cyclicity of Unit digit of 3 is 4 i.e. Unit digit repeats after 4th power hence Unit of 13^4n+3 = Unit of 3^4n+3 = Unit of 3^3 = 7 Cyclicity of Unit digit of 9 is 2 i.e. Unit digit repeats after 2nd power... so Unit digit of 9^2n = Unit digit of 9^2 = 1 Answer: Option D --- GMATinsight | www.GMATinsight.com GMAT FOCUS PREP Expert Youtube: www.Youtube.com/GMATinsight

siddhantvarma · Answer

The key here is to know that the remainder of a number by 10 is determined by it's unit digit. Even if you don't remember it, taking a few examples can also help establish it. 
For example, 134 % 10 = 4 (which is the units digit in 13 4 )
2566 % 10 = 6 (which is the units digit in 256 6 )

So we only need to determine the units digit of the term: (13^4n+3)(9^2n)
Let's rewrite it as (13^4n+3)(3^4n) (Since 9 = 3^2)
Let's find the pattern for 13 first:
n=1 => 13^n => 13 % 10 = 3
n=2 => 13^n => 169 % 10 = 9
n=3 => 13^n => __7 % 10 = 7
n=4 => 13^n => __1 % 10 = 1
n=5 => 13^n => __3 % 10 = 3
...

Notice that beyond 13^2, I don't need to calculate the value. I only used 169 because I remember that 13^2 is 169. I'm only interested in the units digit of 13^n so I just calculated what the units digit of 13^3 and and 13^4 would be. 
Units digit of 13^3 = units digit of 169 => 9 x units digit of 13 => 9 x 3 => 27 => 7. Similarly, you can get unit digits of number of the form 13^n.

Now units digit of 13^4n+3 = units digit of 13^3 => 7.

Similarly, we can find the cyclic nature of 3^4n which would tell us that units digit of 3^4n is 1.

Therefore, unit digit of (13^4n+3)(9^2n) is 7, which will be the remainder when you divide it by 10. Hence, D.

RohanNewDelhi · Answer

1. Find the units digit of 13^(4n+3)
The units digit of any power of 13 is determined entirely by its last digit, 3. Let's look at the repeating pattern (cyclicity) of the powers of 3:

3^1 =  3   
  3^2 =  9   
  3^3 = 2 7   
  3^4 = 8 1   
  3^5 = 24 3  (The cycle of 3, 9, 7, 1 repeats every 4 powers)   
 The exponent in the problem is 4n + 3. This algebraically represents a multiple of 4, with a remainder of 3. Because the units digit pattern resets exactly every 4 steps, the "+ 3" tells us that the value will always land on the 3rd step of the cycle.
The 3rd number in the cycle is 7.
2. Find the units digit of 9^(2n)
Next, let's look at the cyclicity for powers of 9:

9^1 =  9   
  9^2 = 8 1   
  9^3 = 72 9  (The cycle of 9, 1 repeats every 2 powers)   
 The rule for base 9 is straightforward: odd powers end in 9, and even powers end in 1. Since n is a positive integer, the exponent 2n is guaranteed to be an even number.
Therefore, the units digit of 9^(2n) is 1.
3. Combine the parts
To find the units digit of the entire expression, simply multiply the units digits of the two individual parts together:
 7 × 1 = 7 
Since the final units digit is 7, dividing the massive resulting number by 10 will leave a remainder of 7.

For a positive integer n, what is the remainder when (13^4n+3)(9^2n)

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For a positive integer n, what is the remainder when (13^4n+3)(9^2n)