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Guys can anyone explain how to approach such questions If n is a positive integer, what is the remainder when 3^(8n+3) + 2 is divided by 5?

A. 0 B. 1 C. 2 D. 3 E. 4

3^3 +2 / 5 = 4 . Is this approach correct?

4 is indeed the correct answer.

The remainder of 3^1 when divided by 5 is 3 The remainder of 3^2 when divided by 5 is 4 The remainder of 3^3 when divided by 5 is 2 The remainder of 3^4 when divided by 5 is 1

When you reach "1" it means that the period you described is going to repeat over and over.

Therefore:

The remainder of 3^(4k+1) when divided by 5 is 3, with k any non-negative integer The remainder of 3^(4k+2) when divided by 5 is 4, with k any non-negative integer The remainder of 3^(4k+3) when divided by 5 is 2, with k any non-negative integer The remainder of 3^(4k+4) when divided by 5 is 1, with k any non-negative integer

Here: 3^(8n+3) = 3^(4*(2n)+3) so the remainder when divided by 5 is 2

Add 2 to this number and the remainder when divided by 5 becomes 4

81 ^ N will always have a 1 in the units place. so when multiplied by 3, will have 3 in the units place which so when multiplied by 9, will have 7 in the units place

adding two will give a 9 in the units place.

any integer with a 9 in the units place, when divided by 5 will give a remainder of 4

Consider the unit digit of the index of 3, 3^1 is 3 3^2 is 9 3^3 is 7 3^4 is 1

and then repeating 3, 9, 7, 1

So the unit digit of 3^(8n+3) is 7

So the unit digit of 3^(8n+3) +2 is 9

Which gives the reminder of 4 when divided by 5

(E)

Wrong approach:

Here's the way to proove it

4^24/7 find the remainder?

Now as per your approach the answer should be 6. But the actual answer is 1.

Say 4^1/7 = 4 4^2 = 16 4^3= 64

Thus 4^24 can be written as (4^3).(4^3).(4^3)...8 times, Every power of 4^3/7 will give a remainder of 1. 1.1.1.1...8 times divided by 7 will give a remainder of 1.

81 ^ N will always have a 1 in the units place. so when multiplied by 3, will have 3 in the units place which so when multiplied by 9, will have 7 in the units place

adding two will give a 9 in the units place.

any integer with a 9 in the units place, when divided by 5 will give a remainder of 4

Hence D

OA?

Wrong approach:

Here's the way to proove it

4^24/7 find the remainder?

Now as per your approach the answer should be 6. But the actual answer is 1.

Say 4^1/7 = 4 4^2 = 16 4^3= 64

Thus 4^24 can be written as (4^3).(4^3).(4^3)...8 times, Every power of 4^3/7 will give a remainder of 1. 1.1.1.1...8 times divided by 7 will give a remainder of 1.

Wrong approach: Here's the way to proove it 4^24/7 find the remainder?

Now as per your approach the answer should be 6. But the actual answer is 1. Say 4^1/7 = 4 4^2 = 16 4^3= 64

Thus 4^24 can be written as (4^3).(4^3).(4^3)...8 times, Every power of 4^3/7 will give a remainder of 1. 1.1.1.1...8 times divided by 7 will give a remainder of 1.

thanx, i think i see the mistake

could you please state the remainder theorem? cheers

Guys can anyone explain how to approach such questions If n is a positive integer, what is the remainder when 3^(8n+3) + 2 is divided by 5?

A. 0 B. 1 C. 2 D. 3 E. 4

3^3 +2 / 5 = 4 . Is this approach correct?

The way I would tackle this Q is following

If you see 3^n then unit digit gets repeated after every four numbers

3, 9 , 27, 81, 243

Wrong approach:

Here's the way to proove it

4^24/7 find the remainder?

Now as per your approach the answer should be 6. But the actual answer is 1.

Say 4^1/7 = 4 4^2 = 16 4^3= 64

Thus 4^24 can be written as (4^3).(4^3).(4^3)...8 times, Every power of 4^3/7 will give a remainder of 1. 1.1.1.1...8 times divided by 7 will give a remainder of 1.

Therefore 3 ^ (8n + 3) means last digit is 7 which when added by 2 becomes 9. And remainder of any number with unit digit 9 is 4

Well hold on ... where did I say that 4^n gets repeated ?

Each specific number has its own properties e.g. if number has to be divisible by 3 then total of digits should be divisible by 3. This does not mean that if total of digits is divisible by 4 then it is divisible by 4 as well. 4 has its own properties.

I cud not understand your approach. Kindly solve 4^24/7 for the remainder from your approach and post. Probably you have an easier and shorter way.

Rahul you are trying to compare apples with oranges

When divisor is 5 no matter what is tens digit the units digit always determine the remainder. In case of 7 this does not hold true so approach will be different.

Each number's specific properties regarding their squares, remainders etc are applied after seeing a particular Q and is only used to speed up that particular Q and should not be used to solve every Q of that nature.

Take a look at example below

If Q is whether 249678 is divisible by 11 then I would find sum of each even digit and compare with sum of each odd digit. If sum is equal that means number is divisible by 11. Now if Q is whether same number is divisible by 9 then I will sum each digit and see whether sum is divisible by 9 or not. So different approaches allow me to get the solution faster.

A consistent approach will be to divide the number and see which will give you correct answers for all the similar Qs.