Aptitude - Numbers - Discussion

@Abhigyan reddy.

The units digit can be 1, 3, 7 or 9.

Power are not equals then what we do.

Hi I have few questions.

1) In the above problem we have taken the unit digits of every number regardless of of the power.

2) If any such type questions come can we use the same method to solve the problem. Please answer.

Given question is.

{(6374)^1793 x (625)^317 x (341)^491}.

Ans:

NOTE: Unit digit for (4)^odd num is 4.

Unit digit for (4)^even num is 6.
Unit digit for (5)^any num is 5.
Unit digit for (1)^any num is 1.

Now apply the above conditions,

Unit digit of 6374 is 4.
Unit digit of 625 is 5.
Unit digit of 341 is 1.

Just multiply 4*5*1 = 20.

So unit digit of 20 is 0.

Answer is 0.

What will be the remainder if 2^856 is divided by 9?

Can any one mail me shortest trick?

7 is multiplied by 7 107 times, unit place number is?

7*7^107.
i.e 7^108.

We know 7^1 unit place will be 7.

7^2 unit place will be 9.
7^3 unit place will be 3.
7^4 unit place will be 1.

So unit place will be 7, 9, 3, 1 after these 7, 9, 3, 1 again and again.

We have 108 that is divisible by 4 so unit place will be one. You can find it on scientific calculator as well.

Hi guys,

These type of problems are solved by cyclicity rule.

Here this is the problem.

6374)^1793x(625)^317x(341)^491.

First term's unit digit is 4 cylicity for 4 is 4 (4^1) , 6 (4^2) , 4 (4^3) , 6 (4^4). This cycle continues in the same pattern. (Remember I only consider unit digits). So cyclicity is 2.

Now check first term is power 1793 is divisible by 2 or not. Clearly it shows that it will leave reminder 1. Which means unit digit is 4 for first term.

Second term unit digit 5, and last term is 1 so the answer is 4X5X1=20. Unit digit is 0. Note: Make a table for cylicity for numbers 1 to 9. Then it will become very handy.

Thanks. I understood the answer. But when the end no is 3. E.g. (2463) power 323.

Then how to solve explain this?

Answer for 2463^323 may be after considering the cyclicity of 3 i.e., 4 we may get 3 as the answer.