Aptitude - Numbers - Discussion

Discussion Forum : Numbers - General Questions (Q.No. 10)
10.
What is the unit digit in {(6374)1793 x (625)317 x (341491)}?
2
3
5
Answer: Option
Explanation:

Unit digit in (6374)1793 = Unit digit in (4)1793

    = Unit digit in [(42)896 x 4]

    = Unit digit in (6 x 4) = 4

Unit digit in (625)317 = Unit digit in (5)317 = 5

Unit digit in (341)491 = Unit digit in (1)491 = 1

Required digit = Unit digit in (4 x 5 x 1) = 0.

Discussion:
123 comments Page 6 of 13.

Vivek said:   10 years ago
Follow this Rule.

Unit Digit of Base, Powers, Unit digit of product.

0, 1, 5, 6, any powers, same digit.

3, 7, 9, power exactly divisible by 4, 1.

2, 4, 8 " 6.

So in this problem: {(6374)^1793x(625)^317x(341)^491)}.

= (6374)^1793 unit digit is 4.

Apply the above rule: Divide 1793 by 4, we get remainder as 1.

Take the remainder to the power of unit digit's i.e. 4^1.

So we get (6374)^1793 = 4. For (625) ^317 same digit as the unit digit.

So we get (625)^317 = 5. For (341)^(491) same digit as the unit digit.

So we get (341)^(491) = 1.

Finally (4*5*1) = 20.

Thus unit digit is 0 = Answer.

Bipin said:   10 years ago
(6374)^1793.

= Here unit digit is 4 then we can write (4)^1793.

As we know that multiple of 4 will be = 4, 16, 64, 256 = Here unit digit is 4 and 6 continuing coming then we have to take 2 nos. which should come 4 and 6 that's why we took.

= (4)^1793.

= 4*(4)^1792 (we can write it).

= 4*(4^2)^896.

Take only unit digit = 4*16 = 4*6 = 24 = 4.

Same as:

Unit digit in (625) 317 = Unit digit in (5) 317 = 5.

Unit digit in (341) 491 = Unit digit in (1) 491 = 1.

Answer:

Required digit = Unit digit in (4 x 5 x 1) = 0.

Neelam said:   10 years ago
4^2 = 16, 4^3 = 64. Unit digit = 4, when power is 3 (odd).

Unit digit of any number having odd power remains same as the original number has;here 6374^1793. Power is odd so unit digit will be 4, similarly 625^317 & 341^491 unit digits are 5 and 1 respectively, since power is odd.

So the unit digit of overall expression = (4.5.1) = 20 i.e. 0.

Anu said:   10 years ago
Find unit digit of 1^1+2^2+3^3+4^4+5^5+6^6?

Nanditha said:   10 years ago
Answer for 2463^323 may be after considering the cyclicity of 3 i.e., 4 we may get 3 as the answer.

Venugopal said:   1 decade ago
Thanks. I understood the answer. But when the end no is 3. E.g. (2463) power 323.

Then how to solve explain this?

Prasanna Kartik said:   1 decade ago
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.

Dr. pratik sharma said:   1 decade ago
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.

Dinesh sen said:   1 decade ago
7 is multiplied by 7 107 times, unit place number is?

Varun said:   1 decade ago
What will be the remainder if 2^856 is divided by 9?

Can any one mail me shortest trick?


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