Aptitude - Numbers - Discussion

Exp :
7^0 = 1 unit digit = 1.
7^1 = 7 unit digit = 7.
7^2 = 49 unit digit = 9.
7^3 = 343 unit digit = 3.
7^4 = 2401 unit digit = 1.
7^5 = 16807 unit digit = 7.
7^6 = 117649. unit digit = 9.

Look at the unit digit of each exponent then u gonna find the pattern repeats after 4 times (1,7,9,3,then 1,7,9,3 then 1,7,9,3).

Instead of counting until 105 we divide by 4 and the remainder is the new exponent for 7.
in this case 105/4 = 26 1/4 the remainder is 1 so 7^1=7.

If 7^106 106/4=26 2/4 the remainder is 2 then 7^2 = 49 the unit digit is 9 and for sure 7^107 unit digit going to be 3 and 7^108 is 1 and so on.

It's all about cyclicity.

Cyclicity: no of repetitions occurring in a multiple.

Exp :
7^1 = 7
7^2 = 49
7^3 = 343
7^4 = 2401
7^5 = 16807
7^6 = 117649.

So the cyclicity is 4 (no of patterns repeating).

So we must divide exponent by 4.

105/4 we get remainder 1.

Taking remainder 1 to cyclicity 7^1 = 7.

If the remainder is 2 then the answer is 7^2 = 9.

If the remainder is 3 then the answer is 7^3 = 3, and so on.

If we take nth power of a no having unit digit 1 then a no is obtained having unit digit 1. As unit digit of 7 power 4 is 1 so 26th power of a number having unit digit 1 is 1. i.e. (7^4)^26=a no having unit digit 1.

7^105.

Firstly, divide the given power number by 4 (105/4=remainder is 1), then remove the given power and add remainder 1 as power (7^1), now 7 to the power 1 is 7. So, the answer is "7".

if we calculate 7 power 4, we will get unit digit is 1. so 1(unit digit) power anything is there, we will get 1 in unit digit

I too tried the same method @Shadab.

Is the method right or wrong, if yes then why? please explain it.

I don't understand cyclicity.

Why can't we do 7^2^(52) * 7?

By this way we get unit digit 3.

Why we are taking 7 power 4?

My doubt is why we are not taking another number like 7 power 3?

please tell me how to know that unit digit in 7 power 4 whole power 26 is 1..?

What is the unit digit? How can we find?

Please explain with example.