Aptitude - Numbers - Discussion

Because 6 is an actual difference but negative come to take 10 value.

i.e means 10-6 = 4 is the right answer.

How come we take,
7^95 = 7^92+7^3?

What happens if we take, 7^95 = 7^93 + 7^2, How to solve this? Please tell me.

To all values of n = (x*m-a*m) is divisible by (x- a) thus 7-3 = 4 is the Ans.

Step1. Find the period of 7=7^0=1.
7^1=7.
7^2=9.
7^3=3.
So the period of 7 is 4.

Divide the power of 7 is 95 by 4 and find the remainder.

So the remainder is 3.so the 3rd unit digit of the period is 3.
step2: Similarly find the period of 3 and find the remainder. this remainder is 2nd, so the 2nd unit digit of the 3 periods is =9.

Step3: The Final ans=(3-9)=4 hints:3 is small as compare 9. So borrow 1 from left digit.

7^95= 95/4 = remainder 3 = 7^3 = 3 from cyclicity.

3^58= 58/4 = remainder 2 = 3^2 =9 from cyclicity.

NOW, the answer CANNOT be a -ve no, so in order to subtract 9 from 3, we borrow to make it 13 such that= 13-9 = 4 which is a unit digit itself.

I think the answer should be 7.

Make the series of number.

Like in 7^95,
The series will be like 7^1=7,
7^2=49 unit digit is 9,
9*7=63 unit digit is 3,
3*7=21 unit digit is 1,

Now, we get the repeated series as 7,9,3,1,7,9,3,1 in every fifth place we get 7 as power is in term of 5 so required unit digit is 7.

For 3^58.
Series of 3 is 3,9,7,1,3,9,7,1,3
58 term will be 3.
7-3 = 4.

Hi,

If number^K and k is zero then we take the last value of cycle that is in case of 7 the cycle is of 4 7^1 is 7.

7^2 is 49, 7^3 is 343 and 7^4 ends with 1 and 7^5 ends with 7 so we see the cycle starts again after 4 steps this 7 has a cyclicity of 4 and now if we have number^k and k is divisible by 7 we take 7^4 for unit digit.

I'm not understanding it, Someone please help me to get it.

The unit digit of 7^95 is 3 and that of 3^58 is 9.

Let 7^95= ......3
Let 3^58= ......9

Subtracting,
13-9= 4 [Since we carried 1 for 3 from its left-hand side].