The official explanation is a hoax- Bogus in the sense that it is correct but is too short

Here a better understanding

1. Units digit of $\(7^{95}\)$ :

The units digits of powers of 7 follow a cycle:
- $\(7^1=7\)$
- $\(7^2=49 \Longrightarrow 9\)$
- $\(7^3=343 \Longrightarrow 3\)$
- $\(7^4=2401 \Longrightarrow 1\)$
- $\(7^5=16807 \Longrightarrow 7\)$

The cycle is ( $\(7,9,3,1\)$ ), which has a length of 4.
To find the units digit of $\(7^{95}\)$, we divide the exponent 95 by the cycle length 4 : $\(95 \div 4=23\)$ with a remainder of 3 .
The units digit of $\(7^{95}\)$ is the same as the 3rd digit in the cycle, which is 3 .

2. Units digit of $\(3^{58}\)$ :

The units digits of powers of 3 follow a cycle:
- $\(3^1=3\)$
- $\(3^2=9\)$
- $\(3^3=27 \Longrightarrow 7\)$
- $\(3^4=81 \Longrightarrow 1\)$
- $\(3^5=243 \Longrightarrow 3\)$

The cycle is $(\(3,9,7,1)\)$, which has a length of 4 .

To find the units digit of $\(3^{58}\)$, we divide the exponent 58 by the cycle length 4 :
$\(58 \div 4=14\)$ with a remainder of 2 .
The units digit of $\(3^{58}\)$ is the same as the 2 nd digit in the cycle, which is 9 .

3. Units digit of $\(7^{95}-3^{58}\)$ :



Therefore, the units digit of $\(7^{95}-3^{58}$ is $\mathbf{4}\)$.
Compare Quantity A and Quantity B:
Quantity A: 4
Quantity B: 4

Therefore, Quantity A is equal to Quantity B.
The final answer is The two quantities are equal.

I hope this really helps you out

Carcass this is my explanation using the concepts explained by brushmyquant from here https://gre.myprepclub.com/forum/how-to ... 34657.html

7^95 - 3^58

95/4 gives 3 as reminder and 7^3 gives 43 for the tenths and units digit

3^58 = ((3^4)^14) * (3^2) = 81^14 * 9 = 21 * 9 = 89 which is the tenths and units digits for 3^58

43 - 89 = -46



6 is greater than 4 so it should be option A


am i missing something?

Hi

we are not dealing with tenth and unit digits but units digit only.

My explanation above is quite extensive , please read sir

You need to find the cycle of seven. The cycle of 3

And because the



it is pretty simple. and whenever you need to find the unit digit of a \( N^n\) is pretty similar on the GRE

You should read this post by brushmyquant https://gre.myprepclub.com/forum/how-to ... ml#p113694

The one you suggested is a bit different what is tested in the question

All this theory is in our Handout for quant sir also https://gre.myprepclub.com/forum/gre-ma ... 29264.html

We did a lot of efforts to provide to stedunts ALL the theory for the GRE

Carcass
i got it now, i missed a crucial detail
thanks for explanation