Aptitude - Numbers - Discussion

Problem:(7^95-3^58).

pb in the form of (a^n-x^n)
So, (a^n-x^n) divided by (a-x)
From the pb a=7, x=3 (a-x=7-3=4).
So the answer is 4.

Use the concept of cyclicity. It will be very much easy for you!

7^95 = we will see 7^5 (because 5 is unit digit of power) and by the cyclic method, we know, after every 4 times the unit digit repeats itself. Then the unit digit is 7.

And vice versa for 3^58 here also use cyclic method,

The unit digit will come 1.

The as per question:- 7-1 =6 (ans).

= 7(90-5)-3(60-2)
= 7(90)-7(5)-3(60)-3(2)
= 7(9-5)-3(6-2)
= 7(4)-3(4)
= 4(4).
Then, the answer is 4.

It's simple, 7 raise to some power will give us four results at the unit place, that is 1, 3, 7, 9. Same with 3 raise to some power will give us four results at a unit place, that is 1, 3, 7, 9.

From the question 7^95 we get 95/4 (taking 4 as the number of unit places we get) which gives us the remainder 3. Using the same principle 3^58 we get 58/4 which gives us the remainder 2.

Using simplification 7^3-3^2= 343-9= 334.

So, the answer will be 4.

It's simple.

Just take the power cycle of 7 & 3.
A power cycle of 7 gives (7^4=1).
And power cycle of 3 gives (3^4 =1).
Apply the Divisibility rule of 4 (if the last 2 digits of a number are divided by 4 then the whole number is divided by 4).

Here:.

7^95 - 3^58 =?
7^95 = (7^4) *7^ (95/4) = 1*7^3=343.
3^58 = (3^4) *3^ (58/4) = 1*3^2=9.
343-9= 334.
Units Digit value is 4.

Agree, the answer is 4.