Aptitude - Numbers - Discussion

Remember One thing first....As we are finding the Units digit
We are concentrated on (NUMBER =0 to 9 .)

And (NUMBER)^(K)....when K/4 remainder=0 then
Units digit value = NUMBER^0


IF (NUMBER)^(K)...when K/4 leaves some remainder ...then
Units digit value= NUMBER^remainder

REMEMBER THAT REMAINDER IN THESE CASES IS BETWEEN 0 to 3..SO
THERE IS NO PROBLEM IN FINDING POWERS...


AS we are done with the things that need to be understood lets dig into the problem .

FIRST ...7^95..........95/4 remainder is 1 so units digit will
be 7^1=7

SECOND....3^58.......58/4 remainder is 0 so units value is
3^0=1

SO DIFFERENCE IS 7-1=6 WHICH IS THE ANSWER.


CORRECT ME IF I AM WRONG

To find unit digit first we should know the periodic value;

Ex:-for 2

2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 16
2^5 = 32
2^6 = 64 and so on.

Clearly observe that unit digit follows a cyclic form i.e after 4 terms it again repeating. So the period of 2 is 4.

Similarly for 3, 7, 8 also the same period 4.

For 0, 1, 5, 6 period is always the same as itself irrespective of their powers.

So to find a unit digit find the period of that number then divided the power value by the period.

7^95 = 95/4 = remainder 3.
3^58 = 58/4 = remainder 2.

So, 7^3 = 343 & 3^2 = 9.

Since we have (7^95-3^58),
Therefore, 343-9 = 334.
Unit digit is = 4.

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.

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.

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).

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.

Hi guys,

Here the concept is ultimately to make 1 as base so that 1 power anything will be one only.

7^95 = (7^2)^47 X 7 = 9^47 X 7 = ((9^2)23) X 7 X 9 = 1^23 X 63 = 63 (Remember here I am only considering nit digits).

Same as calculate for 3^58 it will be 9, so the answer is 63-9 = 54 = 4 (Unit digit).

Or the simplest way would be like this:

As the power of 7 is odd. Do an odd power multiplication.

Like 7^3 and the power of 3 is even. So do an even power multiplication.

Like 3^2 subtract them and then you will get the units digit.

For example: 7^3-3^2=334. Here units digit is 4. So its that easy.