Discussion :: Numbers - General Questions (Q.No.90)
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90.
What is the unit digit in(795 - 358)?
Answer: Option B
Explanation:
Unit digit in 795 = Unit digit in [(74)23 x 73] = Unit digit in [(Unit digit in(2401))23 x (343)] = Unit digit in (123 x 343) = Unit digit in (343)
= 3
Unit digit in 358 = Unit digit in [(34)14 x 32] = Unit digit in [Unit digit in (81)14 x 32] = Unit digit in [(1)14 x 32] = Unit digit in (1 x 9) = Unit digit in (9)
= 9
Unit digit in (795 - 358) = Unit digit in (343 - 9) = Unit digit in (334) = 4.
So, Option B is the answer.
Ravi said: (Jun 9, 2010)
How do define unitdigit. Explain?
Vamshi said: (Jun 26, 2010)
Unitdigit in 7^95 is 3 and unit digit in 3^58 is 9 So the answer shoudld be 6.
Then how come 4 is the answer, I didnt get.. please explain
Shahid said: (May 20, 2011)
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
Shahid said: (May 20, 2011)
EVERYTHING THAT I EXPLAINED WAS RIGHT ..BUT I UN NOTICINGLY DIVIDED 95/4 whose remainder is 3 and 58/4 remainder is 2 so 7^3=343 and 3^2=9 and the difference is 343-9 where there 334.
So 4 is the answer.
Kala said: (Aug 6, 2011)
7^3=343 and 3^2=9
difference 343-9
ans 4
How? Please explain every step.
Lara said: (Apr 17, 2013)
Idea is simply to express the given number, in such a way that unit's place should be 1. i.e., 7^4 is 2401 and 3^4 is 81.
Kunvar said: (May 27, 2013)
I want to correct you. We are not finding the absolute value of the question. 3-9 = 6 is not the answer as it is -6 and you people don't know the concept of negative remainders and modulo. So the answer 4 is absolutely right.
Prasanna Karthik said: (Jul 8, 2015)
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).
Naveen said: (Jul 15, 2015)
Unit digit should not show in negative digits. So actually total digit in 7^95 is 343.43-9=34.
So 4 is the correct answer.
Jot said: (Jul 22, 2015)
My answer is 6. But how is it 4? Please elaborate clearly.
Cradlerian said: (Aug 26, 2015)
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.
Zahin said: (Nov 13, 2015)
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.
Ali N said: (Jul 10, 2016)
@Cradlerian.
Thanks for your solution.
Madhuri said: (Sep 5, 2016)
(7^95 - 3^58) = 7 - 3 = 4.
Bidyut Ghosh said: (Nov 22, 2016)
Thank you for explaining the solution.
J GOPINATH said: (Jan 27, 2017)
Here as per the logic, it's if subtracted it will be 6 but with the negative symbol so you have borrow 10 from the previous place i. e, ten's place which makes the 3 to be 13 and hence 13-9=4 which the correct answer choice.
Rajat Kumar said: (Apr 23, 2017)
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].
LINGANAYAKA said: (Apr 29, 2017)
I'm not understanding it, Someone please help me to get it.
Arsalan said: (Jun 6, 2017)
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.
Tabassum said: (Jun 15, 2017)
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.
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.
Pranasish said: (Sep 3, 2018)
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.
Snehasish said: (Feb 14, 2019)
To all values of n = (x*m-a*m) is divisible by (x- a) thus 7-3 = 4 is the Ans.
Premnath said: (Mar 4, 2019)
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.
Ranga said: (Jul 29, 2019)
Because 6 is an actual difference but negative come to take 10 value.
i.e means 10-6 = 4 is the right answer.
Reddivari Radha Gayathri said: (Jun 30, 2020)
By using cyclicity: 95/4= rem 3.
58/4= rem 2. So, replace the power with the remainder
= 7^3-3^2.
= 343-9.
= 334.
They asked unit digit so we have to consider the unit digit the unit digit is 4.
So, the Answer is 4.
Kunaal Aarya said: (Oct 21, 2020)
I got it. Thank you all.
Saikiran said: (Dec 2, 2020)
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.
Deepak Sharma said: (Jan 17, 2021)
7^4=2401 here the unit place is 1 similarly,
3^4=81 here also the unit place is 1. So we do; (7^4)^23 * 7^3=(2401)^23 * 343.
=1^23 x 343 = 1 x 343 = 343. (3^4)^14 x 3^2=(81)^14 x 9
=1 x 9= 9.
(343-9) = 334 here unit place is 4. So 4 is the correct answer.
NK Liya said: (Apr 17, 2021)
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.
Abhishek said: (Jul 22, 2021)
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.
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.
