Aptitude - Numbers - Discussion
Discussion Forum : Numbers - General Questions (Q.No. 90)
90.
What is the unit digit in(795 - 358)?
Answer: Option
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.
Discussion:
36 comments Page 2 of 4.
Cradlerian said:
10 years ago
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.
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:
10 years ago
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.
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:
9 years ago
@Cradlerian.
Thanks for your solution.
Thanks for your solution.
Madhuri said:
9 years ago
(7^95 - 3^58) = 7 - 3 = 4.
Bidyut ghosh said:
9 years ago
Thank you for explaining the solution.
J GOPINATH said:
9 years ago
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.
(1)
Rajat Kumar said:
8 years ago
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].
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:
8 years ago
I'm not understanding it, Someone please help me to get it.
Arsalan said:
8 years ago
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.
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:
8 years ago
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.
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.
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