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 3 of 4.
Deepanshu said:
7 years ago
I think the answer should be 7.
Dipankar Mahanta said:
7 years ago
7^95= 95/4 = remainder 3 = 7^3 = 3 from cyclicity.
3^58= 58/4 = remainder 2 = 3^2 =9 from cyclicity.
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.
3^58= 58/4 = remainder 2 = 3^2 =9 from cyclicity.
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:
7 years ago
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.
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:
6 years ago
To all values of n = (x*m-a*m) is divisible by (x- a) thus 7-3 = 4 is the Ans.
Premnath said:
6 years ago
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.
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:
6 years ago
Because 6 is an actual difference but negative come to take 10 value.
i.e means 10-6 = 4 is the right answer.
i.e means 10-6 = 4 is the right answer.
(1)
Reddivari radha gayathri said:
5 years ago
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.
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.
(7)
Kunaal Aarya said:
5 years ago
I got it. Thank you all.
(2)
Saikiran said:
5 years ago
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.
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.
(7)
Deepak sharma said:
5 years ago
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.
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.
(2)
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