Latest TCS Aptitude Question SOLUTION: (32^32^32)/9 will leave a rimender????? 1- 4 2- 7 3- 1 4- 2

• Placement Papers
• /
• TCS
• /
• Numerical Ability
• /
• Number System

TCS Question Papers: TCS Interview Experience(90) TCS Company Profile

(#M40088359) TCS Numerical Ability question Number System Keep an EYE

(32^32^32)/9 will leave a rimender?????
1- 4
2- 7
3- 1
4- 2

Asked In TCS Pranav Bharti (9 years ago)

Solved Anku Srivastav Read Solution (10)

Is this Puzzle helpful?   (11)   (4) Submit Your Solution Number System

Read Solution (10) :

Anku Srivastav  (9 years ago) Best Solution

32 = 2⁵
(2⁵)³² = 2¹⁶⁰
(2¹⁶⁰)³² = 2⁵¹²⁰
You are looking for the remainder of 2⁵¹²⁰ ÷ 9.

2¹ mod 9 = 2
2² mod 9 = 4
2³ mod 9 = 8
2⁴ mod 9 = 7
2⁵ mod 9 = 5
2⁶ mod 9 = 1
2⁷ mod 9 = 2
2⁸ mod 9 = 4
2⁹ mod 9 = 8
2¹⁰ mod 9 = 7
2¹¹ mod 9 = 5
2¹² mod 9 = 1
...
Note that the pattern repeats, and when the exponent is divisible by 6, the remainder is 1.
5120 = 6(853) + 2, therefore
2⁵¹²⁰ mod 9 = 4

Is this solution Helpfull? Yes (35) | No   

karthikeyank  (9 years ago)

ANS IS : 4
=(32^32^32)/9
=(2^5)^32
=(2^160)^32
=(2^5120)/9
2¹ mod 9 = 2
2² mod 9 = 4
2³ mod 9 = 8
2⁴ mod 9 = 7
2⁵ mod 9 = 5
2⁶ mod 9 = 1
2⁷ mod 9 = 2
2⁸ mod 9 = 4
2⁹ mod 9 = 8
2¹⁰ mod 9 = 7
2¹¹ mod 9 = 5
2¹² mod 9 = 1
LOOKING FOR CYCLIC REMAINDER THEOREM
5210/9=REMAINDER IS 8
8TH POSITION VALUE IS 4

Is this solution Helpfull? Yes (12) | No   

ANNU KUMARI  (9 years ago)

32^32^32=(2^5)^32^32=2^5120.
2^5120/9
=((2^3)^1706*2^2)/9
=((9-1)^1706*2^2)/9
=(-1)^1706*4
=4

Is this solution Helpfull? Yes (9) | No   

PRITHA GHOSH  (8 years ago)

the last 2 digits of 2^10 will always be 76.
so the last 2 digits of 2^5120=76.
76/9 = remainder is 4

Is this solution Helpfull? Yes (1) | No   

sree lekha kadiyam  (9 years ago)

please answer to this question??

Is this solution Helpfull? Yes | No (1)  

monika samal  (9 years ago)

answer will be 4

Is this solution Helpfull? Yes | No (3)  

shuvodeep roy  (9 years ago)

ans will be 4

Is this solution Helpfull? Yes | No (2)  

Priya Pandey  (9 years ago)

32/9=5(rem)
5^32^32/9

5/9=5
25/9=7
7*5 / 9=8

5^6 / 9 =1 rem= 1

then 32^32/6= 2^32/6
2/6=2
4/6=4
8/6=2
16/6=4
odd even funda
4th term of cycle that is 5^4/9=4(rem) answaer

Is this solution Helpfull? Yes | No   

kaustubh salunke  (9 years ago)

the remainder will be 4.hence option 1 is the ans.

Is this solution Helpfull? Yes | No (2)  

PURBASHA ROY  (8 years ago)

The Euler no. of 9 is =9 * 2/3 =6
now 32/6 remainder=2 and 32/9 remainder is=5
then (5^2)^32=25^32.
again using Euler no. 32/6 leaves remainder 2.
Then,
(5^2)^2/9 =625/9 leaves remainder 4.

Is this solution Helpfull? Yes | No   

Please Login to Read Solution.