Latest TCS Aptitude Question SOLUTION: (6^1001+7^1001)/17 find the reminder

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(#M40033577) TCS Numerical Ability question apti-reminder Keep an EYE

(6^1001+7^1001)/17
find the reminder

Asked In TCS Raja Bhowmick (10 years ago)

Unsolved Read Solution (8)

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Read Solution (8) :

RAKESH  (10 years ago)

6^1001/17
=(6^2)^500 *6/17
= (17*2+2)^500 *6/17
= 2^500 *6/17
= (2^4)^125 *6/17
= (17-1)^125 *6/17
= -6/17

7^1001/17
=(7^2)^500 *7/17
=(17*3-2)^500 *7/17
= (-2)^500 *7/17
= (2^4)^125 *7/17
= (17-1)^125 *7/17
= -7/17

so, (6^1001+7^1001)/17 => rem(-6/17 + (-7/17))=> rem(-13/17) => rem(4/17)

remainder is 4

Is this solution Helpfull? Yes (25) | No (1)  

Debankan  (10 years ago)

6*6=36 36/17 remainder 2
7*7 =49 49/17 remainder 15

(6^2)^500 * 6 + (7^2)^500 * 7

remainder 2^500 * 6 + 15^500 *7
2^5=32 32/17=15 15^2=225 225/17=4 both remainder
(2^5)^100 *6 + (15^2)^250 *7
15^100 *6 + 4^250 *7
4^50 * 6 + 13^81 *28
13^16 *96 + 16^40 *364
16^8 *96 + 1^20 *364
1^4 *96 +1^20 *364
96+364=460
460/17=27 remainder is 1.

Is this solution Helpfull? Yes (6) | No (4)  

Yogesh Sharma  (10 years ago)

6^1001 + 7^1001
(6^2)^500*6 + (-10)^1001
(2^4)^125*6 + 100^500*(-10)

(-1)^125*6 + (-2)^500*(-10)
-6 + (-2^4)^125*(-10)
-6 + (-1)(-10)
-6 + 10
4 remainder = answer.

Is this solution Helpfull? Yes (3) | No   

jaat  (10 years ago)

according to-->
6/7 has remainder 6
36/7 has remainder 1
216/7 has remainder 6
1296/7 has remainder 1

We see that 6 to an odd power divided by 7 has remainder 6
and 6 to an even power divided by 7 has remainder 1

so 6^1001 mod 7 being an odd power will have a remainder of 6.

We could say that 6 in mod 7 is -1
so -1^1001 mod7 = -1 mod 7....and since 7-1=6 the answer is 6.

now let go to our example

(6^1001+7^1001)/7
= 6^1001/7=6 and 7^1001/7=0
=6+0=1 ans

Is this solution Helpfull? Yes (1) | No (14)  

Arnab Dutta  (10 years ago)

4 remainder will the answer...

Is this solution Helpfull? Yes (1) | No   

Hardik Gujarati  (8 years ago)

6^1+7^1=13/17 so remainder 4

Is this solution Helpfull? Yes (1) | No   

sanju  (10 years ago)

can anyone expln in shrtct mthd plzzzzzzz

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Renuka Parmar  (4 years ago)

remainder 1

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