Latest TCS Aptitude Question SOLUTION: remainder of 2^47/47

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(#M40023907) TCS Numerical Ability question Keep an EYE

remainder of 2^47/47

Asked In TCS MUKESH KUMAR CHAURASIA (10 years ago)

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Ankita Agasti  (10 years ago)

ans 2
a^p-a is divisible by p when p is prime

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ASHISH  (10 years ago)

answers is= 2

Is this solution Helpfull? Yes (6) | No (3)  

Animesh  (10 years ago)

ANS WILL BE=18

Is this solution Helpfull? Yes (5) | No (13)  

rohan chawla  (10 years ago)

ans is 2
it is because 2^23 leaves remainder 1 when divided by 47
so we can express 2^23 as 47x+1
so 2^46= (47x+1)^2 = (47x)^2 + 2*47x + 1
and 2*47 = 2*2^46=2*(47x)^2 + 2*2*47x + 2
we see that first two terms are divisible by 47 but last term i.e. 2 is not ..
so the remainder is 2

Is this solution Helpfull? Yes (5) | No (2)  

Alok  (10 years ago)

mistake earlier
2^47/47 =(2^6)^7*(2^5)/47=((47+17)^7*(2^5))/47
=(17)^7*32/47=17*((17^2)^3)*32/47=17*((282+7)^3)*32/47 as 282 is divisible by 47 so
17*(7)^3*32/47=186592/47=2 hence 2

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Alok  (10 years ago)

lets start
2^47/47 =(2^6)*(2^5)/47=((47+17)^7*(2^5))/47
=(17)^7*32/47=17*((17^2)^3)*32/47=17*((282+7)^3)*32/47 as 282 is divisible by 47 so
17*(7)^3*32/47=186592/47=2 hence 2 is and uph had to do a lot :) thanks bfr

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Priyajit Biswas  (10 years ago)

answers is= 24

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Thomas George  (9 years ago)

2mod47=2
2^2mod47=4
2^4mod47=(2^2)*(2^2)mod47=4*4mod47=16
2^8mod47=(2^4)*(2^4)mod47=16*16mod47=21
2^16mod47=(2^8)*(2^8)mod47=21*21mod47=18
2^32mod47=(2^16)*(2^16)mod47=18*18mod47=42
2^47mod47=(2^32)*(2^8)*(2^4)*(2^2)*2mod47=42*21*16*4*2mod47=2
ANS=2

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Pinak Ranjan Roy  (10 years ago)

2^47/47=3*10^12

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soumith vooda  (8 years ago)

Pls explain it clearly and crctly

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mukul kumar ray  (8 years ago)

2^47/47 =2

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Saagnik Adhikary  (5 years ago)

Apply fermat's little theorem
a^(p-1) is congruent to 1 (mod p)
here a = 2, p = 47
Thus, 2^46 yields remainder 1,
So by property of congruence we can say
2^46 * 2 yields remainder 1 * 2. Answer is 2

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