[Kartik Raghavamurty S]

We must find 3^2002 + 7^2002 + 2002 (mod 29)

Since 29 is a prime, 3^2002 = ((3^28)^71)x 3^14= 1^71 x 3^14 = 3^14 (mod 29). This is by fermat's little theorem

Similarly we get 7^2002 = 7^14(mod 29) and finally 2002=1 mod 29

now our expression is reduced to 3^14  + 7^14 +1 (mod 29)

We can also write this as (9^7 + 49^7) +1(mod29)

since (a+b) is always a factor of a^k + b^k when k is odd, 9+49 i.e 58 is a factor of 9^7 + 49^7

since 58=0 (mod 29), 9^7 + 49^7 = 0 (mod 29)

Therefore our answer is 1

[Kartik Raghavamurty S]

C because all of the integers are equal

[Kartik Raghavamurty S]

divisible by 24 but not always divisible by 48 because the product of 4 consecutive integers (n^3 - n)(n-2)=(n-2)(n-1)(n)(n+1) is divisible by 4!=24

[Kartik Raghavamurty S]

let f(x)=x^1/3 + x^1/2 -1 f(x) is continuous on R and f(64)=13>0 f(1/64)=-5/8 <0 therefore by intermediate value theorem there exists a root between 1/64 and 64

[Kartik Raghavamurty S]

Since rhs is congruent to 3 modulo 4 there are no solutions because the sum of 2 squares can only be congruent to 0 1 or 2 mod 4

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