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Q.

Standard form of a number : Standard form of a number N(>1)  is N=P1n1,P2n2,P3n3......Pknk  where P1,P2,P3,....Pk  are distinct primes n1,n2,......nk  are positive integers.

Number of divisors of a natural number : If N=P1n1,P2n2,P3n3......Pknk  then number of divisors of  N,d(N)=(n1+1)(n2+1)......(nk+1) .

Sum of divisors of natural number (σ(N)) : σ(N)=[P1n1+11P11][P2n2+11P21].......[Pknk+11Pk1]

Product of divisors of N=N12d(N) .

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