Question #6a527

Question

218 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-17T12:49:05

# 98.#

Recall that, if a number (no.) #N# can be factorised using primes

#p_i, 1 le i le n,# as, #N=p_1^(m_1)*p_2^(m_2)*...*p_n^(m_n), where,#

#m_i in NN, 1 le i le n,# then, the no. of factors of #N, i.e., f(N),# are,

#f(N)=(m_1+1)(m_2+1)...(m_n+1).#

Here, #83,000,000=83xx10^6=83xx(2xx5)^6=2^6xx5^6xx83^1.#

#:. f(83,000,000)=(6+1)(6+1)(1+1)=98.#

Enjoy Maths.!