Question

The sum of the exponents of the prime factors in the prime factorization of 1764 is
(a) 3
(b) 4.
(c) 5
(d) 6​

Answer 1

$\underline{\textbf{Given:}}$

$\textsf{1764}$

$\underline{\textbf{To find:}}$

$\textsf{The sum of the exponents of the prime factors in the }$

$\textsf{prime factorization of 1764}$

$\underline{\textbf{Solution:}}$

$\mathsf{Consider,}$

$\mathsf{1764=2{\times}882}$

$\mathsf{1764=2{\times}2{\times}441}$

$\mathsf{1764=2{\times}2{\times}3{\times}147}$

$\mathsf{1764=2{\times}2{\times}3{\times}3{\times}49}$

$\mathsf{1764=2{\times}2{\times}3{\times}3{\times}7{\times}7}$

$\implies\mathsf{1764=2^2{\times}3^2{\times}7^2}$

$\mathsf{Now,}$

$\textsf{Sum of the exponents}$

$\mathsf{=2+2+2}$

$\mathsf{=6}$

$\therefore\textbf{Sum of the exponents of the prime factors is 6}$

$\underline{\textbf{Answer:}}$

$\mathsf{option\;(d)\;is\;correct}$

Answer 2

Answer:6 (d)

Step-by-step explanation:The prime factorization for 1764 is 2^2*3^2*7^2

So if we add the exponents (or count the factors)we will get the ans 6...