Question

Total number of divisors of n = 3⁵. 5⁷. 7⁹ that are of the form 4l + 1, l ³ 0 is equal to-

• A. 240
• B. 30
• C. 120
• D. 15

Answer 1

Answer: (A)

240

Answer 2

$3^{n_{1}}$= (4 – 1 $)^{n_{1}}$= 4l₁ + (–1 $)^{n_{1}}$

$5^{n_{2}}$= (4 + 1$)^{n_{2}}$= 4l₂ + 1

$7^{n_{3}}$= (8 – 1$)^{n_{3}}$=4l₃ + (–1$)^{n_{3}}$

Hence, any positive integer power of 5 will be in the form of 4l₂ + 1. Even power of 3 and 7 will be in the form of 4l + 1 and odd power of 3 and 7 will be in the form of 4l –1 thus required number of divisors = 8 (3.5 + 3.5) = 240