Question

The no. of +ve integral factors of 3⁵.5⁴.7⁷ which are of the form 4m + 1, M ∈ N is

• A. 120
• B. 119
• C. 118
• D. 121

Answer 1

Answer: (B)

119

Answer 2

(x - 1)ⁿ =

$nC_{0}x^{n} -^{n}C_{1}x^{n - 1} +^{n}C_{2}x^{n - 2} + ......... + ( - 1{)^{n - 1}}^{n}C_{n - 1}x + ( - 1)^{n}$.Let $3^{i}5^{j}7^{k}$be a factor.

= (4 - 1)ⁱ (4 + 1)^j (8 - 1)^k

= (4p + (-1)ⁱ) (4q + 1) (4r + (-1)^k).

= 4m + (-1)^{i + k}.

Now (-1)^{i + k} should be equal to 1 ⇒ i + k is even ⇒ i & k are both even or both odd.

No. of ways in which i & k are even is 3 x 4

No. of ways in which i & k are odd is 3 x 4

No. of ways in which j can be chosen is 5.

Since M ∈ N, 1 is not a factor.

∴ Total no. of ways = (2 x 3 x 4 x 5) - 1 = 119.