Math miscellaneousEasy
Question
Number of divisors of 1024 × (225)2 × (25)7 which are of the form 3k + 1 (where k is a whole number) is equal to
Options
A.45
B.60
C.50
D.72
Solution
1024 × (225)2 × (25)7 = 210 × 34 × 518
2 & 5 are of the form 3k + 2.
If two number of the form 3k + 2 are multiplied together than we get a number of the form 3k + 1
⇒ Divisor must contain even powers of 2 & 5 and it must not contain any power of 3.
⇒ Number of such divisors = 6 × 10 = 60 where number of ways to select power of 2 = 6 & number of ways to select power of 5 = 10
2 & 5 are of the form 3k + 2.
If two number of the form 3k + 2 are multiplied together than we get a number of the form 3k + 1
⇒ Divisor must contain even powers of 2 & 5 and it must not contain any power of 3.
⇒ Number of such divisors = 6 × 10 = 60 where number of ways to select power of 2 = 6 & number of ways to select power of 5 = 10
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