Math miscellaneousHard
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
Create a free account to view solution
View Solution FreeMore Math miscellaneous Questions
Consider the system of linear equations:x1 + 2x2 + x3 = 32x1 + 3x2 + x3 = 33x1 + 5x2 + 2x3 = 1The system has...If A = then...The variance of first 50 even natural numbers is...A die is tossed 5 times. Getting an odd number is considered a success. Then the variance of distribution of success is...The system of equations αx + y + z = α - 1, x + αy + z = α - 1, x + y + αz = α - 1 has no ...