Number System-Railway Maths Part V

Perfect squares can't end with 2, 3, 7 and 8.

LCM of 6 and 8 = 24
Numbers which are divisible by both 6 and 8 must be divisible by 24
Total numbers divisible by both 6 and 8 = 100/24 = Quotient = 4

Rational number between 1/4 and 1/2 = (1/4 + 1/2)/2 = 3/8

Let the middle digit be x Then 2x = 18 or x = 9
So, the number is either 396 or 693 and another pair is 594 or 495 Since the number increases on reversing the digits, so the hundred's digits is smaller than the units digit Hence, required number is 396 because difference of 594 and 495 is not 297

Concept → the smallest four digit number is 1000
Now, LCM of 2,3 and 5 = 30
When we divide 1000 by 30 we get 10 as the remainder.
The smallest four digit number which is divisible by 2, 3 and 5 = 1000 -10 +30 =1020, the smallest 4-digit number which when divided by 2, 3 and 5 leaves a remainder of 1 in each case = 1020 + 1 = 1021

Let, number = X, then
A/Q, x² - x = 812
X(x - 1) = 812, 29 × 28 = 812

A/Q,
factors of 2² × 3¹ × 5² × 7¹ are divisible by 50 but not by 100
(i) 25 × 2 (ii) 25 × 2 × 3 (iii) 25 × 2 × 7 (iv) 25 × 2 × 7 × 3
Possible numbers- 4
We will not take '2' twice because it will be a multiple of 100.

120 = 2 × 2 × 2 × 3 × 5 = 2³ × 3 × 5
Number of total factors
= (3 + 1)(1 + 1)(1 + 1) = 4 × 2 × 2 = 16
Number of even factors
= 3 × 2 × 2 = 12
Number of divisors = no. of factors
Number of even divisors = 12
Number of odd divisors = 16 -12 = 4

Sol.249.(d) -0.96, -0.64, 0.24, 0.58, 0.83

LCM of (21, 35, 63) = 315
Number should be divisible by 315
Now, 10000/315 = 31(Quotient)