Numbers - General Questions (7)

This is an A.P. in which a = 6, d = 6 and S_n = 1800

	n	[2 x 6 + (n - 1) x 6] = 1800
	2

 3n (n + 1) = 1800

 n(n + 1) = 600

 n² + n - 600 = 0

 n² + 25n - 24n - 600 = 0

 n(n + 25) - 24(n + 25) = 0

 (n + 25)(n - 24) = 0

 n = 24

Number of terms = 24.

99 = 11 x 9, where 11 and 9 are co-prime.

By hit and trial, we find that 114345 is divisibleby 11 as well as 9. So, it is divisible by 99.

The sum of two odd number is even. So, a + b is even.

(Place value of 6) - (Face value of 6) = (6000 - 6) = 5994

Required sum = (2 + 4 + 6 + ... + 30)

This is an A.P. in which a = 2, d = (4 - 2) = 2 and l = 30.

Let the number of terms be n. Then,

t_n = 30     a + (n - 1)d = 30

 2 + (n - 1) x 2 = 30

 n - 1 = 14

 n = 15

Sn =	n	(a + l)	=	15	x (2 + 30)   = 240.
	2			2

√551 > 22

All prime numbers less than 24 are : 2, 3, 5, 7, 11, 13, 17, 19, 23.

119 is divisible by 7; 187 is divisible by 11; 247 is divisible by 13 and 551 is divisible by 19.

So, none of the given numbers is prime.

Let the two consecutive even integers be 2n and (2n + 2). Then,

(2n + 2)² = (2n + 2 + 2n)(2n + 2 - 2n)

     = 2(4n + 2)

     = 4(2n + 1), which is divisible by 4.