Answer: Option B
Explanation:
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.
Answer: Option C
(Place value of 6) - (Face value of 6) = (6000 - 6) = 5994
This is an A.P. in which a = 6, d = 6 and Sn = 1800
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.
Answer: Option E
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.
Answer: Option A
The sum of two odd number is even. So, a + b is even.
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, tn = 30 a + (n - 1)d = 30 2 + (n - 1) x 2 = 30 n - 1 = 14 n = 15
= (15 x 16) = 240.
Let the given number be 476 xy 0.
Then (4 + 7 + 6 + x + y + 0) = (17 + x + y) must be divisible by 3.
And, (0 + x + 7) - (y + 6 + 4) = (x - y -3) must be either 0 or 11.
x - y - 3 = 0 y = x - 3
(17 + x + y) = (17 + x + x - 3) = (2x + 14)
x= 2 or x = 8.
x = 8 and y = 5.
987 = 3 x 7 x 47
So, the required number must be divisible by each one of 3, 7, 47
553681 --> (Sum of digits = 28, not divisible by 3)
555181 --> (Sum of digits = 25, not divisible by 3)
555681 --> is divisible by 3, 7, 47.
Let X be the number and Y be the quotient. Then,
X = 357 x Y + 39
= (17 x 21 x Y) + (17 x 2) + 5
= 17 x (21Y + 2) + 5)
Required remainder = 5.
Clearly, 4864 is divisible by 4.
So, 9P2 must be divisible by 3. So, (9 + P + 2) must be divisible by 3.
P = 1.
Sn =