If n is a whole number greater than 1 then, n2(n2 - 1) is always divisible by.
Options:
60
24
48
12
Put n = 2,
n2(n2 - 1) = 12 which is divisible by 12.
Put n = 3,
n2(n2 - 1) = 72 which is divisible by 12.
Put n = 4,
n2(n2 - 1) = 240 which is divisible by 12.
So, we can say that n2(n2 - 1) is always divisible by 12.
Find the remainder when 919 + 6 is divided by 8.
4
5
6
7
Given,
N = 919 + 6
= (1+8)19 + 6
N/8 = [(1+8)19 + 6]/8
= (1 + 6)/8
= 7/8
So, remainder = 7
If N is divided by 56 it gives 29 as remainder. Find the remainder when N is divided by 8.
3
As per question,
N = 56q + 29
= (8x7q) + (8x3) + 5
= 8(7q + 3) + 5
So, remainder = 5
Find the sum of first 6 multiples of 5.
105
100
95
90
Given, n = 6 and x = 5
Sum of first n multiples of x = x[n(n+1)/2]
= 5[6(6+1)/2]
= 105
Find the sum of first 5 odd numbers.
21
23
25
27
Given, n = 5
Sum of first n odd numbers = n2
= 52
= 25
Find the sum of first 5 even numbers.
28
30
32
34
Sum of first n even numbers = n(n+1)
= 5(5+1)
= 30
Find the sum of the cubes of the first 5 natural numbers.
223
224
225
226
Sum of the cubes of first n natural numbers = [n(n+1)/2]2
= [5(5+1)/2]2
= 225
Find the sum of the squares of first 10 natural numbers.
375
385
395
365
Given, n = 10
Sum of the sqaures of first n natural numbers = n(n+1)(2n+1)/6
= 10(10+1)(20+1)/6
= 385
Find the remainder when 41000 is divided by 7.
1
2
41000/7
= ((42)500)/7
= ((16)500)/7
= ((14 + 2)500)/7
= ((2)500)/7
= (22 x (23)166)/7
= (4 x (8)166)/7
= (4 x (7 + 1)166)/7
= (4 x (1)166)/7
= 4
If the sum and difference of the digits of a two digit number is 14 and 2 respectively. Find the product of the digits.
46
50
Insufficient data
Let the two digits be x and y.
Given, x + y = 14 and x - y = 2
Therefore, x = 8 and y = 6
So, product = xy = 8x6 = 48