Maths-

General

Easy

Question

# If a, b, c are the three natural numbers such that a + b + c is divisible by 6, then a^{3} + b^{3} + c^{3} must be divisible by -

- 6
- 12
- 24
- None of these

## The correct answer is: 6

### We have (a^{3} – a) + (b^{3} – b) + (c^{3} – c)

= (a – 1) (a) (a + 1) + (b – 1) (b) (b + 1) + (c – 1) (c) (c + 1)

Since (x – 1) x (x + 1) is divisible both 2 & 3, it is divisible by 6. Thus,

6 | {(a^{3} – a) + (b^{3} – b) + (c^{3} – c)}

6 | (a^{3} + b^{3} + c^{3}) as 6|(a + b + c).

### Related Questions to study

maths-

### For x R, let [x] denote the greatest integer x, then value of++ +…+is -

### For x R, let [x] denote the greatest integer x, then value of++ +…+is -

maths-General

maths-

### The number of integral solutions of the equation x + y + z = 24 subjected to conditions that 1 x 5, 12 y 18, z –1

### The number of integral solutions of the equation x + y + z = 24 subjected to conditions that 1 x 5, 12 y 18, z –1

maths-General

maths-

### Ravish write letters to his five friends and address the corresponding envelopes. In how many ways can the letters be placed in the envelopes so that at least two of them are in the wrong envelopes -

### Ravish write letters to his five friends and address the corresponding envelopes. In how many ways can the letters be placed in the envelopes so that at least two of them are in the wrong envelopes -

maths-General

maths-

### In how many ways can we get a sum of at most 17 by throwing six distinct dice -

### In how many ways can we get a sum of at most 17 by throwing six distinct dice -

maths-General

maths-

### The number of non negative integral solutions of equation 3x + y + z = 24

### The number of non negative integral solutions of equation 3x + y + z = 24

maths-General

maths-

### Sum of divisors of 2^{5 }·3^{7 }·5^{3 }· 7^{2} is –

### Sum of divisors of 2^{5 }·3^{7 }·5^{3 }· 7^{2} is –

maths-General

maths-

### The length of the perpendicular from the pole to the straight line is

### The length of the perpendicular from the pole to the straight line is

maths-General

maths-

### The condition for the lines and to be perpendicular is

### The condition for the lines and to be perpendicular is

maths-General

Maths-

### If f : R →R; f(x) = sin x + x, then the value of (f^{-1} (x)) dx, is equal to

### If f : R →R; f(x) = sin x + x, then the value of (f^{-1} (x)) dx, is equal to

Maths-General

maths-

### The polar equation of the straight line with intercepts 'a' and 'b' on the rays and respectively is

### The polar equation of the straight line with intercepts 'a' and 'b' on the rays and respectively is

maths-General

maths-

### The polar equation of the straight line parallel to the initial line and at a distance of 4 units above the initial line is

### The polar equation of the straight line parallel to the initial line and at a distance of 4 units above the initial line is

maths-General

maths-

### The polar equation of axy is

### The polar equation of axy is

maths-General

maths-

### If x, y, z are integers and x 0, y 1, z 2, x + y + z = 15, then the number of values of the ordered triplet (x, y, z) is -

### If x, y, z are integers and x 0, y 1, z 2, x + y + z = 15, then the number of values of the ordered triplet (x, y, z) is -

maths-General

Maths-

### If then the equation whose roots are

### If then the equation whose roots are

Maths-General

Maths-

### Let p, q {1, 2, 3, 4}. Then number of equation of the form px^{2} + qx + 1 = 0, having real roots, is

Here we used the concept of quadratic equations and solved the problem. We also understood the concept of discriminant and used it in the solution to find the intervals. Therefore, total 7 seven solutions are possible so 7 equations can be formed.

### Let p, q {1, 2, 3, 4}. Then number of equation of the form px^{2} + qx + 1 = 0, having real roots, is

Maths-General

Here we used the concept of quadratic equations and solved the problem. We also understood the concept of discriminant and used it in the solution to find the intervals. Therefore, total 7 seven solutions are possible so 7 equations can be formed.