Quiz 6

288 = 2⁵ * 3². So it has 6 * 3 = 18 factors. Or, there are 9 ways of writing this number as a product of two positive integers.

Let us list these down.
1 * 288, 2 * 144, 3 * 96, 4 * 72, 6 * 48, 8 * 36, 9 * 32, 12 * 24 and 16 * 18

Now, this is where the question gets interesting. If a, b are integers either a + b and a – b have to be both odd or a + b and a – b have to be both even. So, within this set of possibilities
1 * 288, 3 * 96 and 9 * 32 will not result in integer values of a, b. So, there are 6 sets of numbers that work for us.
Moving on to these six sets; let us start with one example and see how many possibilities we can generate from this.

Let us consider the set 2 * 144.
Let us solve for this for a, b being natural numbers first, then we will extend this to integers.
When a, b are natural numbers
a + b > a – b
So, a + b = 144, a – b = 2; a = 73 and b = 71

Now, if a = 73, b = 71 holds good. We can see that a = 73, b = –71 also holds good. a = –73, b = 71 works and so does a = –73, b = –71.

There are 4 possibilities.
a = 73, b = 71
a = 73, b = –71
a = –73, b = 71
a = –73, b = –71
So, for each of the 6 products remaining, we will have 4 possibilities each. Total number of (a, b) that will satisfy this equation = 6 * 4 = 24.

The question is "How many pairs of integer (a, b) are possible such that a² – b² = 288?"

Hence the answer is "24"

 

Let a, b, c be the roots of this cubic equation

a + b + c = 5
ab + bc + ca = 8
abc = 4
This happens when a = 1, b = 2 and c = 2 {This is another approach to solving cubic equations.}

The other approach is to use polynomial remainder theorem.
If you notice, sum of the coefficients = 0

=> P(1) = 0
=> (x – 1) is a factor of the equation. Once we find one factor, we can find the other two by dividing the polynomial by (x–1) and then factorising the resulting quadratic equation.
(x – 1) (x – 2) (x – 2) > 0

Let us call the product (x – 1)(x – 2)(x – 2) a black box.
If x is less than 1, the black box is a –ve number.
If x is between 1 and 2, the black box is a +ve number.
If x is greater than 2, the black box is a +ve number.
Since we are searching for the regions where black box is a +ve number, the solution is as follows:

1 < x < 2 OR x > 2

The question is "Solve the inequality x³ – 5x² + 8x – 4 > 0."

Hence the answer is "(1, 2) ∪ (2, ∞)"

Let f(x) = x⁴ – ax³ + bx² – cx + 8

f(1) = 4
1 – a + b – c + 8 = 4
–a + b – c = - 5 ...... Eqn (1)
f(–1) = 3
1 + a + b + c + 8 = 3
a + b + c = - 6 ...... Eqn (2)
Adding Equation (1) and (2) we have
2b = -11,
or b = -5.5

The question is "x⁴ – ax³ + bx² – cx + 8 = 0 divided by x – 1 leaves a remainder of 4, divided by x + 1 leaves remainder 3, find b."

Hence the answer is "-5.5"

This Expression is same as (1² + 2² + 3² + 4² + 5² …….30² + 31²) – (2² + 4² + 6² ….30²)

= (1² + 2² + 3² + 4² + 5² …….30² + 31²) – 4(1²+ 2² + 3² ….15²)

1² + 2² + 3² +.......... + n² = (n(n+1)(2n+1))/6, for all n > 1.

Expresssion = (31 * 32 * 63)/6 - (4 * 15 * 16 * 31)/6
= (31 * 32 * 33)/6 = 31 * 11 * 16 = 5456

The question is "What is the sum of 1² + 3² + 5² …….31²?"

Hence the answer is "5456"

When f(x) is divided by x-a , the remainder = f(a),
Here f(x) = x⁴ + 5x³ – 3x² + 4x + 3
x + a = x + 2
a = -2
Therefore remainder = f(-2) = (-2)⁴ + 5(-2)³ – 3 (-2)² + 4(-2) + 3
F(-2) = 16 - 40 - 12 - 8 + 3
F(-2) = -41
Hence remainder is -41

The question is "What is the remainder when x⁴ + 5x³ – 3x² + 4x + 3 is divided by x + 2?"

Hence the answer is "-41"

Let p, q, r, s be the roots of the equation
p + q + r + s = 8
pqrs = 16
This happens only when p = q = r = s = 2
(p+q+r+s)4(p+q+r+s)4 = 2.
√(pqrs) = 2.
Arithemtic mean is equal to Geometric Mean. This is possible only when all the numbers are equal. (Note that the question says all roots are positive and real)
p = q = r = s = 2
pq + pr + ps + qr + qs + rs = a
24 = a
pqr + pqs + prs+ qrs = b
32 = b
So, a – b = 24 – 32 = – 8

The question is "If x⁴ – 8x³ + ax² – bx + 16 = 0 has positive real roots, find a – b."

Hence the answer is "-8"

 

Let f(x) = 4x³ + ax² – bx + 3.
Remainder on division of f(x) by x - 2 = f(2)
Or, f(2) = 32 + 4a – 2b + 3 = 2
4a – 2b = -33 ---- (1)
Remainder on division of f(x) by x + 3 = f(-3)
Or, f(-3) = –108 + 9a + 3b + 3 = 3
9a + 3b = 108
Dividing by 3, the above equation becomes
3a + b = 36. Multiplying this equation by 2, we get
6a + 2b = 72 ---- (2)
Adding equations (1) and (2), we have 10a = 39
a = 3.9
4(3.9) – 2b = –33
b = 24.3
f(x) = 4x³ + 3.9x² – 24.3x + 3
Remainder on division of f(x) by x + 2 = f(-2)
f(–2) = 4(–8) + 3.9(4) – 24.3(–2) + 3 = 35.2

The question is "4x³ + ax² – bx + 3 divided by x – 2 leaves remainder 2, divided by x + 3 leaves remainder 3. Find remainder when it is divided by x + 2"

Hence the answer is "35.2"

Let us say, N = 3²⁰⁰ – 5¹⁰⁰
N = 9¹⁰⁰ – 5¹⁰⁰.
This is a multiple of 9 - 5. Hence, 4, 2 and 1 are factors of the dividend.
N can also be written as 81⁵⁰ – 25⁵⁰.
This is a multiple of 81 - 25 = 56. Hence, 56, 7, 8 are factors of the dividend.
Dividend = (81²)²⁵ – (25²)²⁵
This is a multiple of 81² - 25², which is equal to (81 - 25) (81 + 25) = 56 × 106 = 7 × 8 × 2 × 53
Hence, 53 is also a factor of the dividend. The ‘2’ from this could combine with the 8 of the previous factor to mean that 16 is also a factor of the dividend.
The one exception of this list is ‘12’. Why? 3²⁰⁰ is a multiple of 3. 5¹⁰⁰ is not a multiple of 3. If a non–multiple of 3 is subtracted from a multiple of 3, the result will be a non–multiple of 3.
Hence, 3 is not a factor of the dividend. => 12 is not a factor of this number.
So, three of the numbers are factors.

The question is "How many of the following are factors of 3²⁰⁰ – 5¹⁰⁰?"

Hence the answer is "3"

p + q + z = 18
pq + qz + zp = b
pqz = c
According to the question, (pqz)^1/3 = 6. Hence, pqz = c = 216.
This is possible only when p = q= z = 6. {Arithmetic mean of p, q and z = 6, GM is also 6. This is possible only when all three numbers are equal.}
So, pq + qz + zp = 36 * 3 = 108 = b.
Hence, b = 108.

The question is "x³ – 18x² + bx – c = 0 has positive real roots, p, q and z. If geometric mean of the roots is 6, find b."

Hence the answer is "108"

Choice C is the correct answer.

(3x + y)³ = 27x³ + 3(3x)²y + 3(3x)y² + y³ = 27x³ + 27x²y + 9xy² + y³
So, the given expression is (3x + y)³ – 9x²y + 3xy²
= (3x + y)³ – 3xy ( 3x – y)
= (3 × 4 – 8)³ – {3 * 4 * -8 (12 + 8)}
= 4³ + 12 * 8 * 20 = 64 + 1920 = 1984

The question is "What is the value of 27x³ + 18x²y + 12xy² + y³ when x = 4, y = – 8? "

Hence the answer is "1984"