Algebra - Number Theory - Previous Year CAT/MBA Questions

Answer: 217

Text Explanation :

Answer: 31

Text Explanation :

Answer: Option C

Text Explanation :

Answer: Option D

Text Explanation :

Answer: 7

Text Explanation :

Answer: Option A

Text Explanation :

Answer: Option A

Text Explanation :

Answer: Option A

Video Explanation

Text Explanation :

Given, x² + (x – 2y - 1)² = -4y(x + y)

⇒ x² + (x – 2y - 1)² = -4yx - 4y² 

⇒ x² + 4yx + 4y² + (x – 2y - 1)² = 0

⇒ (x² + 2 × x × 2y + (2y)²) + (x – 2y - 1)² = 0

⇒ (x + 2y)² + (x – 2y - 1)² = 0

Sum of squares of two number can be 0 only when both the numbers are 0.

∴ x + 2y = 0 and x – 2y – 1 = 0

⇒ x – 2y = 1

Hence, option (a).

Answer: Option C

Video Explanation

Text Explanation :

168 = 2³ × 21 = 2³ × 3 × 7
1134 = 2 × 567 = 2 × 3⁴ × 7 

Now (1134)ⁿ = 2ⁿ × 34ⁿ × 7ⁿ 
Since 168 (2³ × 3 × 7) completely divides 1134_n (2ⁿ × 34ⁿ × 7ⁿ)
⇒ Power of 2 in 1134_n ≥ Power of 2 in 168
⇒ n ≥ 3
Similarly, we can check for power of 3 and power of 7 and we get the least value of n = 3.

Now (168)^m = 2^3m × 3^m × 7^m 
Since 1134ⁿ (2ⁿ × 3⁴ⁿ × 7ⁿ = 2³ × 3¹² × 7³) completely divides 168m (2^3m × 3^m × 7^m)
⇒ Power of 3 in 168^m ≥ Power of 3 in 1134ⁿ 
⇒ m ≥ 12
Similarly, we can check for power of 2 and power of 7 and we get the least value of m = 12.

∴ n + m = 3 + 12 = 15

Hence, option (c).

Concept:

Factors

Answer: Option B

Video Explanation

Text Explanation :

For any natural numbers m, n and k, such that k divides both m + 2n and 3m + 4n, k must be a common divisor of

k divides m + 2n ⇒ m + 2n = a × k   ...(1)
k divides 3m + 4n ⇒ 3m + 4n = b × k   ...(2)

(1) × 3 - (2)
⇒ 3m + 6n - (3m + 4n) = 3ak - bk
⇒ 2n = (3a - b)k
∴ 2n is divisible by k

(2) - (1) × 2
⇒ 3m + 4n - 2(m + 2n) = bk - 2ak
⇒ m = (b - 2a)k
∴ m is divisible by k

∴ 2n and m are divisible by k
⇒ k is common divisor of 2n and m.

Hence, option (b).

Answer: 15

Video Explanation

Text Explanation :

Positive integers having exactly two distinct factors other than 1 and itself will be of the form a × b or a³, where a and b are prime numbers.

Case 1: a × b < 50. Possible combinations are:
If one of them is 2, the other prime number can be 3, 5, 7, 11, 13, 17, 19, 23 i.e., 8 possibilities.
If one of them is 3, the other prime number can be 5, 7, 11, 13 i.e., 4 possibilities.
If one of them is 5, the other prime number can be 7 i.e., 1 possibility.
∴ 13 cases.

Case 2: a³. Possible values of a are 2 and 3 i.e., numbers are 8 and 27.
∴ 2 cases.

∴ Total 13 + 2 = 15 such numbers are possible.

Hence, 15.

Concept:

Number of Factors

Answer: Option B

Video Explanation

Text Explanation :

If p² + q² - 29 = 2pq - 20 = 52 - 2pq, then the difference between the maximum and minimum possible value of (p³ - q³) is

Given, p² + q² - 29 = 2pq - 20
⇒ p² + q² - 2pq = 29 - 20
⇒ (p - q)² = 9
⇒ p - q = ± 3

Also given, p² + q² - 29 = 52 - 2pq
⇒ p² + q² + 2pq = 81
⇒ (p + q)² = 81
⇒ p + q = ± 9

Case 1: p - q = + 3 and p + q = + 9
Solving these 2 equations we get, p = 6 and q = 3
∴ p³ - q³ = 216 - 27 = 189

Case 2: p - q = - 3 and p + q = + 9
Solving these 2 equations we get, p = 3 and q = 6
∴ p³ - q³ = 27 - 216 = - 189

Case 3: p - q = + 3 and p + q = - 9
Solving these 2 equations we get, p = - 3 and q = - 6
∴ p³ - q³ = (-27) - (-216) = 189

Case 4: p - q = - 3 and p + q = - 9
Solving these 2 equations we get, p = - 6 and q =  - 3
∴ p³ - q³ = - 216 - (-27) = - 189

∴ Highest possible value of p³ - q³ = 189 least possible value of p³ - q³ = - 189.

∴ Required difference = 189 - (-189) = 378

Hence, option (b).

Answer: Option C

Video Explanation

Text Explanation :

A system of linear equations a₁x + b₁y = c₁ and a₂x + b₂y = c₂ has infinite solutions if

$\frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}$ = $\frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}$ = $\frac{{\mathrm{c}}_1}{{\mathrm{c}}_2}$

For, x + y = 4 and (a + 5)x + (b² -15)y = 8b to have infinite solution

⇒ $\frac{1}{\mathrm{a}+5}$ = $\frac{1}{{\mathrm{b}}^2-15}$ = $\frac{4}{8\mathrm{b}}$   ...(1)

⇒ 8b = 4(b² -15)

⇒ 4b² - 8b - 60 = 0

⇒ b² - 2b - 15 = 0

⇒ (b - 5)(b + 3) = 0

⇒ b = 5 or -3

From (1): ⇒ a + 5 = b² - 15
⇒ a = b² - 20

∴ a = 5 when b = 5 ⇒ ab = 25
or, a = -11 when b = -3 ⇒ ab = 33

∴ Maximum value of ab = -11 × -3 = 33

Hence, option (c).

Answer: Option C

Video Explanation

Text Explanation :

8^m = 2^3m and 8ⁿ = 2³ⁿ 

Now, 2^3m < 2^x < 2³ⁿ 

We have to find least possible value of (m + n) such that there are 41 possible values of x.

Least possible of m = 1, hence we get, 2³ < 2^x < 2³ⁿ 

Now, x can be any interger from 4 till 44 (41 values).

∴ Least possible value of 3n = 45, hence n = 15.

∴ Least possible value of m + n = 1 + 15 = 16.

Hence, option (c).

Answer: 468

Video Explanation

Text Explanation :

A number having 15 factors can be of the form:

a¹⁴ or a⁴ × b² 

Case 1: a¹⁴ 
The smallest possible such number = 2¹⁴ = 1024 × 16 > 16000

Case 2: a⁴ × b² 
The smallest possible such number = 2⁴ × 3²  = 16 × 9 = 144

The second smalled possible such number = 3⁴ × 2²  = 81 × 4 = 324

∴ The required sum = 144 + 324 = 468.

Hence, 468.

Answer: Option C

Video Explanation

Text Explanation :

Given a and b are natural numbers.

a² + ab + a = 14
⇒ a(a + b + 1) = 14 = 1 × 14 or 2 × 7

Case 1: a = 1 and a + b + 1 = 14
⇒ b = 12
a = 1 and b = 12 does not satisfy b² + ba + b = 28

Case 2: a = 2 and a + b + 1 = 7
⇒ b = 4
a = 2 and b = 4 satisfies b² + ba + b = 28

∴ 2a + b = 4 + 4 = 8

Hence, option (c).

Answer: 82

Video Explanation

Text Explanation :

Given, 3^k + 4^k + 5^k 
put k = 1, we have 3^k + 4^k + 5^k = 3 + 4 + 5 = 12
put k = 2, we have 3^k + 4^k + 5^k = 9 + 16 + 25 = 50

The only common factor between 12 and 50 is 2.

3^k + 4^k + 5^k is always even for any value of k, and hence always divisible by 2.

∴ A = 2

Given, 4^k + 3(4^k) + 4^k+2
= 4^k(1 + 3 + 4²)
= 4^k × 20
For k ≥ 1, this number is always divisible by 4 × 20 = 80
∴ B = 80

⇒ A + B = 2 + 80 = 82.

Hence, 82.

Answer: 34

Video Explanation

Text Explanation :

x, y and z are natural numbers.

⇒ xy + yz = 19
⇒ y(x + z) = 19 = 1 × 19
∴ y = 1 and x + z = 19 [x + z ≥ 2, since they are natural numbers]   …(1)

Also, yz + xz = 51
⇒ z(1 + x) = 51 = 3 × 17 or 1 × 51

Case 1: z = 3 and 1 + x = 17
⇒ x = 16
But then x + z = 19
Hence, this case is accepted.
⇒ xyz = 16 × 1 × 3 = 84

Case 2: z = 17 and 1 + x = 3
⇒ x = 2
But then x + z = 19
Hence, this case is accepted.
⇒ xyz = 2 × 1 × 17 = 34

Case 3: z = 1 and 1 + x = 51
⇒ x = 50
But then x + z = 51
Hence, this case is rejected.

Case 4: z = 51 and 1 + x = 1
⇒ x = 0
But x should be a natural number.
Hence, this case is rejected.

∴ Lowest possible value of xyz = 34 [Case 2].

Hence, 34.

Answer: Option B

Video Explanation

Text Explanation :

Given, a + 2b = 6.
⇒ a + b = 6 - b 

∴ (a + b) will be maximum when b is least. Least value of b can be 0, since b cannot be negative.
⇒ (a + b)_max = 6

∴ (a + b) will be minimum when b is highest. Highest value of b can be 3, since a cannot be negative.
⇒ (a + b)_min = 3

⇒ Average of highest and lowest values of (a + b) = $\frac{3+6}{2}$ = 4.5

Hence, option (b).

Answer: Option C

Video Explanation

Text Explanation :

For (15,000)! to be completely divisible by (n!)!, n! ≤ 15,000

Now we know, 
6! = 720
7! = 5,040
8! = 40,320

∵ n ≤ 15,000, highest value n can take is 7

Hence, option (c).

Answer: 150

Video Explanation

Text Explanation :

The number of students will be of the form = LCM(9, 10, 12, 25) × k + 4 = 900k + 4

The number of students is less than 5000 and also completely divisible by 11. This is possible when k = 2 and hence the number the students = 1804.

Maximum number of groups of 12 students that can be formed = Quotient of [1804 / 12] = 150

Hence, 150.

Answer: 4195

Video Explanation

Text Explanation :

Let the thousands, hundreds, tens and units digits be a, b, c and d.

Given, sum of its digits in the thousands, hundreds and tens places is 14
a + b + c = 14   …(1)
The sum of its digits in the hundreds, tens and units places is 15
b + c + d = 15   …(2)

(2) – (1) ⇒ a = d – 1   …(3)

The tens place digit is 4 more than the units place digit ⇒ c = d + 4   …(4)

For the number to be highest possible d should also be highest possible.

Highest possible of d is 5 (from (4)).
∴ highest possible value of c = 9 and that of a = 4

From (1), b = 14 – a – c = 1

∴ The highest such number is 4195

Hence, 4195.

Answer: Option A

Video Explanation

Text Explanation :

Given x² – xy – x = 22 and y² – xy + y = 34.

Adding both the equations

⇒ x² + y² – 2xy - (x – y) = 56

⇒ (x - y)² – (x - y) = 56

⇒ (x - y)(x - y - 1) = 56

Let (x – y) = a [a > 0 since x > y]

⇒ a(a - 1) = 56

⇒ a² – a – 56 = 0

⇒ (a – 8)(a + 7) = 0

⇒ a = 8 or -7 (rejected)

∴ a = x - y = 8

Hence, option (a).

Answer: Option B

Video Explanation

Text Explanation :

Let the two sides of the rectangle be ‘l’ and ‘b’ ft.

⇒ l × b = 60,000

Cost of fencing = 200 × l + 100 × (l + 2b) = 100 (3l + 2b)

To minimize cost we need to minimize 3l + 2b

Now, we know AP ≥ GP

⇒ $\frac{3l+2b}{2}$ ≥ $\sqrt{3l\times 2b}$
⇒ 3l + 2b ≥ 2 × √$\sqrt{6\times 60,000}$
⇒ 3l + 2b ≥ 1200

∴ Minimum cost of fencing = 100 × 1200 = 1,20,000.

Hence, option (b).

Answer: 21

Video Explanation

Text Explanation :

Let the digits of the 3-digit number be p, q, & r.
∴ 2 < p × q × r < 7
⇒ p × q × r can take the values 3, 4, 5, or 6.

Let's start with prime numbers 3 & 5.
Since they are prime, they can't be split, and hence if one of p, q or r is 3, the remaining two should be 1.

So, the possible combinations are

If p × q × r = 3
The number can be {113, 131, 311}

If p × q × r = 5
The number can be {115, 151, 511}

If p × q × r = 4
4 can be written as 1 × 2 × 2 or 1 × 1 × 4. 
Therefore, the possible combinations of p, q, r are {122, 212, 221, 114, 141, 411}

If p × q × r = 6
6 can be written as 1 × 3 × 2 or 1 × 1 × 6. 
Therefore, the possible combinations of p, q, r are {123, 132, 213, 231, 312, 321, 116, 161, 611}

Therefore, the total number of possibilities are 3 + 3 + 6 + 9 = 21.
Hence, 21.