# Algebra - Number Theory - Previous Year CAT/MBA Questions

You can practice all previous year CAT questions from the topic Algebra - Number Theory. This will help you understand the type of questions asked in CAT. It would be best if you clear your concepts before you practice previous year CAT questions.

**CAT 2021 QA Slot 2 | Algebra - Number Theory**

For a 4-digit number, the sum of its digits in the thousands, hundreds and tens places is 14, the sum of its digits in the hundreds, tens and units places is 15, and the tens place digit is 4 more than the units place digit. Then the highest possible 4-digit number satisfying the above conditions is

Answer: 4195

**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.

Workspace:

**CAT 2021 QA Slot 2 | Algebra - Number Theory**

Consider the pair of equations: x^{2} – xy – x = 22 and y^{2} – xy + y = 34. If x > y, then x – y equal.

- A.
8

- B.
7

- C.
6

- D.
4

Answer: Option A

**Explanation** :

Given x^{2} – xy – x = 22 and y^{2} – xy + y = 34.

Adding both the equations

⇒ x^{2} + y^{2} – 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

⇒ a^{2} – a – 56 = 0

⇒ (a – 8)(a + 7) = 0

⇒ a = 8 or -7 (rejected)

∴ a = x - y = 8

Hence, option (a).

Workspace:

**CAT 2021 QA Slot 2 | Algebra - Number Theory**

Three positive integers x, y and z are in arithmetic progression. If y − x > 2 and xyz = 5(x + y + z), then z − x equals

- A.
12

- B.
8

- C.
14

- D.
10

Answer: Option C

**Explanation** :

Let the three integers x, y and z be (a - d), a, (a + d) respectively.

[(a – d), a and (a + d) are all positive integers]

Since y – x > 2, hence d > 2.

Given, xyz = 5(x + y + z)

⇒ (a – d) × a × (a + d) = 5 × 3a

⇒ (a – d)(a + d) = 15

Here (a – d) and (a + d) are positive integers

∴ We need to write 15 as product of 2 positive integers. This can be done in two ways, 1 × 15 or 3 × 5

Hence, (a, d) is either (8, 7) or (4, 1).

Since d > 0 hence, (4, 1) is rejected.

∴ (a, d) = (8, 7)

∴ z – x = (a + d) – (a - d) = 2d = 14

Hence, option (c).

Workspace:

**CAT 2021 QA Slot 3 | Algebra - Number Theory**

The cost of fencing a rectangular plot is ₹ 200 per ft along one side, and ₹ 100 per ft along the three other sides. If the area of the rectangular plot is 60000 sq. ft, then the lowest possible cost of fencing all four sides, in INR, is

- A.
160000

- B.
120000

- C.
100000

- D.
90000

Answer: Option B

**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).

Workspace:

**CAT 2020 QA Slot 1 | Algebra - Number Theory**

How many 3-digit numbers are there, for which the product of their digits is more than 2 but less than 7?

Answer: 21

**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.

Workspace:

**CAT 2020 QA Slot 1 | Algebra - Number Theory**

If a, b and c are positive integers such that ab = 432, bc = 96 and c < 9, then the smallest possible value of a + b + c is

- A.
49

- B.
56

- C.
46

- D.
59

Answer: Option C

**Explanation** :

Given ab = 432, bc = 96 and c < 9.

To get the minimum value for a + b + c, the two bigger numbers should be as close as possible.

So possible values are

a = 36, b = 12, c = 8 ⇒ Sum = 58

a = 27, b = 16, c = 6 ⇒ Sum = 49

a = 18, b = 24, c = 4 ⇒ Sum = 46

Least possible value = 46

Hence, option (c).

Workspace:

**CAT 2020 QA Slot 2 | Algebra - Number Theory**

For real x, the maximum possible value of $\frac{x}{\sqrt{1+{x}^{4}}}$ is

- A.
1/2

- B.
1

- C.
1/√2

- D.
1/√3

Answer: Option A

**Explanation** :

$\frac{x}{\sqrt{1+{x}^{4}}}$ = $\frac{1}{\sqrt{\frac{1+{x}^{4}}{{x}^{2}}}}$ = $\frac{1}{\sqrt{\frac{1}{{x}^{2}}+{x}^{2}}}$

We know x2 is positive and sum of a positive number and its reciprocal is always ≥ 2, i.e.,

$\frac{1}{{x}^{2}}+{x}^{2}$ ≥ 2.

⇒ $\frac{1}{\sqrt{\frac{1}{{x}^{2}}+{x}^{2}}}$ will be maximum when 1$\frac{1}{{x}^{2}}+{x}^{2}$ is minimum i.e., 2.

∴ Maximum value of $\frac{1}{\sqrt{\frac{1}{{x}^{2}}+{x}^{2}}}=\frac{1}{\sqrt{2}}$

Hence, option (a).

Workspace:

**CAT 2020 QA Slot 2 | Algebra - Number Theory**

If x and y are non-negative integers such that x + 9 = z, y + 1 = z and x + y < z + 5, then the maximum possible value of 2x + y equals

Answer: 23

**Explanation** :

Given,

x = z – 9 and

y = z – 1

Also, x + y < z + 5

⇒ (z – 9) + (z – 1) < z + 5

⇒ z < 15

∴ Highest integral value of z is 14.

Now, we have to calculate maximum value of 2x + y

= 2(z - 9) + (z - 1)

= 3z – 19

= 3 × 14 – 19 = 23

Hence, 23.

Workspace:

**CAT 2020 QA Slot 2 | Algebra - Number Theory**

If x and y are positive real numbers satisfying x + y = 102, then the minimum possible value of 2601$\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)$ is

Answer: 2704

**Explanation** :

Let’s try to calculate the minimum possible value of $\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)$

⇒ $\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)$ = $\frac{1+x+y+xy}{xy}$ = $\frac{103}{xy}$ + 1

Minimum value of this expression will be when xy is maximum.

Now, when sum of two number is constant their product is maximum when both the number are equal to each other.

∴ x = y = 51

∴ Least possible value of 2601$\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)$ = 2601 × $\frac{52}{51}\times \frac{52}{51}$ = 2704

Hence, 2704.

Workspace:

**CAT 2020 QA Slot 2 | Algebra - Number Theory**

The number of pairs of integer (x, y) satisfying x ≥ y ≥ - 20 and 2x + 5y = 99 is

Answer: 17

**Explanation** :

Given, 2x + 5y = 99

⇒ x = (99 - 5y)/2

For x to be an integer, y has to be odd.

Possible value of x and y

** x y**

97 -19

92 -17

…

17 13

∴ Total 17 possible values.

Hence, 17.

Workspace:

**CAT 2020 QA Slot 3 | Algebra - Number Theory**

Let m and n be natural numbers such that n is even and 0.2 < $\frac{\mathrm{m}}{20},\frac{\mathrm{n}}{\mathrm{m}},\frac{\mathrm{n}}{11}$ < 0..5. Then m – 2n equals

- A.
4

- B.
3

- C.
2

- D.
1

Answer: Option D

**Explanation** :

We know, 0.2 < m/20 <0.5 ⇒ 4 < m < 10

Values m can take = {5, 6, 7, 8, 9}

We know, 0.2 < n/11 <0.5 ⇒ 2.2 < n < 5.5

Values n can take = {3, 4, 5}

Since n is even, the only possible value of n is 4.

Also, 0.2 < $\frac{\mathrm{m}}{20},\frac{\mathrm{n}}{\mathrm{m}},\frac{\mathrm{n}}{11}$ < 0.5 ⇒ m/5 < n < m/2

Since n = 4 ⇒ 20 > m > 8

The only possible value of m is 9.

∴ m – 2n = 9 – 2 × 4 = 1.

Hence, option (d).

Workspace:

**CAT 2020 QA Slot 3 | Algebra - Number Theory**

Let N, x and y be positive integers such that N = x + y, 2 < x < 10 and 14 < y < 23. If N > 25, then how many distinct values are possible for N?

Answer: 6

**Explanation** :

Given, 2 < x < 10

x can take any of the values from {3, 4, 5, 6, 7, 8, 9}

Also, 14 < y < 23

y can take any of the values from {15, 16, 17, 18, 19, 20, 21, 22}

The highest value N (i.e., x + y) can take = 9 + 22 = 31 (when x = 9; y = 22)

The lowest value N (i.e., x + y) can take = 3 + 15 = 18 (when x = 3; y = 15)

But, N = x + y > 25. Hence the different values of x + y are {31, 30, 29, 28, 27, 26}.

Hence, x + y, and thereby N can take 6 distinct values.

Hence, 6.

Workspace:

**CAT 2020 QA Slot 3 | Algebra - Number Theory**

How many of the integers 1, 2, … , 120, are divisible by none of 2, 5 and 7?

- A.
42

- B.
41

- C.
40

- D.
43

Answer: Option B

**Explanation** :

There are total 120 numbers given.

Integers not divisible by 2, 5 and 7 = 120 – (numbers that are divisible by at least one of 2, 5 and 7)

We know, n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) + n(A ∩ B ∩ C).

n(x) = number of multiples of x.

∴ n(2 ∪ 5 ∪ 7) = n(2) + n(5) + n(7) - n(2 ∩ 5) - n(5 ∩ 7) - n(7 ∩ 2) + n(2 ∩ 5 ∩ 7)

⇒ n(2 ∪ 5 ∪ 7) = n(2) + n(5) + n(7) - n(10) - n(35) - n(14) + n(70)

⇒ n(2 ∪ 5 ∪ 7) = 60 + 24 + 17 - 12 - 3 - 8 + 1 = 79.

Now, Integers not divisible by 2, 5 and 7 = 120 – (numbers that are divisible by at least one of 2, 5 and 7)

= 120 – 79 = 41

∴ Out of 120 numbers given, 41 numbers are not divisible by any of 2, 5 and 7.

Hence, option (b).

Workspace:

**CAT 2019 QA Slot 1 | Algebra - Number Theory**

The product of two positive numbers is 616. If the ratio of the difference of their cubes to the cube of their difference is 157 : 3, then the sum of the two numbers is

- A.
58

- B.
85

- C.
50

- D.
95

Answer: Option C

**Explanation** :

Let the two numbers be x and y.

So, xy = 616 and (x^{3} − y^{3})/(x − y)^{3} = 157/3.

3(x^{3} − y^{3}) = 157(x − y)^{3}

∴ 3(x^{3} − y^{3}) =157 [x^{3} − y^{3} − 3xy(x − y)]

∴ 154(x^{3} − y^{3}) = 3 × 157 × xy(x − y)

∴ 154(x^{3} − y^{3}) = 3 × 157 × 616 × (x − y) [Using xy = 616]

∴ (x − y)(x^{2} + y^{2} + xy) = 1884(x − y)

∴ (x^{2} + y^{2} + xy) = 1884 (If x ≠ y)

Adding xy both sides, we get;

(x^{2}+ y^{2} + 2xy) = 1884 + xy = 1884 + 616 = 2500.

∴ x + y = 50.

Hence option 3.

Workspace:

**CAT 2019 QA Slot 2 | Algebra - Number Theory**

Let a, b, x, y be real numbers such that a^{2} + b^{2} = 25 , x^{2} + y^{2} = 169 and ax + by = 65. If k = ay - bx, then

- A.
k = 0

- B.
k > 513

- C.
k = 513

- D.
0 < k ≤ 513

Answer: Option A

**Explanation** :

Given: a^{2} + b^{2} = 25 and x^{2} + y^{2} = 169

We know 5^{2} = 25 and 13^{2} = 169

Multiply both equations to get (a^{2} + b^{2}) (x^{2} + y^{2}) = 25 × 169

(a^{2} + b^{2}) × (x^{2} + y^{2}) = 4225

We know, 6225 = 65^{2}

We also know that ax + by = 65

So, numerically (Not algebraically),

(a^{2} + b^{2}) × (x^{2} + y^{2}) = (ax + by)^{2}

Expanding the equation,

⇒ (ax)^{2} + (ay)^{2} + (bx)^{2} + (by)^{2} = (ax)^{2} + (by)^{2} + 2axby

⇒ (ay)^{2} + (bx)^{2} = 2axby

⇒ (ay)^{2} + (bx)^{2} - 2axby = 0

This is of the form, (p - q)^{2}

(ay - bx)^{2} = 0

⇒ ay - bx = 0 = k

Hence, option (1).

Workspace:

**CAT 2019 QA Slot 2 | Algebra - Number Theory**

How many factors of 2^{4} × 3^{5} × 10^{4} are perfect squares which are greater than 1?

Answer: 44

**Explanation** :

N = 2^{4} × 3^{5} × 10^{4 }= 2^{4} × 3^{5} × 2^{4} × 5^{4} = 2^{8} × 3^{5} × 5^{4}

Now, for any number to be a perfect square, it must have even powers.

So, if we consider 2^{8}, only powers of 0, 2, 4, 6, 8 can lead to perfect squares i.e. 5 ways

Now,

If we consider 3^{5}, only powers of 0, 2, 4 can lead to perfect squares i.e. 3 ways

If we consider 5^{4}, only powers of 0, 2, 4 can lead to perfect squares i.e. 3 ways

So, total number of possibilities = 5 × 3 × 3 ways = 45 ways.

Since we need to find the number of factors greater than 1,

Required number of ways = 45 -1 = 44 ways

Hence, 44.

Workspace:

**CAT 2019 QA Slot 2 | Algebra - Number Theory**

What is the largest positive integer such that $\frac{({n}^{2}+7n+12)}{({n}^{2}-n-12)}$ is also positive integer?

- A.
6

- B.
8

- C.
16

- D.
12

Answer: Option D

**Explanation** :

Given : $\frac{({n}^{2}+7n+12)}{({n}^{2}-n-12)}$

$\frac{({n}^{2}+7n+12)}{({n}^{2}-n-12)}$ = $\frac{({n}^{2}+4n+3n+12)}{({n}^{2}-4n+3n-12)}$ = $\frac{(n+4)(n+3)}{(n-4)(n+3)}$

Now, n cannot be equal to -3, since denominator cannot be 0

⇒ $\frac{({n}^{2}+7n+12)}{({n}^{2}-n-12)}$ = $\frac{(n+4)}{(n-4)}$ = $\frac{(n-4+8)}{(n-4)}$ = $1+\frac{8}{(n-4)}$

For $\frac{({n}^{2}+7n+12)}{({n}^{2}-n-12)}$ to be an integer, 8/(n-4) should also be an integer.

Largest value of n – 4 = 8

∴ largest possible value of n = 12.

Hence, option (4).

Workspace:

**CAT 2019 QA Slot 2 | Algebra - Number Theory**

How many pairs (m,n) of positive integers satisfy the equation m^{2} + 105 = n^{2}?

Answer: 4

**Explanation** :

We know that m^{2} + 105 = n^{2}

105 = n^{2} - m^{2}

105 = (n + m) × (n - m)

105 can be written as 3 × 5 × 7, which can be written as product of two numbers in 4 ways i.e. (105 × 1), (35 × 3), (21 × 5), (15 × 7)

Solving each of these 4 cases will give corresponding values of m and n.

Hence, 4.

Workspace:

**CAT 2018 QA Slot 1 | Algebra - Number Theory**

The number of integers x such that 0.25 < 2^{x} < 200, and 2x + 2 is perfectly divisible by either 3 or 4, is

Answer: 5

**Explanation** :

For x = (−2), 2^{x} = 0.25 and x = 7 is the largest value that x can take for 2^{x} ≤ 200

For x = (−2) and (−1), (2^{x} + 2) is not an integer.

For x = 0, (2^{x} + 2) = 3 i.e., divisible by 3.

For x = 1, (2^{x} + 2) = 4 i.e., divisible by 4.

For x = 2, (2^{x} + 2) = 6 i.e., divisible by 3.

For x = 3, (2^{x} + 2) = 10 i.e., neither divisible by 3 nor by 4.

For x = 4, (2^{x} + 2) = 18 i.e., divisible by 3.

For x = 5, (2^{x} + 2) = 34 i.e., neither divisible by 3 nor by 4.

For x = 6, (2^{x} + 2) = 66 i.e., divisible by 3.

For x = 7, (2^{x} + 2) = 130 i.e., neither divisible by 3 nor by 4.

Thus, for x = 0, 1, 2, 4 and 6, 2^{x} + 2 is perfectly divisible by either 3 or 4.

Hence, 5.

Workspace:

**CAT 2018 QA Slot 1 | Algebra - Number Theory**

While multiplying three real numbers, Ashok took one of the numbers as 73 instead of 37. As a result, the product went up by 720. Then the minimum possible value of the sum of squares of the other two numbers is

Answer: 40

**Explanation** :

Let X and Y are the two real numbers.

X × Y × 73 − X × Y × 37 = 720

X × Y × (73 – 37) = 720

X × Y × 36 = 720

X × Y = 20

AM(X^{2}, Y^{2}) ≥ GM(X^{2}, Y^{2})

∴ $\frac{{X}^{2}+{Y}^{2}}{2}$ ≥ $\sqrt{{X}^{2}\times {Y}^{2}}$

∴ (X^{2} + Y^{2}) ≥ 40

Answer : 40

Alternately,

We know, X × Y = 40

Also, (X - Y)^{2} ≥ 0 [Square of any number is always greater than or equal to zero]

∴ X^{2} + Y^{2} - 2XY ≥ 0

∴ X^{2} + Y^{2} ≥ 2XY

∴ X^{2} + Y^{2} ≥ 40

Workspace:

**CAT 2018 QA Slot 2 | Algebra - Number Theory**

How many two-digit numbers, with a non-zero digit in the units place, are there which are more than thrice the number formed by interchanging the positions of its digits?

- A.
6

- B.
8

- C.
7

- D.
5

Answer: Option A

**Explanation** :

If ‘x’ is the digit in the ten’s place and ‘y’ is the digit in the unit’s place of the two digit number, we have the following:

$10x+y>3(10y+x)\phantom{\rule{0ex}{0ex}}\therefore x>\frac{29}{7}y$

When y = 1, x = 5, 6, 7, 8 or 9 (Total 5 values)

When y = 2, x = 9 (Total 1 value)

Therefore the required answer is 5 + 1 = 6.

Hence option 1.

Workspace:

**CAT 2018 QA Slot 2 | Algebra - Number Theory**

If A = {6^{2n} -35n -1: n = 1,2,3,...} and B = {35(n - 1) : n = 1,2,3,...} then which of the following is true?

- A.
Every member of A is in B and at least one member of B is not in A

- B.
At least one member of A is not in B

- C.
Every member of B is in A.

- D.
Neither every member of A is in B nor every member of B is in A

Answer: Option A

**Explanation** :

For n = 1, A_{1} = 0

For n = 2, A_{2} = 1225 = 35 × 35

Also, 6^{2n}- 35n - 1 = (36n-1) - 35n.

The term in the bracket 36^{n} - 1 is always divisible by (36 - 1)or by 35. Similarly 35n is also always divisible by 35. Therefore each term in the set A is divisible by 35. However, not all positive multiples of 35 are present in set A.

For n = 1, B_{1} = 35(1 - 1) = 0

For n = 2, B_{2} = 35(2 - 1) = 35 and so on.

Thus all whole number multiples of 35 are present in set B.

Thus every member of set A is present in every member of set B but at least one member of set B is not present in set A.

Hence, option 1.

Workspace:

**CAT 2017 QA Slot 1 | Algebra - Number Theory**

If x + 1 = x^{2} and x > 0, then 2x^{4} is:

- A.
6 + 4√5

- B.
3 + 5√5

- C.
5 + 3√5

- D.
7 + 3√5

Answer: Option D

**Explanation** :

x + 1 = x^{2}

⇒ x^{2 }- x – 1 = 0

⇒ x = (1+√5)/2 (∵ x > 0)

Now, squaring the original equation, we get

(x +1)^{2} = x^{4}

⇒ x^{2} + 1 + 2x = x^{4}

⇒ x^{4} = x + 1 + 1 + 2x (Since, x^{2} = x + 1)

⇒ x^{4} = 3x + 2

⇒ 2x^{4} = 6x + 4

⇒ 2x^{4} = 6((1+√5)/2) + 4

⇒ 2x^{4} = 7 + 3√5

Alternately,

We know, x = (1+√5)/2

⇒ x ≈ 1.62

∴ 2x^{4} ≈ 13.8

option I ≈ 14.8,

option II ≈ 11.6,

option IV ≈13.6

Hence, option 4.

Workspace:

**CAT 2017 QA Slot 2 | Algebra - Number Theory**

If the product of three consecutive positive integers is 15600 then the sum of the squares of these integers is

- A.
1777

- B.
1785

- C.
1875

- D.
1877

Answer: Option D

**Explanation** :

Since 15600 is product of 3 consecutive we can start by taking the cube of an integer closest to 15600, which is actually 15625

Now $\sqrt[3]{15625}=25$

Dividing 15600 by 25 we get 624

624 can be expressed as a product of 24 and 26. So the 3 consecutive positive integers are 24, 25 and 26

24^{2} + 25^{2 }+ 25^{2} = 576 + 625 + 676

= 1877

Hence, option 4.

Workspace:

**CAT 2017 QA Slot 2 | Algebra - Number Theory**

How many different pairs (a, b) of positive integers are there such that a ≤ b and $\frac{1}{a}+\frac{1}{b}=\frac{1}{9}$?

Answer: 3

**Explanation** :

Given $\frac{1}{a}+\frac{1}{b}=\frac{1}{9}$

$\frac{b+a}{ab}=\frac{1}{9}$

⇒ 9b + 9a = ab

⇒ ab + 9b – 9a + 81 = 81

⇒ b(a – 9) – 9(a – 9) = 81

⇒ (a – 9) = (b – 9) = 81

Now 81 as a product of 2 positive integers can be written as 1 × 81, 3 × 21 or 9 × 9

If a - 9 = 1 ⇒ a = 10. Also b - 9 = 81

⇒ b = 90

Further, if a – 9 = 3, ⇒ a = 12 Also b – 9

= 27 ⇒ b = 36

If a – 9 = 9 ⇒ a = 18. Also b – 9 = 9 ⇒ b = 18

∴ 3 pairs of number i.e., (10, 90), (12, 36) and (18, 18) will satisfy the given equation.

Answer: 3

Workspace:

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