However, it’s not only important for students to understand the concepts and topics but they need to gain a clear idea about the nature of the exam. This is where the question paper steps in. Apart from 2021 candidates, the IOQM 2021 question paper provided here will help future aspirants to understand the trend of questions and paper pattern in a much better way. Ultimately, it will enable each candidate in listing out the most important topics, develop better time management skills and plan their preparation accordingly. Students can find below the question paper of IOQM below.

**Question 1. Let ABCD be a trapezium in which AB||CD and AB = 3CD. Let E be the midpoint of the diagonal BD. If [ABCD] = n × [CDE], what is the value of n? (Here [Γ] denotes the area of the geometrical figure Γ.)**

Answer: 8

In trapezium ABCD

AB = 3CD

And E is midpoint of BD

ar (ABCD) = n × ar (CDE)

16 a = n × 2a

n = 16a / 2a

n = 8

**Question 2. A number N in base 10, is 503 in base b and 305 in base b + 2. What is the product of the digits of N?**

Answer: 64

Number N in base 10

503 in base b and 305 in base (b + 2)

Convert in base 10,

5 × b^{2} + 3 = 3 (b + 2)^{2} + 5

5b^{2} + 3 = 3 (b^{2} + 4b + 4) + 5

5b^{2} + 3 = 3b^{2} + 12b + 12 + 5

2b^{2} – 12b – 14 = 0

b^{2} – 6b – 7 = 0

b^{2} – 7b + b – 7 = 0

b(b – 7) + 1 (b – 7) = 0

(b – 7) (b + 1) = 0

b = 7, – 1 rejected

N = 5b^{2} + 3 or 3(b + 2)^{2} + 5

N = 5(7)^{2} + 3

= 248

Product of digits of N = 2 × 4 × 8 = 64

**Question 3. If **

**\(\begin{array}{l}\sum_{k=1}^{n}\frac{2k+1}{(k^{2}+k)^{2}} = 0.9999\end{array} \)**

**then determine the value of N.**

Answer: 99

n + 1 = 100

n = 99

**Question 4. Let ABCD be a rectangle in which AB + BC + CD = 20 and AE = 9, where E is the midpoint of the side BC. Find the area of the rectangle. **

Answer: 19

Area of rectangle ABCD =a × 2b = 2ab

In ∆ABE, a^{2} + b^{2} = 81

and a + 2b + a = 20

∴ a + b = 10 ⇒ a^{2} + b^{2} + 2ab = 100

⇒ 81 + 2ab = 100

⇒ 2ab = 19

⇒ ar(ABCD) = 19.

**Question 5. Find the number of integer solutions to ||x| – 2020| < 5.**

Answer: 18

||x| – 2020| < 5.

⇒ -5 < |x| – 2020 < 5

⇒2015 < |x| < 2025

∴ x = ±2016, ± 2017,…,± 2024

So, the total number of integer solutions = 18

**
\(\begin{array}{l}cos \: C = \frac{6^{2}+4^{2}+5^{2}}{2\times 6\times 4}\end{array} \)
\(\begin{array}{l}cos \: C = \frac{36+16+25}{48}\end{array} \)
\(\begin{array}{l}cos \: C = \frac{27}{48} = \frac{9}{16}\end{array} \)
\(\begin{array}{l}\frac{360^{o}}{gcd(105^{o}, 60^{o})} = \frac{360^{o}}{15^{o}} = 24\end{array} \)
\(\begin{array}{l}\frac{x^{2} + 2\Delta }{x} = \frac{y^{2} + 2\Delta }{y}\end{array} \)
\(\begin{array}{l}\frac{A+5}{A+B+C+15}\times|N| = \frac{6}{7}\times \frac{|N|}{2}\end{array} \)
\(\begin{array}{l}\frac{A+5}{A+B+C+15}\times|N| = \frac{3}{7}\end{array} \)
\(\begin{array}{l}\frac{C+5}{A+B+C+15}\times N = 2\left ( \frac{C}{A+B+C} \right )\times N\end{array} \)
\(\begin{array}{l}\frac{C+5}{35} = \frac{2C}{20}\end{array} \) [ ∵ A = B + C = 10 ⇒ A + B + C = 20]
\(\begin{array}{l}\frac{B+5}{A+B+C+15}\times N = \frac{B}{A+B+C}N – 1000\end{array} \)
\(\begin{array}{l}\frac{13N}{35} = \frac{8N}{20} – 1000\end{array} \)
\(\begin{array}{l}\frac{13N}{35} = \frac{2N}{5} – 1000\end{array} \)
\(\begin{array}{l}\frac{13N}{35} = \frac{2N – 5000}{5}\end{array} \)
\(\begin{array}{l}2 = \frac{|10-16-m|}{\sqrt{1+m^{2}}}\end{array} \) (condition of tangency)
\(\begin{array}{l}\left | \frac{-16}{m_{1}} + \frac{-16}{m_{2}} \right |\end{array} \)
\(\begin{array}{l}= 16\frac{|m_{1}-m_{2}|}{m_{1}m_{2}}\end{array} \)
\(\begin{array}{l}\frac{16\sqrt{81-64}}{4}\end{array} \)
\(\begin{array}{l}\frac{\left \langle N \right \rangle}{N} = \frac{91.120.143.180}{100.121.144.169} = \frac{21}{22}\end{array} \)
\(\begin{array}{l}\frac{\left \langle N \right \rangle}{N} = \frac{21}{22}\end{array} \)
**

**Question 6. What is the least positive integer by which 2 ^{5}.3^{6}.4^{3}.5^{3}.6^{7} should be multiplied so that the product is a perfect square?**

Answer: 15

2^{5} × 3^{6 }× 4^{3} × 5^{3} × 6^{7}

= 2^{5} × 3^{6} × 2^{3} × 2^{3} × 5^{3} × 2^{7} × 3^{7}

= 2^{18} × 3^{13} × 5^{3}

To make this product perfect square, all the powers of the prime factor should be even.

Hence, the number should be 3 and 5.

So, the required number is 3 × 5 = 15

**Question 7. Let ABC be a triangle with AB = AC. Let D be a point on the segment BC such that BD = 48 × 1/61 and DC = 61. Let E be a point on AD such that CE is perpendicular to AD and DE = 11. Find AE.**

Answer: 25

Given,

BD = 48 × 1/61

DC = 61

By Pythagoras theorem,

CE = 60

60^{2} + x^{2} = a^{2}

By Stewart theorem,

a^{2} × 61 + a^{2} × 48 × 1/61 = (48 × 1/61 + 61) {48 × 61 + 1 + (x + 11)^{2}}

⇒ a^{2} = 2929 + (x + 11)^{2}

3600 + x^{2} = x^{2} + 22x + 121 + 2929

⇒ 22x = 550

⇒ x = 25

AE = 25

**Question 8. A 5-digit number (in base 10) has digits k, k + 1, k + 2, 3k, k + 3 in that order, from left to right. If this number is m ^{2} for some natural number m, find the sum of the digits of m.**

Answer: 15

According to the question,

10^{4}(k) + 10^{3}(k + 1) + 10^{2} (k + 2) + 10(3k) + k + 3 = m^{2}

10000 k + 1000 k + 1000 + 100k + 200 + 31 k + 3 = m^{2}

11131 k + 1203 = m^{2} …..(1)

Given that 3k is a digit.

3k is digit when k ≤ 3

k = 1, 2, 3

If we put k = 1, 2 and 3 in equation (1),

we will get a perfect square number.

Only for k = 3

11131(3) + 1203 = m^{2}

34596 = m^{2}

m = 186

Sum of digits of m is 1 + 8 +6 = 15

**Question 9. Let ABC be a triangle with AB=5, AC=4, BC=6. The internal angle bisector of C intersects the side AB at D. Points M and N are taken on sides BC and AC, respectively, such that DM‖AC and DN‖BC. If (MN) ^{2} = p/q where p and q are relatively prime positive integers then what is the sum of the digits of |p-q|?**

Answer: 2

AD / DB = 4 / 6 = 2 / 3

DM || AC

BD / AB = BM / BC

3 / 5 = BM / 6

BM = 18 / 5 = 3.6

CM = 6 – 3.6 = 2.4

DN || BC

AD / AB = AN / AC

2 / 5 = AN / 4

8 / 5 = AN

⇒ AN = 1.6

MN^{2} = (2.4)^{2} + (2.4)^{2} – 2 × 2.4 × 2.4 cos C

MN^{2} = 5.76 + 5.76 – 2 × 5.76 × 9 / 16

MN^{2 } = 11.52 (1 – 9 / 16)

= 11.52 × 7/16

MN^{2} = 0.72 × 7 = 5.04 = 504 / 100 = 126 / 25 / = p / q

|p-q|=101

Sum of the digits in |p-q| is 2

**Question 10. Five students take a test on which any integer score from 0 to 100 inclusive is possible. What is the largest possible difference between the median and the mean of the scores? (The median of a set of scores is the middlemost score when the data is arranged in increasing order. It is exactly the middle score when there is an odd number of scores and it is the average of the two middle scores when there is an even number of scores.)**

Answer: 40

Let numbers (marks) be

0, 0, 100, 100, 100

Median = 100

Mean = 300 / 5 = 60

Difference of (median – mean)

= 100 – 60

= 40

**Question 11. Let X={-5,-4,-3,-2,-1, 0, 1, 2, 3, 4, 5} and**

**S = {(a,b)∈X × X : x ^{2} + ax + b and x^{3} + bx + a have at least a common real zero}. **

**How many elements are there in S?**

Answer: 24

If there is a common root, then a^{2}x + a(b + 1) must be a factor of x^{2} + ax + b.

So, x = – (b+1 / a) is a root of x^{2} + ax + b = 0.

⇒ (b + 1 / a)^{2} – a (b + 1) a + b = 0

⇒ a = ± (b + 1)

a + b = –1 or b – a + 1 = 0

S = {(–1, 0), (0, –1), (–2, 1), (1, – 2), (2, – 3), (–4, 3), (3, –4), (4, –5), (2, 1), (3, 2), (4, 3), (5, 4), (–1, –2,), (–2, –3), (–3, –4), (–4, –5), (1, 0)}

{(0, –2), (0, –3), (0,–4), (0, –5) (0, 0)} are also the solutions.

Therefore, the total number of elements in S is 24.

**Question 12. Given a pair of concentric circles, chords AB, BC, CD,…… of the outer circle are drawn such that they all touch the inner circle. If ∠ABC=75°, how many chords can be drawn before returning to the starting point? **

Answer: 24

Given, AB, BC and CD be the chords of the outer circle.

Since, AB, BC and CD touch the inner circle,

Hence, length of chords must be equal i.e., AB = BC = CD.

So, central angle =105° (∵ ∠NOM + ∠ABC = 180°)

So, total number of chords =

**Question 13. Find the sum of all positive integers n for which is a perfect square.**

Answer: 6

|2^{n} + 5^{n} – 65| = m^{2} (let)

If n = 2, then m^{2} = |4 + 25 – 65| = |–36| = 6^{2}

If n ≥ 3, then

m^{2} = |2^{n} + 5(5^{n–1} – 13)|

If n = 4, then

m^{2} = 576 = 24^{2}

If n ≥ 5, then

m^{2} = 2^{n }+ 5pq (25 – 13)

= 2^{n} + 5pq 12

= 2^{n} + α.pq 60 (where α∈ I^{+})

So, for n = 1, 3, 5, 7, ….. not possible as the unit digit is 2 and 8.

Also, for n = 2k

m^{2} = 4^{k} + 5 (5^{2k–1} – 13)

⇒ (m – 2^{k}) (m + 2^{k}) = 5 (5^{2k–1} – 13)

Here, the last 2 digits always end with 60 in RHS. So, not possible for n = 6, 8, 10, …..

∴ n = 2 and 4

⇒ Sum of n= 6

**Question 14. The product 55 × 60 × 65 is written as the product of five distinct positive integers. What is the least possible value of the largest of these integers?**

Answer: 20

55 = 5 × 11

60 = 5 × 12 = 5 × 2 × 2 × 3

65 = 5 × 13

∴ 55 × 60 × 65

= 5 × 11 × 5 × 2 × 2 × 3 × 5 × 13

= 11 × 25 × 12 × 13 × 5

or

2 × 15 × 10 × 10 × 13 × 5

or

5 × 11 × × 15 × 13

So, the least possible value of the largest of these integers is 20.

**Question 15. Three couples sit for a photograph in 2 rows of three people each such that no couple is sitting in the same row next to each other or in the same column one behind the other. How many arrangements are possible?**

Answer: 96

6 places

6 × 2 × 4 × 1 × 2 × 1 = 96

Alternate Solution :

Case I

Wives and Husbands in different rows.

Derangements of three couples = 3! (1–1 / 1! + 1 / 2! – 1 / 3! = 2

Now arrangements = (3! × 2) × 2 = 24 ways

{Arrangement of husbands in rows = 3!, the arrangement of wives = 2!, rows can be exchanged in 2 ways}

Case II

Two husbands and wives in one row.

Let H_{1}, H_{2}, W_{3 }be in the same row, these can be selected in 3 ways.

Now, again derangements will happen in 2 ways and arrangements = (3! × 2) × 2 = 24 (as in case I)

Total arrangements of case II = 3 × 24 = 72

Therefore, total number of arrangements = 24 + 72 = 96.

**Question 16. The sides x and y of a scalene triangle satisfy x + 2Δ / x = y + 2Δ / y, where Δ is the area of the triangle. If x = 60, y = 63, what is the length of the largest side of the triangle?**

Answer: 87

x^{2}y + 2Δy = xy^{2} + 2Δx

xy(x – y) = 2Δ(x – y)

xy = 2Δ

Let z be the side of a triangle.

Also, area (Δ) = ½ xysin z

Δ = ½ .2Δ.sin z

⇒ sin z = 1

⇒ z = 90°

Right angle triangle

Hence, z = √x^{2} + y^{2}

= √60^{2} + 63^{2}

z = 87

where z is the side of a triangle.

**Question 17. How many two digit numbers have exactly 4 positive factors? (Here 1 and the number n are also considered as factors of n.)**

Answer: 30

It will be either of the form a^{3} or a × b (where a and b both are prime numbers)

Now, two digits cubes = 27, (only)

And of the a × b form

2 and (5 to 47) = total primes = 13

3 and (5 to 31) = total primes = 9

5 and (7 to 19) = total primes = 5

And 7 and (11, 13) = total primes = 2

∴Total number = 13 + 9 + 5 + 2 + 1 = 30

**Question 18. If **

**\(\begin{array}{l}\sum_{k=1}^{40}\sqrt{1+\frac{1}{k^{2}}+\frac{1}{(k+1)^{2}}} = a + \frac{b}{c}\end{array} \)**

**where a, b, c ∈N, b < c, gcd (b, c) = 1, then what is the value of a + b?**

Answer: 80

From this, put k = 1, 2, 3 ….40 we get

We get,

a = 40, b = 40, c = 41

a + b = 80

**Question 19. Let ABCD be a parallelogram. Let E and F be midpoints of AB and BC respectively. The lines EC and FD intersect in P and form four triangles APB, BPC, CPD and DPA. If the area of the parallelogram is 100 sq. units, what is the maximum area in sq. units of a triangle among these four triangles?**

Answer: 40

a + b = 25

b + c = 25

2a + 2b + c = d = 100

ΔEQA ≅ ΔEBC

⇒ [ECQ] = a + 2b

ΔFPC ~ ΔPDQ

⇒ Area of ΔFCP = h_{1}x / 2

h_{1} / h_{2} = x / 4x = ¼

⇒ h_{2} = ⅕ h

⇒ [FCP] = hx / 10 = 2hx / 20 × 100 / 20 = 5

⇒ a = 15

b = 5

c = 20

d = 100 – 2a – 2b – c = 40

**Question 20. A group of women working together at the same rate can build a wall in 45 hours. When the work started, all the women did not start working together. They joined the work over a period of time, one by one, at equal intervals. Once at work, each one stayed till the work was complete. If the first woman worked 5 times as many hours as the last woman, for how many hours did the first woman work.**

Answer: 75

Let total women who work together be N.

Now, N woman can complete the total work in 45 hours.

The amount of work completed by one woman in 1 hour = 1/45 N

We are given that :

First women worked 5 times as compared to the last woman.

Since, the last women would have worked an equal interval, the total number of women working is 4.

Output at x hours = x / 45N

On substituting N = 4,

Let the second woman join after x hours.

Output at x hours = x / 180

When third women joined, the output = 2x / 180 + x / 180

When fourth women joined, the output = 3x / 180 + 2x / 180 + x / 180

After x hours, since the 4^{th} woman joined, the wall will be built and ready.

Hence, the output = 5x / 180 + 4x / 180+ 3x / 180 + 2x / 180 + x / 180 = 1 (completion of wall)

12x = 180

x = 180 / 12 = 15 hours

Hereby we know, each woman joined after a time interval of 15 hours.

Hence, the time spent by the first woman = 5 × x = 75 hours.

**Question 21. A total fixed amount of N thousand rupees is given to three persons A, B, C, every year, each being given an amount proportional to her age. In the first year, A got half the total amount. When the sixth payment was made, A got six-seventh of the amount that she had in the first year; B got Rs. 1000 less than that she had in the first year; and C got twice of that she had in the first year. Find N.**

Answer: 35

Let their age at beginning be A, B, C

Given : A received half of the total amount

⇒ a / A + B+ C × |N| = |N| / 2

⇒ 2A = A + B + C ⇒ A = B + C …(i)

In sixth payment,

A got 6 / 7 of amount she received in the first year

7A + 35 = 3A + 3B + 3 C + 45

4A – 3 (B + C) = 10

From equation (1),

4A – 3 (A) = 10 ⇒ A = 10 years

C received twice the amount it received in the first year.

20C + 100 = 70 C ⇒ 50C = 100 ⇒ C = 2

B = 8

In the sixth payment, B received 1000 less than she got in the first year.

65 N = 70 N – 35 × 5000

5 N = 35 × 5000

N = 35000

Therefore, the answer is 35.

**Question 22. In triangle ABC, let P and R be the feet of the perpendicular from A onto the external and internal bisectors of ∠ABC, respectively; and let Q and S be the feet of the perpendiculars from A onto the internal and external bisectors of ∠ACB, respectively. If PQ = 7, QR = 6 and RS = 8, what is the area of triangle ABC?**

Answer: 84

Let M, N be the midpoints of AB, AC.

Join PM and NS

We have,

M is circumcentre of right angled ΔAPB

⇒ ∠MAP = ∠MPA = ∠B/2 (obviously)

∵ ∠MBP = 𝜋 / 2 – ∠B / 2 and ∠MPB = 𝜋 / 2 – ∠B / 2

∴ ∠BMP = π – (∠MBP + ∠MPB)

= 𝜋 – (𝜋 / 2 – ∠B / 2 + 𝜋 / 2 – ∠B / 2)

∠BMP = ∠B = ∠MBC

⇒ PM || BC

Similarly, NS || BC and MN || BC.

⇒ P, M, N, S are collinear.

Similarly, we can see MR || BC and NQ || BC

∴ M, Q, R, N are collinear.

Hence, PMQRNS is a straight line.

∵ AC = b

⇒ CS = b sin c / 2

CQ = b cos c / 2

Now in right angled triangle CQS

∴ QS = √CS^{2} + CQ^{2}

= √(b sin C/2)^{2} + (c/2)^{2}

QS = b = QR + RS

b = 6 + 8 = 14 = AC

Similarly, c = 7 + 6 = 13

a = 7 + 8 = 15

Using Heron’s formula,

Area = √ 21 × 7 × 8 × 6 = 84

**Question 23. The incircle Γ of a scalene triangle ABC touches BC at D, CA at E and AB at F. Let r _{A} be the radius of the circle inside ABC which is tangent to Γ and the sides AB and AC. Define r_{B} and r_{C} similarly. If r_{A} = 16, r_{B} = 25 and r_{C} = 36, determine the radius of Γ.**

Answer: 74

r_{A} = 16

r_{B} = 25

r_{C} = 36

R is radius of incircle Γ

As we know,

r = √r_{A}.r_{B} + √r_{B}.r_{C} + √r_{C}.r_{A}

= √16.25 + √25.36 + √36.16

= 20 + 30 + 24

= 74

**Question 24. A light source at the point (0, 16) in the coordinate plane casts light in all the directions. A disc (a circle along with its interior) of radius 2 with center at (6, 10) casts a shadow on the X axis. The length of the shadow can be written in the form m √n where m, n are positive integers and n is square-free. Find m+n.**

Answer: 21

Equation of rays through P (0, 16) with slope m are

y – 16 = mx …(i)

Since, it is tangent to circle,

∴

4 (1+m^{2}) = |6 + 6m^{2}| = 36 (1 + m^{2} + 2m)

9m^{2} + 18m + 9 = m^{2} + 1

8m^{2} + 18m + 8 = 0

4m^{2} + 9m + 4 = 0 …(ii)

The line (i) cuts the x-axis at x = -16 / m

∴ Length of shadow, AB =

=

= 4 √ 17

Therefore, m + n = 21

**Part-A-2**

**Question 25. For a positive integer n, let ⟨n⟩ denote the perfect square integer closest to n. For example, ⟨74⟩ = 81, ⟨18⟩ = 16. If N is the smallest positive integer such that **

⟨91⟩.⟨120⟩.⟨143⟩.⟨180⟩.⟨N⟩ = 91.120.143.180.N

Find the sum of the squares of the digits of N.

Answer: 56

⟨91⟩.⟨120⟩.⟨143⟩.⟨180⟩.⟨N⟩ = 91.120.143.180.N

⇒ 100.121.144.169⟨N⟩ = 91.120.143.180.N

⇒

⇒

⇒ 22〈N〉 = 21.N

22 × (462) = 21 × (462)

Ν = 462

Sum of squares of digits of

N = 4^{2} + 6^{2} + 2^{2} = 56

**Question 26. In the figure given below, 4 of the 6 disks are to be colored black and 2 are to be colored white. Two colourings that can be obtained from one another by a rotation or a reflection of the entire figure are considered the same.**

**There are only four such colourings for the given two colors, as shown in figure 1. In how many ways can we colour the 6 disks such that 2 are coloured black, 2 are colored white, 2 are colored blue with the given condition of identification?**

Answer: 18

First 2 disks can be colored black in 4 ways.

Now, in (I) two disks can be coloured white and two blue in 4! / 2!2! = 6 ways.

In (II), (III) and (IV) the two more colors can be put in the following ways.

Total number of ways = 6 + 12 = 18.

**Question 27. A bug travels in the coordinate plane moving only along the lines that are parallel to the x-axis or y-axis. Let A = (–3, 2) and B(3, –2). Consider all possible paths of the bug from A to B of length at most 14. How many points with integer coordinates lie on at least one of these paths.**

Answer: 87

|x – 3| + |x + 3| + |y – 2| + |y +2| ≤ 14

The bug can travel from A to B by only taking horizontal and vertical steps in total 10 steps only.

Therefore, the number of integer points = 35

But, since the maximum path length is 14, the bug can also go off track by maximum 2 units which gives rise to 28 + 20 more integer points.

There also 4 more integer points are lying at the corners where the bug can go circular.

Therefore, the total number of points = 35 + 28 + 20 + 4 = 87

**Question 28. A natural number n is said to be good if n is the sum of r consecutive positive integers, for some r > 2. Find the number of good numbers in the set {1, 2,…, 100}.**

Answer: 93

n = k + k + 1 +⋯ + k + r – 1

⇒ n = rk + (1 + 2 + 3+⋯ + r – 1)

⇒ n = rk + r(r-1 ) / 2

⇒ 2n = r(2k + r–1)

When r is odd, 2k + r – 1 is even and when r is even, 2k + r – 1 is odd.

The above equality can hold good for n (taking r ≥ 2) except numbers of the form 2^{k}

i.e. n = 2, 4, 8, 16, 32, 64.

Also, n cannot be equal to 1.

Therefore, the total number of good numbers n are 100 – 7 = 93.

**Question 29. Positive integers a, b, c satisfy ab / a – b = c. What is the largest possible value of a + b + c not exceeding 99?**

Answer: 99

Ab / a–b = c ⇒ a – b / ab = 1 / c

⇒ 1 / b – 1 / a = 1 / c

⇒ 1 / b = 1 / a + 1 / c

⇒ 1 / b = a + c / ac

⇒ ac = ab + bc

⇒ (a – b) (c – b) = b^{2} …(1)

∵ a + b + c ≤ 99,

we have

a > b and c > b

Possible value of a, b, c is

a = 27, b = 18, c = 54 which satisfy (1).

**Question 30. Find the number of pairs (a, b) of natural numbers such that b is a 3-digit number, a + 1 divides b – 1 and b divides a ^{2} + a + 2.**

Answer: 16

Since, b divides a^{2} + a + 2

∴ a^{ 2} + a + 2 = bk, where k ∈ I

∵ (a + 1) divides (b – 1)

(a + 1) ℓ = b – 1, where ℓ ∈ I

∴ (a + 1) ℓ + 1 = b

a^{2} + a + 2 = ((a + 1) ℓ + 1)k

a^{2} + a + 2 = ℓk((a + 1) + k

a(a + 1) + 2 = ℓk(a + 1) + k

On comparing,

k = 2, kℓ = a ⇒ 2ℓ = a

⇒ b = a (a + 1) / 2 + 1

b is a three digit number when a = 14, 15, ….., 44

(∵ a = 2ℓ, a must be an even number)

So, a = 14, 16, 18, …. , 44

∴ Hence, number of pairs (a, b) = 16.

**
**

## Comments