Samacheer Kalvi 10th Maths Exercise 2.1 Solutions

10th Maths Exercise 2.1 solutions (Chapter 2 Numbers and Sequences) with step-by-step answers for all 13 questions, as per the 2026-27 Samacheer Kalvi textbook.

Answers at a glance9 answers · full steps below

Q1All positive integers of the form $3q+2$ , that is $2, 5, 8, 11, 14, \ldots$
Q2Number of completed rows $= 25$ and $7$ flower pots are left over.
Q6 (i)HCF $(340, 412) = 4$
Q6 (ii)HCF $(867, 255) = 51$
Q6 (iii)HCF $(10224, 9648) = 144$
Q6 (iv)HCF $(84, 90, 120) = 6$
Q7The largest number is $174$ .
Q8 $d = 4$ , $x = 2$ , $y = -1$
Q9Remainder $= 6$

Q1. Find all positive integers, when divided by $3$ leaves remainder $2$ .

By Euclid's division lemma, $a = 3q + r$ where $0 \le r < 3$ .
Here the remainder is $r = 2$ .
So the numbers are of the form $a = 3q + 2$ , where $q = 0,1,2,3,\ldots$
Putting $q = 0,1,2,3,\ldots$ we get $2, 5, 8, 11, 14, \ldots$

Answer

All positive integers of the form $3q+2$ , that is $2, 5, 8, 11, 14, \ldots$

Q2. A man has $532$ flower pots. He wants to arrange them in …

A man has $532$ flower pots. He wants to arrange them in rows such that each row contains $21$ flower pots. Find the number of completed rows and how many flower pots are left over.

Here $a = 532$ (total pots), $b = 21$ (pots in each row).
By Euclid's division lemma, $a = bq + r$ , $0 \le r < b$ .
$532 = 21 \times 25 + 7$
So quotient $q = 25$ and remainder $r = 7$ .

Answer

Number of completed rows $= 25$ and $7$ flower pots are left over.

Q3. Prove that the product of two consecutive positive integers is divisible by …

Prove that the product of two consecutive positive integers is divisible by $2$ .

Show steps (8)

Let the two consecutive positive integers be $n$ and $n+1$ .
Their product is $n(n+1)$ .
Case (i): $n$ is even, so $n = 2k$ for some integer $k$ .
Then $n(n+1) = 2k(n+1) = 2m$ , where $m = k(n+1)$ .
Case (ii): $n$ is odd, so $n = 2k+1$ for some integer $k$ .
Then $n+1 = 2k+2 = 2(k+1)$ .
So $n(n+1) = (2k+1)\cdot 2(k+1) = 2m$ , where $m = (2k+1)(k+1)$ .
In both cases the product is of the form $2m$ .

Answer

The product of two consecutive positive integers is always divisible by $2$ .

Q4. When the positive integers $a$ , $b$ and $c$ are divided by …

When the positive integers $a$ , $b$ and $c$ are divided by $13$ , the respective remainders are $9,7$ and $10$ . Show that $a+b+c$ is divisible by $13$ .

Show steps (6)

By Euclid's division lemma,
$a = 13q_1 + 9$ , $b = 13q_2 + 7$ , $c = 13q_3 + 10$ .
Adding, $a+b+c = 13(q_1+q_2+q_3) + 26$
$= 13(q_1+q_2+q_3) + 13 \times 2$
$= 13(q_1+q_2+q_3+2)$
This is a multiple of $13$ .

Answer

$a+b+c = 13(q_1+q_2+q_3+2)$ , hence $a+b+c$ is divisible by $13$ .

Q5. Prove that square of any integer leaves the remainder either $0$ or …

Prove that square of any integer leaves the remainder either $0$ or $1$ when divided by $4$ .

Let $x$ be any integer. Then $x$ is either even or odd.
Case (i): $x$ is even, so $x = 2k$ .
$x^2 = 4k^2 = 4q + 0$ , where $q = k^2$ . Remainder is $0$ .
Case (ii): $x$ is odd, so $x = 2k+1$ .
$x^2 = 4k^2 + 4k + 1 = 4(k^2+k) + 1 = 4q + 1$ , where $q = k^2+k$ . Remainder is $1$ .

Answer

The square of any integer is of the form $4q$ or $4q+1$ , so the remainder on division by $4$ is either $0$ or $1$ .

Q6. Use Euclid's Division Algorithm to find the Highest Common Factor (HCF) of

(i)

$340$ and $412$

Show steps (8)

$412 = 340 \times 1 + 72$
$340 = 72 \times 4 + 52$
$72 = 52 \times 1 + 20$
$52 = 20 \times 2 + 12$
$20 = 12 \times 1 + 8$
$12 = 8 \times 1 + 4$
$8 = 4 \times 2 + 0$
The remainder is zero, so the last divisor $4$ is the HCF.

Answer

HCF $(340, 412) = 4$

(ii)

$867$ and $255$

$867 = 255 \times 3 + 102$
$255 = 102 \times 2 + 51$
$102 = 51 \times 2 + 0$
The remainder is zero, so the last divisor $51$ is the HCF.

Answer

HCF $(867, 255) = 51$

(iii)

$10224$ and $9648$

$10224 = 9648 \times 1 + 576$
$9648 = 576 \times 16 + 432$
$576 = 432 \times 1 + 144$
$432 = 144 \times 3 + 0$
The remainder is zero, so the last divisor $144$ is the HCF.

Answer

HCF $(10224, 9648) = 144$

(iv)

$84$ , $90$ and $120$

First find the HCF of $84$ and $90$ .
$90 = 84 \times 1 + 6$
$84 = 6 \times 14 + 0$ , so HCF $(84, 90) = 6$ .
Now find the HCF of $6$ and $120$ .
$120 = 6 \times 20 + 0$ , so HCF $(6, 120) = 6$ .

Answer

HCF $(84, 90, 120) = 6$

Q7. Find the largest number which divides $1230$ and $1926$ leaving remainder $12$ …

Find the largest number which divides $1230$ and $1926$ leaving remainder $12$ in each case.

Show steps (8)

Since the remainder is $12$ in each case, the required number divides
$1230 - 12 = 1218$ and $1926 - 12 = 1914$ .
So we find the HCF of $1218$ and $1914$ .
$1914 = 1218 \times 1 + 696$
$1218 = 696 \times 1 + 522$
$696 = 522 \times 1 + 174$
$522 = 174 \times 3 + 0$
The last divisor is $174$ .

Answer

The largest number is $174$ .

Q8. If $d$ is the Highest Common Factor of $32$ and $60$ …

If $d$ is the Highest Common Factor of $32$ and $60$ , find $x$ and $y$ satisfying $d = 32x + 60y$ .

Show steps (9)

Find the HCF of $32$ and $60$ by Euclid's Division Algorithm.
$60 = 32 \times 1 + 28$
$32 = 28 \times 1 + 4$
$28 = 4 \times 7 + 0$
So $d = 4$ .
Write $4$ in the form $32x + 60y$ :
$4 = 32 - 28 = 32 - (60 - 32)$
$4 = 32 \times 2 + 60 \times (-1)$
Comparing with $d = 32x + 60y$ , we get $x = 2$ , $y = -1$ .

Answer

$d = 4$ , $x = 2$ , $y = -1$

Q9. A positive integer when divided by $88$ gives the remainder $61$ . …

A positive integer when divided by $88$ gives the remainder $61$ . What will be the remainder when the same number is divided by $11$ ?

Let the positive integer be $a$ . Then $a = 88q + 61$ for some integer $q$ .
Since $88 = 11 \times 8$ and $61 = 11 \times 5 + 6$ ,
$a = 11 \times 8q + 11 \times 5 + 6$
$a = 11(8q + 5) + 6$
So when $a$ is divided by $11$ , the remainder is $6$ .

Answer

Remainder $= 6$

Q10. Prove that two consecutive positive integers are always coprime.

Show steps (6)

Let the two consecutive positive integers be $n$ and $n+1$ .
Let $d$ be their Highest Common Factor.
Then $d$ divides $n$ and $d$ divides $n+1$ .
So $d$ divides their difference $(n+1) - n = 1$ .
The only positive divisor of $1$ is $1$ , so $d = 1$ .
Since their HCF is $1$ , they are coprime.

Answer

HCF of $n$ and $n+1$ is $1$ , so two consecutive positive integers are always coprime.

Quick check

If $A = 2^{65}$ and $B = 2^{64} + 2^{63} + 2^{62} + \cdots + 2^{0}$ which of the following is true?
1. (1) $B$ is $2^{64}$ more than $A$
2. (2) $A$ and $B$ are equal
3. (3) $B$ is larger than $A$ by $1$
4. (4) $A$ is larger than $B$ by $1$
Show answer
Answer: (4) $A$ is larger than $B$ by $1$
1. $B = 2^0 + 2^1 + \cdots + 2^{64}$ is a G.P. with $a = 1$ , $r = 2$ , $n = 65$ .
2. $B = \dfrac{1(2^{65} - 1)}{2 - 1} = 2^{65} - 1$ .
3. $A - B = 2^{65} - (2^{65} - 1) = 1$ .
Question 1 of 5
The sum of the exponents of the prime factors in the prime factorization of $1729$ is
1. (1) $1$
2. (2) $2$
3. (3) $3$
4. (4) $4$
Show answer
Answer: (3) $3$
1. $1729 = 7 \times 13 \times 19 = 7^1 \times 13^1 \times 19^1$ .
2. Sum of exponents $= 1 + 1 + 1 = 3$ .
Question 2 of 5
If $6$ times of $6^{\text{th}}$ term of an A.P. is equal to $7$ times the $7^{\text{th}}$ term, then the $13^{\text{th}}$ term of the A.P. is
1. (1) $0$
2. (2) $6$
3. (3) $7$
4. (4) $13$
Show answer
Answer: (1) $0$
1. $6t_6 = 7t_7 \Rightarrow 6(a+5d) = 7(a+6d)$ .
2. $6a + 30d = 7a + 42d \Rightarrow a + 12d = 0$ .
3. $t_{13} = a + 12d = 0$ .
Question 3 of 5
An A.P. consists of $31$ terms. If its $16^{\text{th}}$ term is $m$ , then the sum of all the terms of this A.P. is
1. (1) $16\,m$
2. (2) $62\,m$
3. (3) $31\,m$
4. (4) $\dfrac{31}{2}\,m$
Show answer
Answer: (3) $31\,m$
1. $t_{16} = a + 15d = m$ .
2. $S_{31} = \dfrac{31}{2}\,[2a + 30d] = \dfrac{31}{2} \times 2(a + 15d)$ .
3. $= 31(a+15d) = 31m$ .
Question 4 of 5
Using Euclid's division lemma, if the cube of any positive integer is divided by $9$ then the possible remainders are
1. (1) $0, 1, 8$
2. (2) $1, 4, 8$
3. (3) $0, 1, 3$
4. (4) $1, 3, 5$
Show answer
Answer: (1) $0, 1, 8$
1. Any positive integer is of the form $3q$ , $3q+1$ or $3q+2$ .
2. $(3q)^3 = 27q^3 = 9(3q^3) \Rightarrow$ remainder $0$ .
3. $(3q+1)^3 = 27q^3+27q^2+9q+1 = 9(3q^3+3q^2+q)+1 \Rightarrow$ remainder $1$ .
4. $(3q+2)^3 = 27q^3+54q^2+36q+8 = 9(3q^3+6q^2+4q)+8 \Rightarrow$ remainder $8$ .
Question 5 of 5

Chapter 2 formulas

Euclid's division lemma
$a = bq + r,\ 0 \le r < |b|$
Fundamental theorem of arithmetic
$N = p_1^{a_1} p_2^{a_2} p_3^{a_3} \cdots p_k^{a_k}\ \text{(unique except for order)}$
Arithmetic Progression
$a,\ a+d,\ a+2d,\ a+3d,\ \ldots$
n-th term of an A.P.
$t_n = a + (n-1)d$
Sum to first n terms of an A.P.
$S_n = \frac{n}{2}\left[2a + (n-1)d\right]$
Sum of an A.P. when the last term l is given
$S_n = \frac{n}{2}\left[a + l\right]$
Geometric Progression
$a,\ ar,\ ar^2,\ \ldots,\ ar^{n-1}$
n-th term of a G.P.
$t_n = ar^{n-1}$
Sum to first n terms of a G.P. (r ≠ 1)
$S_n = \frac{a(r^n - 1)}{r - 1}$
Sum to first n terms of a G.P. when r = 1
$S_n = na$
Sum to infinite terms of a G.P. (−1 < r < 1)
$S_\infty = \frac{a}{1 - r}$
Sum of first n natural numbers
$1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}$
Sum of squares of first n natural numbers
$1^2 + 2^2 + 3^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}$
Sum of cubes of first n natural numbers
$1^3 + 2^3 + 3^3 + \cdots + n^3 = \left[\frac{n(n+1)}{2}\right]^2$
Sum of first n odd natural numbers
$1 + 3 + 5 + \cdots + (2n-1) = n^2$

Answers checked against the Samacheer Kalvi textbook · Updated 28 September 2026

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Q1. Find all positive integers, when divided by 333 leaves remainder 222 .

Q2. A man has 532532532 flower pots. He wants to arrange them in …

Q3. Prove that the product of two consecutive positive integers is divisible by …

Q4. When the positive integers aaa , bbb and ccc are divided by …

Q5. Prove that square of any integer leaves the remainder either 000 or …

Q6. Use Euclid's Division Algorithm to find the Highest Common Factor (HCF) of

(i)

(ii)

(iii)

(iv)

Q7. Find the largest number which divides 123012301230 and 192619261926 leaving remainder 121212 …

Q8. If ddd is the Highest Common Factor of 323232 and 606060 …

Q9. A positive integer when divided by 888888 gives the remainder 616161 . …

Q10. Prove that two consecutive positive integers are always coprime.

Quick check

Q1. Find all positive integers, when divided by $3$ leaves remainder $2$ .

Q2. A man has $532$ flower pots. He wants to arrange them in …

Q4. When the positive integers $a$ , $b$ and $c$ are divided by …

Q5. Prove that square of any integer leaves the remainder either $0$ or …

Q7. Find the largest number which divides $1230$ and $1926$ leaving remainder $12$ …

Q8. If $d$ is the Highest Common Factor of $32$ and $60$ …

Q9. A positive integer when divided by $88$ gives the remainder $61$ . …