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Samacheer Kalvi 10th Maths Exercise 2.2 Solutions

10th Maths Exercise 2.2 solutions (Chapter 2 Numbers and Sequences) with step-by-step answers for all 9 questions, as per the 2026-27 Samacheer Kalvi textbook.

Answers at a glance9 answers · full steps below
  1. Q14n4^n ends with the digit 6 for all even natural numbers nn (i.e. n=2,4,6,…n=2,4,6,\dots)
  2. Q2There is no value of mm; 2n×5m2^n\times 5^m always ends in 00, never in 55
  3. Q3HCF =10101=10101
  4. Q4a=9,  b=3a=9,\;b=3
  5. Q5p1=2,  p2=3,  p3=5,  p4=7p_1=2,\;p_2=3,\;p_3=5,\;p_4=7 and x1=3,  x2=4,  x3=2,  x4=1x_1=3,\;x_2=4,\;x_3=2,\;x_4=1
  6. Q6HCF =34=34, LCM =2040=2040
  7. Q7999720999720
  8. Q836473647
  9. Q925202520

Q1. For what values of natural number nn , 4n4^n can end with …

For what values of natural number nn, 4n4^n can end with the digit 6?

  1. 4n=(22)n=22n4^n=(2^2)^n=2^{2n}, so 4n4^n is always even.

  2. List the last digits: 41=4,  42=16,  43=64,  44=256,…4^1=4,\;4^2=16,\;4^3=64,\;4^4=256,\dots

  3. The last digit is 66 for n=2,4,6,…n=2,4,6,\dots and 44 for n=1,3,5,…n=1,3,5,\dots

  4. So the unit digit is 66 exactly when nn is even.

Answer

4n4^n ends with the digit 6 for all even natural numbers nn (i.e. n=2,4,6,…n=2,4,6,\dots)

Q2. If mm , nn are natural numbers, for what values of mm …

If mm, nn are natural numbers, for what values of mm, does 2n×5m2^n \times 5^m ends in 5?

  1. For natural numbers m,nm,n the number 2n×5m2^n\times 5^m has 22 as a factor.

  2. So 2n×5m2^n\times 5^m is always an even number.

  3. Also 55 is a factor, so the number is divisible by 1010 and hence ends in 00.

  4. A number ending in 55 is odd, which is impossible here.

Answer

There is no value of mm; 2n×5m2^n\times 5^m always ends in 00, never in 55

Q3. Find the HCF of 252525252525 and 363636363636 .

  1. 252525=10101×25252525=10101\times 25

  2. 363636=10101×36363636=10101\times 36

  3. 10101=3×7×13×3710101=3\times 7\times 13\times 37

  4. 25=5225=5^2 and 36=22×3236=2^2\times 3^2 have no common factor, so HCF(25,36)=1(25,36)=1.

  5. Hence HCF =10101×1=10101\times 1.

Answer

HCF =10101=10101

Q4. If 13824=2a×3b13824 = 2^a \times 3^b then find aa and bb .

  1. Factorise 1382413824 into primes:

  2. 13824=2×6912=22×3456=23×1728=24×864=25×43213824=2\times 6912=2^2\times 3456=2^3\times1728=2^4\times864=2^5\times432

  3. =26×216=27×108=28×54=29×27=2^6\times216=2^7\times108=2^8\times54=2^9\times27

  4. 27=3327=3^3, so 13824=29×3313824=2^9\times 3^3.

  5. Comparing with 2a×3b2^a\times 3^b gives a=9,  b=3a=9,\;b=3.

Answer

a=9,  b=3a=9,\;b=3

Q5. If p1x1×p2x2×p3x3×p4x4=113400p_1^{x_1} \times p_2^{x_2} \times p_3^{x_3} \times p_4^{x_4} = 113400 where p1,p2,p3,p4p_1, p_2, p_3, p_4 are primes in ascending order and x1,x2,x3,x4x_1, x_2, x_3, x_4 are …

If p1x1×p2x2×p3x3×p4x4=113400p_1^{x_1} \times p_2^{x_2} \times p_3^{x_3} \times p_4^{x_4} = 113400 where p1,p2,p3,p4p_1, p_2, p_3, p_4 are primes in ascending order and x1,x2,x3,x4x_1, x_2, x_3, x_4 are integers, find the value of p1,p2,p3,p4p_1, p_2, p_3, p_4 and x1,x2,x3,x4x_1, x_2, x_3, x_4.

Show steps (6)

  1. Factorise 113400113400:

  2. 113400=23×14175113400=2^3\times 14175

  3. 14175=34×17514175=3^4\times 175

  4. 175=52×7175=5^2\times 7

  5. So 113400=23×34×52×71113400=2^3\times 3^4\times 5^2\times 7^1.

  6. Comparing with p1x1×p2x2×p3x3×p4x4p_1^{x_1}\times p_2^{x_2}\times p_3^{x_3}\times p_4^{x_4} (primes in ascending order).

Answer

p1=2,  p2=3,  p3=5,  p4=7p_1=2,\;p_2=3,\;p_3=5,\;p_4=7 and x1=3,  x2=4,  x3=2,  x4=1x_1=3,\;x_2=4,\;x_3=2,\;x_4=1

Q6. Find the LCM and HCF of 408408 and 170170 by applying the …

Find the LCM and HCF of 408408 and 170170 by applying the fundamental theorem of arithmetic.

  1. 408=23×3×17408=2^3\times 3\times 17

  2. 170=2×5×17170=2\times 5\times 17

  3. HCF == product of common primes with least powers =2×17=34=2\times 17=34

  4. LCM == product of all primes with highest powers =23×3×5×17=2040=2^3\times 3\times 5\times 17=2040

Answer

HCF =34=34, LCM =2040=2040

Q7. Find the greatest number consisting of 66 digits which is exactly divisible …

Find the greatest number consisting of 66 digits which is exactly divisible by 24,15,3624,15,36?

  1. 24=23×3,  15=3×5,  36=22×3224=2^3\times 3,\;15=3\times 5,\;36=2^2\times 3^2

  2. LCM =23×32×5=360=2^3\times 3^2\times 5=360

  3. The greatest 6-digit number is 999999999999.

  4. Divide: 999999÷360999999 \div 360 gives quotient 27772777 and remainder 279279.

  5. Required number =999999−279=999720=999999-279=999720.

Answer

999720999720

Q8. What is the smallest number that when divided by three numbers such …

What is the smallest number that when divided by three numbers such as 3535, 5656 and 9191 leaves remainder 77 in each case?

  1. The number leaves remainder 77 in each case, so (number −7-7) is divisible by 35,56,9135,56,91.

  2. 35=5×7,  56=23×7,  91=7×1335=5\times 7,\;56=2^3\times 7,\;91=7\times 13

  3. LCM =23×5×7×13=3640=2^3\times 5\times 7\times 13=3640

  4. Smallest such number =3640+7=3647=3640+7=3647.

Answer

36473647

Q9. Find the least number that is divisible by the first ten natural …

Find the least number that is divisible by the first ten natural numbers.

  1. The first ten natural numbers are 1,2,3,4,5,6,7,8,9,101,2,3,4,5,6,7,8,9,10.

  2. Required number is their LCM.

  3. 4=22,  8=23,  9=32,  10=2×54=2^2,\;8=2^3,\;9=3^2,\;10=2\times5

  4. Highest powers: 23,  32,  5,  72^3,\;3^2,\;5,\;7

  5. LCM =8×9×5×7=2520=8\times 9\times 5\times 7=2520

Answer

25202520

Quick check

  1. If A=265A = 2^{65} and B=264+263+262+⋯+20B = 2^{64} + 2^{63} + 2^{62} + \cdots + 2^{0} which of the following is true?

    1. (1) BB is 2642^{64} more than AA
    2. (2) AA and BB are equal
    3. (3) BB is larger than AA by 11
    4. (4) AA is larger than BB by 11
    Show answer

    Answer: (4) AA is larger than BB by 11

    1. B=20+21+⋯+264B = 2^0 + 2^1 + \cdots + 2^{64} is a G.P. with a=1a = 1, r=2r = 2, n=65n = 65.

    2. B=1(265−1)2−1=265−1B = \dfrac{1(2^{65} - 1)}{2 - 1} = 2^{65} - 1.

    3. A−B=265−(265−1)=1A - B = 2^{65} - (2^{65} - 1) = 1.

    Question 1 of 5

  2. The sum of the exponents of the prime factors in the prime factorization of 17291729 is

    1. (1) 11
    2. (2) 22
    3. (3) 33
    4. (4) 44
    Show answer

    Answer: (3) 33

    1. 1729=7×13×19=71×131×1911729 = 7 \times 13 \times 19 = 7^1 \times 13^1 \times 19^1.

    2. Sum of exponents =1+1+1=3= 1 + 1 + 1 = 3.

    Question 2 of 5

  3. If 66 times of 6th6^{\text{th}} term of an A.P. is equal to 77 times the 7th7^{\text{th}} term, then the 13th13^{\text{th}} term of the A.P. is

    1. (1) 00
    2. (2) 66
    3. (3) 77
    4. (4) 1313
    Show answer

    Answer: (1) 00

    1. 6t6=7t7⇒6(a+5d)=7(a+6d)6t_6 = 7t_7 \Rightarrow 6(a+5d) = 7(a+6d).

    2. 6a+30d=7a+42d⇒a+12d=06a + 30d = 7a + 42d \Rightarrow a + 12d = 0.

    3. t13=a+12d=0t_{13} = a + 12d = 0.

    Question 3 of 5

  4. An A.P. consists of 3131 terms. If its 16th16^{\text{th}} term is mm, then the sum of all the terms of this A.P. is

    1. (1) 16 m16\,m
    2. (2) 62 m62\,m
    3. (3) 31 m31\,m
    4. (4) 312 m\dfrac{31}{2}\,m
    Show answer

    Answer: (3) 31 m31\,m

    1. t16=a+15d=mt_{16} = a + 15d = m.

    2. S31=312 [2a+30d]=312×2(a+15d)S_{31} = \dfrac{31}{2}\,[2a + 30d] = \dfrac{31}{2} \times 2(a + 15d).

    3. =31(a+15d)=31m= 31(a+15d) = 31m.

    Question 4 of 5

  5. Using Euclid's division lemma, if the cube of any positive integer is divided by 99 then the possible remainders are

    1. (1) 0,1,80, 1, 8
    2. (2) 1,4,81, 4, 8
    3. (3) 0,1,30, 1, 3
    4. (4) 1,3,51, 3, 5
    Show answer

    Answer: (1) 0,1,80, 1, 8

    1. Any positive integer is of the form 3q3q, 3q+13q+1 or 3q+23q+2.

    2. (3q)3=27q3=9(3q3)⇒(3q)^3 = 27q^3 = 9(3q^3) \Rightarrow remainder 00.

    3. (3q+1)3=27q3+27q2+9q+1=9(3q3+3q2+q)+1⇒(3q+1)^3 = 27q^3+27q^2+9q+1 = 9(3q^3+3q^2+q)+1 \Rightarrow remainder 11.

    4. (3q+2)3=27q3+54q2+36q+8=9(3q3+6q2+4q)+8⇒(3q+2)^3 = 27q^3+54q^2+36q+8 = 9(3q^3+6q^2+4q)+8 \Rightarrow remainder 88.

    Question 5 of 5
Chapter 2 formulas
  • Euclid's division lemma

    a=bq+r, 0≤r<∣b∣a = bq + r,\ 0 \le r < |b|

  • Fundamental theorem of arithmetic

    N=p1a1p2a2p3a3⋯pkak (unique except for order)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, …a,\ a+d,\ a+2d,\ a+3d,\ \ldots

  • n-th term of an A.P.

    tn=a+(n−1)dt_n = a + (n-1)d

  • Sum to first n terms of an A.P.

    Sn=n2[2a+(n−1)d]S_n = \frac{n}{2}\left[2a + (n-1)d\right]

  • Sum of an A.P. when the last term l is given

    Sn=n2[a+l]S_n = \frac{n}{2}\left[a + l\right]

  • Geometric Progression

    a, ar, ar2, …, arn−1a,\ ar,\ ar^2,\ \ldots,\ ar^{n-1}

  • n-th term of a G.P.

    tn=arn−1t_n = ar^{n-1}

  • Sum to first n terms of a G.P. (r ≠ 1)

    Sn=a(rn−1)r−1S_n = \frac{a(r^n - 1)}{r - 1}

  • Sum to first n terms of a G.P. when r = 1

    Sn=naS_n = na

  • Sum to infinite terms of a G.P. (−1 < r < 1)

    S∞=a1−rS_\infty = \frac{a}{1 - r}

  • Sum of first n natural numbers

    1+2+3+⋯+n=n(n+1)21 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}

  • Sum of squares of first n natural numbers

    12+22+32+⋯+n2=n(n+1)(2n+1)61^2 + 2^2 + 3^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}

  • Sum of cubes of first n natural numbers

    13+23+33+⋯+n3=[n(n+1)2]21^3 + 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+⋯+(2n−1)=n21 + 3 + 5 + \cdots + (2n-1) = n^2

Answers checked against the Samacheer Kalvi textbook · Updated

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