Rd Sharma 2019 2020 for Class 8 Maths Chapter 5 - Playing With Numbers

Answer:

$$\begin{gathered} \mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that}\overline{ 35a64} \mathrm{is} \mathrm{a} \mathrm{multiple} \mathrm{of} 3. \\ \therefore (3+5+a+6+4) \mathrm{is} \mathrm{a} \mathrm{multiple} \mathrm{of} 3. \\ \therefore (a+18) \mathrm{is} \mathrm{a} \mathrm{multiple} \mathrm{of} 3. \\ \therefore (a+18)=0, 3, 6, 9, 12, 15, 18, 21... \\ \mathrm{But} a \mathrm{is} \mathrm{a} \mathrm{digit} \mathrm{of} \mathrm{number} \overline{ 35a64 } . \mathrm{So}, a \mathrm{can} \mathrm{take} \mathrm{value} 0, 1, 2, 3, 4...9. \\ a+18=18 \Rightarrow \mathrm{a}=0 \\ a+18=21 \Rightarrow \mathrm{a}=3 \\ a+18=24 \Rightarrow \mathrm{a}=6 \\ a+18=27 \Rightarrow \mathrm{a}=9 \\ \therefore a=0, 3, 6, 9 \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that} \overline{18\mathrm{x}71} \mathrm{is} \mathrm{a} \mathrm{multiple} \mathrm{of} 3. \\ \therefore (1+8+\mathrm{x}+7+1) \mathrm{is} \mathrm{a} \mathrm{multiple} \mathrm{of} 3. \\ \therefore (17+x) \mathrm{is} \mathrm{a} \mathrm{multiple} \mathrm{of} 3. \\ \therefore 17+x=0, 3, 6, 9, 12, 15, 18, 21... \\ \mathrm{But} x \mathrm{is} \mathrm{a} \mathrm{digit}. \mathrm{So}, \mathrm{x} \mathrm{can} \mathrm{take} \mathrm{values} 0,1,2,3,4...9. \\ 17+x=18 \Rightarrow \mathrm{x}=1 \\ 17+x=21 \Rightarrow \mathrm{x}=4 \\ 17+x=24 \Rightarrow \mathrm{x}=7 \\ x=1,4,7 \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that} \overline{ 66784\mathrm{x}} \mathrm{is} \mathrm{a} \mathrm{multiple} \mathrm{of} 9. \\ \mathrm{Therefore}, (6+6+7+8+4+\mathrm{x}) \mathrm{is} \mathrm{a} \mathrm{multiple} \mathrm{of} 9. \\ \mathrm{And}, \\ (31+\mathrm{x}) \mathrm{is} \mathrm{a} \mathrm{multiple} \mathrm{of} 9. \\ \mathrm{Possible} \mathrm{values} \mathrm{of} (31+\mathrm{x}) \mathrm{are} 0, 9, 18, 27, 36, 45,... \\ \mathrm{But} \mathrm{x} \mathrm{is} \mathrm{a} \mathrm{digit}. \mathrm{So}, \mathrm{x} \mathrm{can} \mathrm{only} \mathrm{take} \mathrm{value} 0, 1, 2, 3, 4,...9. \\ \therefore 31+\mathrm{x}=36 \\ \Rightarrow \mathrm{x}=36-31 \\ \Rightarrow \mathrm{x}=5 \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that} \overline{67\mathrm{y}19} \mathrm{is} \mathrm{a} \mathrm{multiple} \mathrm{of} 9. \\ \therefore (6+7+\mathrm{y}+1+9) \mathrm{is} \mathrm{a} \mathrm{multiple} \mathrm{of} 9. \\ \therefore (23+\mathrm{y}) \mathrm{is} \mathrm{a} \mathrm{multiple} \mathrm{of} 9. \\ 23+\mathrm{y}=0, 9, 18, 27, 36... \\ \mathrm{But} \mathrm{x} \mathrm{is} \mathrm{a} \mathrm{digit}. \mathrm{So}, \mathrm{x} \mathrm{can} \mathrm{take} \mathrm{values} 0,1,2,3,4...9. \\ 23+\mathrm{y}=27 \\ \Rightarrow \mathrm{y}=4 \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{Sum} \mathrm{of} \mathrm{the} \mathrm{digits} \mathrm{at} \mathrm{odd} \mathrm{places} = 3+2= 5 \\ \mathrm{Sum} \mathrm{of} \mathrm{the} \mathrm{digit} \mathrm{at} \mathrm{even} \mathrm{place} = x \\ \therefore \mathrm{Sum} \mathrm{of} \mathrm{the} \mathrm{digit} \mathrm{at} \mathrm{even} \mathrm{place}-\mathrm{Sum} \mathrm{of} \mathrm{the} \mathrm{digits} \mathrm{at} \mathrm{odd} \mathrm{places}=(x-5) \\ \because (\mathrm{x}-5) \mathrm{must} \mathrm{be} \mathrm{multiple} \mathrm{by} 11. \\ \therefore \mathrm{Possible} \mathrm{values} \mathrm{of} (x-5) \mathrm{are} 0, 11, 22, 33... \\ \mathrm{But} \mathrm{x} \mathrm{is} \mathrm{a} \mathrm{digit}; \mathrm{therefore} x \mathrm{must} \mathrm{be} 0, 1 ,2, 3...9. \\ \therefore x-5=0 \\ \Rightarrow x=5 \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{A} \mathrm{natural} \mathrm{number} \mathrm{is} \mathrm{divisible} \mathrm{by} 4 \mathrm{if} \mathrm{the} \mathrm{number} \mathrm{formed} \mathrm{by} \mathrm{its} \mathrm{digits} \mathrm{in} \mathrm{units} \mathrm{and} \mathrm{tens} \mathrm{places} \mathrm{is} \mathrm{divisible} \mathrm{by} 4. \\ \therefore \overline{98215x2} \mathrm{will} \mathrm{be} \mathrm{divisible} \mathrm{by} 4 \mathrm{if} \overline{x2} \mathrm{is} \mathrm{divisible} \mathrm{by} 4. \\ \therefore \overline{x2 }=10x + 2 \\ x \mathrm{is} \mathrm{a} \mathrm{digit}; \mathrm{therefore} \mathrm{possible} \mathrm{values} \mathrm{of} x \mathrm{are} 0, 1, 2, 3...9. \\ \overline{x2 }=2, 12, 22, 32, 42, 52, 62, 72, 82, 92 \\ \mathrm{The} \mathrm{numbers} \mathrm{that} \mathrm{are} \mathrm{divisible} \mathrm{by} 4 \mathrm{are} 12, 32, 52, 72, 92. \\ \mathrm{Therefore}, \mathrm{the} \mathrm{values} \mathrm{of} x \mathrm{are} 1, 3, 5, 7, 9. \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{A} \mathrm{number} \mathrm{is} \mathrm{divisible} \mathrm{by} 11, \mathrm{if} \mathrm{the} \mathrm{difference} \mathrm{of} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{its} \mathrm{digits} \mathrm{at} \mathrm{odd} \mathrm{places} \mathrm{and} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{its} \mathrm{digits} \mathrm{at} \mathrm{even} \mathrm{places} \mathrm{is} \mathrm{either} 0 \mathrm{or} \mathrm{a} \mathrm{multiple} \mathrm{of} 11. \\ \mathrm{Sum} \mathrm{of} \mathrm{digits} \mathrm{at} \mathrm{odd} \mathrm{places}-\mathrm{Sum} \mathrm{of} \mathrm{digits} \mathrm{at} \mathrm{even} \mathrm{places} \\ =(6 + \mathrm{x} + 9)-(7 + 1) \\ =(15 + \mathrm{x})-8=\mathrm{x} + 7 \\ \therefore \mathrm{x} + 7 =11 \\ \Rightarrow \mathrm{x} = 4 \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{If} \mathrm{a} \mathrm{natural} \mathrm{number} \mathrm{is} \mathrm{divided} \mathrm{by} 5, \mathrm{it} \mathrm{has} \mathrm{the} \mathrm{same} \mathrm{remainder} \mathrm{when} \mathrm{its} \mathrm{unit} \mathrm{digit} \mathrm{is} \mathrm{divided} \mathrm{by} 5. \\ \mathrm{Here}, \mathrm{the} \mathrm{unit} \mathrm{digit} \mathrm{of} 981547 \mathrm{is} 7. \mathrm{When} 7 \mathrm{is} \mathrm{divided} \mathrm{by} 5, \mathrm{remainder} \mathrm{is} 2. \\ \mathrm{Therefore}, \mathrm{remainder} \mathrm{will} \mathrm{be} 2 \mathrm{when} 981547 \mathrm{is} \mathrm{divided} \mathrm{by} 5. \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{Sum} \mathrm{of} \mathrm{the} \mathrm{digits} \mathrm{of} \mathrm{the} \mathrm{number} 51439786 = 5 + 1 + 4 + 3 + 9 + 7 + 8 + 6 = 43 \\ \mathrm{The} \mathrm{remainder} \mathrm{of} 51439786, \mathrm{when} \mathrm{divided} \mathrm{by} 3, \mathrm{is} \mathrm{the} \mathrm{same} \mathrm{as} \mathrm{the} \mathrm{remainder} \mathrm{when} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{the} \mathrm{digits} \mathrm{is} \mathrm{divided} 3. \\ \mathrm{When} 43 \mathrm{is} \mathrm{divided} \mathrm{by} 3, \mathrm{remainder} \mathrm{is} 1. \\ \mathrm{Therefore}, \mathrm{when} 51439786 \mathrm{is} \mathrm{divided} \mathrm{by} 3, \mathrm{remainder} \mathrm{will} \mathrm{be} 1. \end{gathered}$$

Answer:

$$\begin{gathered} 798 =\mathrm{A} \mathrm{multiple} \mathrm{of} 11 + (\mathrm{Sum} \mathrm{of} \mathrm{its} \mathrm{digits} \mathrm{at} \mathrm{odd} \mathrm{places} - \mathrm{Sum} \mathrm{of} \mathrm{its} \mathrm{digits} \mathrm{at} \mathrm{even} \mathrm{places}) \\ 798 =\mathrm{A} \mathrm{multiple} \mathrm{of} 11 + (7 + 8 - 9) \\ 798 =\mathrm{A} \mathrm{multiple} \mathrm{of} 11 + (15 - 9) \\ 798 =\mathrm{A} \mathrm{multiple} \mathrm{of} 11 + 6 \\ \mathrm{Therefore}, \mathrm{the} \mathrm{remainder} \mathrm{is} 6. \end{gathered}$$

Answer:

$$\begin{gathered} 928174653= \mathrm{A} \mathrm{multiple} \mathrm{of} 11 + (\mathrm{Sum} \mathrm{of} \mathrm{its} \mathrm{digits} \mathrm{at} \mathrm{odd} \mathrm{places} -\mathrm{Sum} \mathrm{of} \mathrm{its} \mathrm{digits} \mathrm{at} \mathrm{even} \mathrm{places}) \\ 928174653= \mathrm{A} \mathrm{multiple} \mathrm{of} 11 + \left\{\right(9 + 8 + 7 + 6 + 3) - (2 + 1 + 4 + 5\left)\right\} \\ 928174653= \mathrm{A} \mathrm{multiple} \mathrm{of} 11 + (33 - 12) \\ 928174653= \mathrm{A} \mathrm{multiple} \mathrm{of} 11 + 21 \\ 928174653= \mathrm{A} \mathrm{multiple} \mathrm{of} 11 + (11 \times 1 + 10) \\ 928174653= \mathrm{A} \mathrm{multiple} \mathrm{of} 11 + 10 \\ \mathrm{Therefore}, \mathrm{the} \mathrm{remainder} \mathrm{is} 10. \end{gathered}$$

Answer:

(i) 10
Every number with the structure (4n + 2) is an example of a number that is divisible by 2 but not by 4.

(ii) 15
Every number with the structure (6n + 3) is an example of a number that is divisible by 3 but not by 6.

(iii) 28
Every number with the structure (8n + 4) is an example of a number that is divisible by 4 but not by 8.

(iv) 8
Every number with the  structure (32n + 8), (32n + 16) or (32n + 24) is an example of a number that is divisible by 4 and 8 but not by 32.

Answer:

(i) False
Every number with the structures (9n + 3) or (9n + 6) is divisible by 3 but not by 9. Example: 3, 6, 12 etc.
(ii) True
(iii) False
Every number with the structure (8n + 4) is divisible by 4 but not by 8. Example: 4, 12, 20 etc.
(iv) True
(v) False
Example: 24 is divisible by both 3 and 6 but it is not divisible by 18.
(vi) True
(vii) False
Example: 5 divides 10, which is a sum of 3 and 7. However, it neither divides 3 nor 7.
(viii) True
(ix) False
Example: 4 and 9 are co-prime numbers but both are composite numbers too.
(x) True

Answer:

$$\begin{gathered} \mathrm{Two} \mathrm{possible} \mathrm{values} \mathrm{of} \mathrm{A} \mathrm{are}: \\ \left(\mathrm{i}\right) \mathrm{If} 7 + \mathrm{B} \le 9 \\ \therefore 3 + \mathrm{A} = 9 \\ \Rightarrow \mathrm{A} = 6 \\ \mathrm{But} \mathrm{if} \mathrm{A} = 6, 7 + \mathrm{B} \mathrm{must} \mathrm{be} \mathrm{larger} \mathrm{than} 9. \mathrm{Hence}, \mathrm{it} \mathrm{is} \mathrm{impossible}. \\ \left(\mathrm{ii}\right) \mathrm{If} 7 + \mathrm{B} \ge 9 \\ \therefore 1 + 3 + \mathrm{A} = 9 \\ \Rightarrow \mathrm{A} = 5 \\ \mathrm{If} \mathrm{A} = 5 \mathrm{and} 7 + \mathrm{B} = 5, \mathrm{B} \mathrm{must} \mathrm{be} 8 \\ \therefore \mathrm{A} = 5, \mathrm{B} = 8 \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{Two} \mathrm{possibilities} \mathrm{of} \mathrm{A} \mathrm{are}: \\ \left(\mathrm{i}\right) \mathrm{If} \mathrm{B} + 7 \le 9, \mathrm{A} = 6 \\ \mathrm{But} \mathrm{clearly}, \mathrm{if} \mathrm{A} = 6, \mathrm{B} + 7 \ge 9; \mathrm{it} \mathrm{is} \mathrm{impossible} \\ \left(\mathrm{ii}\right) \mathrm{If} \mathrm{B} + 7 \ge 9, \mathrm{A} = 5 \mathrm{and} \mathrm{B} + 7 = 5 \\ \mathrm{Clearly}, \mathrm{B} = 8 \\ \therefore \mathrm{A} = 5, \mathrm{B} = 8 \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{If} 1 + \mathrm{B} = 0 \\ \mathrm{Surely}, \mathrm{B} = 9 \\ \mathrm{If} 1 + \mathrm{A} + 1 = 9 \\ \mathrm{Surely}, \mathrm{A} = 7 \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{B} + 1 = 8, \mathrm{B} = 7 \\ \mathrm{A} + \mathrm{B} = 1, \mathrm{A} + 7 = 1, \mathrm{A} = 4 \\ \mathrm{So}, \mathrm{A} = 4, \mathrm{B} = 7 \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{A} + \mathrm{B} = 9 \mathrm{as} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{two} \mathrm{digits} \mathrm{can} \mathrm{never} \mathrm{be} 19 \\ 2 + \mathrm{A} = 0, \mathrm{A} \mathrm{must} \mathrm{be} 8 \\ \mathrm{A} + \mathrm{B} = 9, 8 + \mathrm{B} = 9, \mathrm{B} = 1 \\ \mathrm{So}, \mathrm{A} = 8, \mathrm{B} = 1 \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{If} \mathrm{A} + \mathrm{B} = 8, \mathrm{A} + \mathrm{B} \ge 9 \mathrm{is} \mathrm{possible} \mathrm{only} \mathrm{if} \mathrm{A} = \mathrm{B} = 9 \\ \mathrm{But} \mathrm{from} 7 + \mathrm{B} = \mathrm{A}, \mathrm{A} = \mathrm{B} = 9 \mathrm{is} \mathrm{impossible} \\ \mathrm{Surely}, \mathrm{A} + \mathrm{B} = 8, \mathrm{A} + \mathrm{B} \le 9 \\ \mathrm{So}, \mathrm{A} + 7 = 9, \mathrm{Surely} \mathrm{A} = 2 \\ 7 + \mathrm{B} = \mathrm{A}, 7 + \mathrm{B} = 2, \mathrm{B} = 5 \\ \mathrm{So}, \mathrm{A} = 2, \mathrm{B} = 5 \end{gathered}$$

Answer:

$$\begin{gathered} 0 \mathrm{is} \mathrm{the} \mathrm{only} \mathrm{unit} \mathrm{digit} \mathrm{number}, \mathrm{which} \mathrm{gives} \mathrm{the} \mathrm{same} 0 \mathrm{at} \mathrm{the} \mathrm{unit} \mathrm{digit} \mathrm{when} \mathrm{multiplied} \mathrm{by} 4. \mathrm{So}, \mathrm{the} \mathrm{possible} \mathrm{value} \mathrm{of} \mathrm{B} \mathrm{is} 0. \\ \mathrm{Similarly}, \mathrm{for} \mathrm{A} \mathrm{also}, 0 \mathrm{is} \mathrm{the} \mathrm{only} \mathrm{possible} \mathrm{digit}. \\ \mathrm{But} \mathrm{then} \mathrm{A}, \mathrm{B} \mathrm{and} \mathrm{C} \mathrm{will} \mathrm{all} \mathrm{be} 0. \\ \mathrm{And} \mathrm{if} \mathrm{A}, \mathrm{B} \mathrm{and} \mathrm{C} \mathrm{become} 0, \mathrm{these} \mathrm{numbers} \mathrm{cannot} \mathrm{be} \mathrm{of} \mathrm{two}-\mathrm{digit} \mathrm{or} \mathrm{three}-\mathrm{digit}. \\ \mathrm{Therefore}, \mathrm{both} \mathrm{will} \mathrm{become} \mathrm{a} \mathrm{one}-\mathrm{digit} \mathrm{number}. \\ \mathrm{Thus}, \mathrm{there} \mathrm{is} \mathrm{no} \mathrm{solution} \mathrm{possible}. \end{gathered}$$

Answer:

(i) Clearly, 69 and 96 are two numbers such that one can be obtained be reversing the digits of the other. Therefore, when the sum of 69 and 96 is divided by 11, we get 15 (sum of the digits) as quotient.

(ii) Clearly, 69 and 96 are two numbers such that one can be obtained be reversing the digits of the other. Therefore, when the sum of 69 and 96 is divided by 15 (sum of the digits), we obtain 11 as quotient.

Answer:

Answer:

$$\begin{gathered} \mathrm{The} \mathrm{sum} \mathrm{of} (985+859+598) \mathrm{when} \mathrm{divided} \mathrm{by}: \\ \left(\mathrm{i}\right) 111 \\ Q\mathrm{uotient} = (9+8+5)=22 \\ \left(\mathrm{ii}\right) 22, \mathrm{i}.\mathrm{e}, (9+8+5) \\ Q\mathrm{uotient} = 111 \\ \left(\mathrm{iii}\right) 37 (= \frac{111}{3}) \\ Q\mathrm{uotient} = 3(9+8+5)=66 \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{If}\overline{ \mathrm{abc}}-\overline{\mathrm{acb}} \mathrm{is} \mathrm{divided} \mathrm{by} 9, \mathrm{the} \mathrm{quotient} \mathrm{is} (\mathrm{b}-\mathrm{c}). \\ \therefore \mathrm{If} (985-958) \mathrm{is} \mathrm{divided} \mathrm{by} 9, \mathrm{quotient}=(8-5)=3 \end{gathered}$$