Question 11 Mark
$20x3$ is a multiple of $3$ if the digit $x$ is ______ or ______ or ______.
Answer$20x3$ is a multiple of $3$ if the digit $x$ is $1$ or $4$ or $7.$
Solution:
We know that, if a number is a multiple of $3,$ then the sum of its digits is again a multiple of $3, i.e.\ 2 + 0 + x + 3$ is a multiple of $3.\ x + 5 = 0, 3, 6, 9, 12, 15$ But, $x$ is a digit of the number $ 20\ x\ 3.\ x$ can take values $0, 1, 2, 3, 9. = x + 5 = 6$ or $9$ or $12$ Hence, $x = 1$ or $4$ or $7$
View full question & answer→Question 21 Mark
The difference of three-digit number and the number obtained by putting the digits in reverse order is always divisible by $9$ and ___________.
AnswerThe difference of three-digit number and the number obtained by putting the digits in reverse order is always divisible by $9$ and $11.$
Solution:
Let abc be a three-digit number, then we have
$abc - cba = (100a + 10b + c) - (100c + 10b + a)$
$= (100a - a) + (c - 10c) = 99a - 99c$
$= 99(a - c) = 9 \times 11 \times (a - c)$
Hence, $abc - cba$ is always divisible by $9,11$ and $(a - c).$
View full question & answer→Question 31 Mark
If $AB \times 4 = 192,$ then $A + B = 7.$
AnswerHere, $B \times 4$ is a two-digit number whose unit’s digit is $2.$
Therefore, the value of $B$ is either $3$ or $8.$
But $B = 3$ is not possible as $A \times 4 + 1 \neq 19$ for any value of $A$ between $0$ to $9.$
$B = 8$ and then $A = 4$
Hence, $A + B = 12$
View full question & answer→Question 41 Mark
Number $7N + 1$ will leave remainder $1$ when divided by $7.$
AnswerGiven, a number of the form $7N + 1 = x ($say$)$
Here, we observe that $*$ is a number which is one more than a multiple of $7.$
i.e. when $x$ is divided by $7,$ it leaves the remainder $1.$
View full question & answer→Question 51 Mark
If a $3-$digit number $abc$ is divisible by $11,$ then _________ is either $0$ or multiple of $11.$
AnswerIf a $3-$digit number abc is divisible by $11,$ then $(a + b) - b$ is either $0$ or multiple of $11.$
solution:
Since, $abc$ is divisible by $11,$ the difference of sum of its digits at odd places and that of even places is either zero or multiple of $11,$
i.e $(a + c) - b$ is either zero or multiple of $11.$
View full question & answer→Question 61 Mark
$\underline{\ \ \ \text{A}\ \ \ \ \text{B}\\\text{-}\ \ \text{B}\ \ \ \ \text{7}\ }\\\underline{\ \ \ \ 4\ \ \ \ 5\ }$
AnswerIn one’s column $B - 7 = 5$
Clearly, $12 - 7 = 5$ so $B = 2$
View full question & answer→Question 71 Mark
If $B \times B = AB,$ then either $A = 2, B = 5$ or $A =$ ______, $B =$ ______.
AnswerIf $B \times B = AB,$ then either $A = 2, B = 5$ or $A = 3, B = 6.$
Solution:
Here, $B \times B$ is a two-digit number, whose unit digit is $B,$
therefore the value of $B$ is either $5$ or $6.$
If $B = 5$, then $A = 2$ and if $B = 6$ then $A = 3$
View full question & answer→Question 81 Mark
If a number $a$ is divisible by $b,$ then it must be divisible by each factor of $b.$
AnswerGiven, $a$ is divisible by $b.$ Let $b = b1.\ p2,$ where $p1$ and $p2$ are primes. Since, a is divisible by $b,\ a$ is a multiple of $b$ i.e. $a = mb$ $a = m. p1. p2$ or $a = cp2 = dp1,$ where $c = mp1, d = mp2 = a$ is a multiple of $p1$ as well as $?2.$ Hence, $a$ is divisible by each factor $b.$
View full question & answer→Question 91 Mark
A number is divisible by $11$ if the differences between the sum of digits at its odd places and that of digits at the even places is either $0$ or divisible by ___________.
AnswerA number is divisible by $11$ if the differences between the sum of digits at its odd places and that of digits at the even places is either $0$ or divisible by $11.$
Solution:
By test of divisibility by $1, 1,$ we know that, a number is divisible by $11,$ if the sum of digits at odd places and even places are equal or differ by a number, which is divisible by $11.$
View full question & answer→Question 101 Mark
$3134673$ is divisible by $3$ and ______.
Answer$3134673$ is divisible by $3$ and $9$ as sum of the digits, $3 + 1 + 3 + 4 + 6 + 7 + 3 = 27$ is divisible by both $3$ and $9$
View full question & answer→Question 111 Mark
If $AB + 7C = 102, $ where $B ≠ 0, C ≠ 0,$ then $A + B + C = 14.$
Answer Here, $B + C$ is either $2$ or a two- digit number whose one’s digit is $2.$ If $B = C = 1$, lf $8 = 5, C = 7, A = 2$ and $A + B + C = 2 + 5 + 7 = 14$
View full question & answer→Question 121 Mark
$3x\ 5$ is divisible by $9$ if the digit $x$ is __________.
Answer$3x\ 5$ is divisible by $9$ if the digit $x$ is $1.$
Solution:
Since, the number $3x\ 5$ is divisible by $9,$ then the sum of its digits is also divisible by $9.$ i.e. $3 + x + 5$ is divisible by $9.$ $= x + 8$ can take values $9,18, 27, … $But $x$ is a digit of the number $3\ x\ 5$, so $x = 1.$
View full question & answer→Question 131 Mark
The sum of a two–digit number and the number obtained by reversing the digits is always divisible by __________.
AnswerThe sum of a two–digit number and the number obtained by reversing the digits is always divisible by $11.$
Solution:
Let ab be any two-digit number, then the number obtained by reversing its digits is $ba.$
Now, $ab + ab = (10a + b) + (10b + a)$
$= 11a + 11b = 11(a + b)$
Hence, $ab + ab$ is always divisible by $11$ and $(a + b).$
View full question & answer→Question 141 Mark
A three-digit number $abc$ is divisible by $5$ if $c$ is an even number.
AnswerBy the test of divisibility by $5,$ we know that if a number is divisible by $5,$ then its one’s digit will be either $0$ or $5,$ i.e. the numbers ending with the digits $0$ or $5$ are divisible by $5.$
View full question & answer→Question 151 Mark
The difference of a two–digit number and the number obtained by reversing its digits is always divisible by_________.
AnswerThe difference of a two–digit number and the number obtained by reversing its digits is always divisible by $9.$
Solution:
Let ab be any two-digit number,
then we have $ab - ba = (10a + b) - (10b + a)$
$= 9a - 9b = 9(a - b)$
Hence, $ab - ba$ is always divisible by $9$ and $(a - b).$
View full question & answer→Question 161 Mark
If $A \times 3 = 1A,$ then $A =$ ________.
AnswerIf $A \times 3 = 1A,$ then $A = 5.$
Solution:
Here, $3 \times A$ is a two-digit number whose unit digit is $A.$
$A$ can take any value between $0$ to $9,$
but only $A = 5$ satisfies the product Hence.
View full question & answer→Question 171 Mark
$1 x 35$ is divisible by $9$ if $x =$ _______.
Answer$1\ x \ 35$ is divisible by $9$ if $x = 0.$
Solution:
If $1\ x\ 35$ is divisible by $9,$
then the sum of its digit is also divisible by $9.$
i.e., $1 + x + 3 + 5$ is divisible by $9.$
$9 + x$ can takes values $0, 9, 18, 27$
$ 9 + x = 9$ or $18\ x = 0$ or $9$
View full question & answer→Question 181 Mark
If$\ \ \ \ \text{B}\ \ \ \ \ 1\\\ \ \ \ \underline{{\times}\ \ \ \ \text{B}}\\\ \ \ \ 4\ \ \ \ \ 9\text{B }$ then $B$ ______.
AnswerIf$\ \ \ \ \text{B}\ \ \ \ \ 1\\\ \ \ \ \underline{{\times}\ \ \ \ \text{B}}\\\ \ \ \ 4\ \ \ \ \ 9\text{B }$ then $B\ 7.$
Solution:
Here, $B$ can take values from $B \times B = 49$ is not satisfied
View full question & answer→Question 191 Mark
If $\ \ \ \ \text{A}\ \ \ \ \text{B}\\\ \ \ \ \underline{{\times}\ \ \ \ \text{B}}\\\ \ \ \ 9\ \ \ \ \ 6$ then $A =$ _______ and $B =$ _______ .
AnswerIf $\ \ \ \ \text{A}\ \ \ \ \text{B}\\\ \ \ \ \underline{{\times}\ \ \ \ \text{B}}\\\ \ \ \ 9\ \ \ \ \ 6$ then $A = 2$ and $B = 4.$
Solution:
Here, $B \times B = 6,$
therefore $B$ can take values either $4$ or $6.$
For $B = 4, A \times B + 1 = 9 A \times 4 + 1 = 9$
$A = 2$ Hence, $A = 2, B = 4$
View full question & answer→Question 201 Mark
If $\ \ \ \ \ 2\ \ \ \ \text{B}\\+\underline{\text{A}\ \ \ \ \text{B}}\\\ \ \ \ 8\ \ \ \ \ \text{A}$ then $A =$ _______ and $B =$ _______ .
AnswerIf $\ \ \ \ \ 2\ \ \ \ \text{B}\\+\underline{\text{A}\ \ \ \ \text{B}}\\\ \ \ \ 8\ \ \ \ \ \text{A}$ then $A = 6$ and $B = 3 .$
Solution:
Here, $B$ can take values from $0$ to $9.$
For $B=0, A=0$ which does not fit in tens column.
For $B = 1, A = 2,$ which does not fit in tens column For $B = 3,$
$A = 6$ which satisfies the tens column.
Hence, $A = 6$ and $B = 3.$
View full question & answer→Question 211 Mark
A two-digit number $??$ is always divisible by $2$ if b is an even number.
Answer By the test of divisibility by $2,$ we know that a number is divisible by $2,$ if it’s unit digit is even.
View full question & answer→Question 221 Mark
A three-digit number $abc$ is divisible by $6$ if $c$ is an even number and $a + b + c$ is a multiple of $3.$
AnswerIf a number is divisible by $6,$ then it is divisible by both $2$ and $3.$
Since, $abc$ is divisible by $6,$
it is also divisible by $2$ and $3.$
Therefore, $c$ is an even number and the sum of digits is divisible by $3,$ i.e. multiple of $3.$
View full question & answer→Question 231 Mark
If the digit $1$ is placed after a $2-$digit number whose tens is $t$ and ones digit is $u,$ the new number is ______.
AnswerIf the digit $1$ is placed after a $2-$digit number whose tens is $t$ and ones digit is $u,$ the new number is $tu1.$
Solution:
Given, a two-digit number whose ones digit isu in and tens digit isu. If the digit $1$ is placed after this number, then the next number will be $tu1.$
View full question & answer→Question 241 Mark
If $213x\ 27$ is divisible by $9,$ then the value of $x$ is $0.$
AnswerGiven, $213\ x\ 27$ is divisible by $9,$ so sum of its digits is also divisible by $9.$
i.e. $21 + 3 + x + 2 + 7 - 0, 9, 18, 27, 36,$
$= x + 15 = 0, 9, 18, 27, 36,$
$= x + 15 = 18 [x$ is a digit of a number$] = x = 3$
View full question & answer→Question 251 Mark
A four-digit number abcd is divisible by $4$ if ab is divisible by $4.$
AnswerAs we know that, if a number is divisible by $4,$
then the number formed by its digits in unit’s and ten’s place is divisible by $4.$
View full question & answer→Question 261 Mark
If $N ÷ 5$ leaves remainder $3$ and $N ÷ 2$ leaves remainder $0,$ then $N ÷ 10$ leaves remainder $4.$
Answer$N + 5$ leaves remainder $3.$
$N = 5n + 3,$ where $n = 0, 1, 2, 3, $ Now, it is also given $n + 2$ leaves remainder $0.$ So, $N$ must be an even number. But $N = 5n + 3$ i.e. sum of two terms whose second term is odd. So, for $N$ should be even it is necessary that $5n$ must be odd. Which is possible, when $n = 1, 3, 5$ So, in this case value of $N$ should be $N = 8, 18, 28, 38$ i.e. $N = 10n + 8, n = Q 1, 2, 3$ When $N ÷ 10$ leaves remainder 8 always
View full question & answer→Question 271 Mark
Number of the form $3N + 2$ will leave remainder $2$ when divided by $3.$
AnswerLet $x = 3N + 2.$ Then, it can be written as. $x = ($a multiple of $3) + 2$ i.e. $x$ is a number which is $2$ more than a multiple of $3$ i.e. $x$ is a number, which when divided by $3,$ leaves the remainder $2.$
View full question & answer→Question 281 Mark
A four-digit number abcd is divisible by $11,$ if $d + b =$ ________ or _______.
AnswerA four-digit number abcd is divisible by $11,$ if $d + b = a + c$ or $12(a + b).$
Solution:
We know that, a number is divisible by $11,$ if the difference between the sum of digits at odd places and the sum of its digits at even places is either $0$ or a multiple of $11.$ Hence, abcd is divisible by $11,$ if $(d + b) − (a + c) = 0, 11, 22, 33, = d + b = a + c$ or $d + b = 12(a + c)$
View full question & answer→