3 Mark Question

Question 13 Marks

If is a digit of the number $\overline{66784\text{x}}$ such that it is divisible by 9, find the possible values of x.

Answer

$\because$ The number $\overline{66784\text{x}}$ is divisible by 9
$\therefore$ The sum of its digits will also be divisible by 9
⇒ 6 + 6 + 7 + 8 + 4 + x is divisible by 9
⇒ 31 + x is divisible by 9
Sum greater than 31, are 36, 45, 54.....
which are divisible by 9
$\therefore$ Values are divisible by 9
$\therefore$ x = 5

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Question 23 Marks

Given that the number $\overline{67\text{y}19}$ is divisible by 9, where y is a digit, what are the possible valuse of y?

Answer

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

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Question 33 Marks

If x denotes the digit at hundreds place of the number $\overline{67\text{x}19}$ such that the number is divisible by 11. Find all possible values of x.

Answer

$\because$ The number $\overline{67\text{x}19}$ is divisible by 11
$\therefore$ The difference of the sums its alternate digits will be 0 or divisible by 11
$\therefore$ Difference of (9 + x + 6) and (1 + 7) is zero divisible by 11
⇒ 15 + x - 8 = 0, or multiple of 11,
7 + x = 0 ⇒ = -7, which in not possible
$\therefore$ 7 + x = 11, 7 + x = 22 etc,
⇒ x = 11 - 7 = 4, x 2 = 22 - 7
⇒ x = 15 which is not a digit
$\therefore$ x = 4

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Question 43 Marks

Find the quotient when difference of 985 and 958 is divided by 9.

Answer

The numbers of three digits are 985 and 958 in which tens and ones digits are reversed, then.
$\overline{\text{abc}}-\overline{\text{acb}}=9(\text{b}-\text{c})$
985 - 958 = 9(8 - 5) = 9 × 3 i. e., it is divisible by 9, then quotient = b - c = 8 - 5 = 3

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Question 53 Marks

Solve Cryptarithms.
$\ \ \ \ \ \ \ \ \ \ \ 3\ \ \ \ 7\\\ \ \ \ \ \underline{+\text{A}\ \ \ \ \text{B}\ }\\\ \ \ \ \ {\ \ \ \ \ \ 9\ \ \ \ \text{A}\ }$

Answer

Two possible values of A are:

If $7+\text{B}\leq9,3+\text{A}=9$

$\therefore\text{A}=6$

But if A = 6, 7 + B must be larger than 9.

Hence, it is impossible.

If $7+\text{B}\geq9$

$\therefore1+3+\text{A}=9$

$\Rightarrow\text{A}=5$

If A = 5 and 7 + B = 5,

B must be 8

$\therefore$ A = 5, B = 8

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Question 63 Marks

If $\overline{3\text{x}2}$ is a multiple of 11, where v is a digit, what is the value of x?

Answer

$\because$ The number $\overline{3\text{x}2}$ is multiple of 11
$\therefore$ It is divisible by 11
$\therefore$ Difference of the sum of its alternate digits is zero or multiple of 11
$\therefore$ Difference of (2 + 3) and x is zero or multiple of 11
⇒ If x - (2 + 3) = 0 ⇒ x - 5 = 0
Then x = 5

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Question 73 Marks

If sum of the number 985 and two other numbers by arranging the digits of 985 in cyclic order is divided by 111, 22 and 37 respectively. Find the quotient in each case.

Answer

The given number is 985
The other two numbers by arranging its digits in cyclic order, will be 859, 598 of the form,
$\overline{\text{abc}},\overline{\text{bca}},\overline{\text{cba}}$
Therefore,
If 985 + 598 + 598 is divided by 111, then quotient will be a + 6 + c = 9 + 8 + 5 = 22
If this sum is divided by 22, then quotient = 111
And if it is divided by 37, then quotient = 3(a + 2 + c) = 3(22) = 66

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Question 83 Marks

Solve Cryptarithms.
$\ \ \ \ \ \ \ \ \ \ \ \text{A}\ \ \ \ \text{B}\\\ \ \ \ \ \underline{+\ \ \ 3\ \ \ \ \ 7\ }\\\ \ \ \ \ {\ \ \ \ \ \ 9\ \ \ \ \text{A}\ }$

Answer

Two possibilities of A are:

If $\text{B}+7<9,$

$\text{A}=6$

But clearly, if A = 6,

$\text{B}+7\geq9;$

It is impossible

If $\text{B}+7\geq9,$

A = 5 and B + 7 = 5

Clearly, B = 8

$\therefore$ A = 5, B = 8

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Question 93 Marks

If x is a digit such that the number $\overline{18\text{x}71}$ is divisible by 3, find possible values of x.

Answer

$\because$ The number $\overline{18\text{x}71}$ is divisible by 3
$\therefore$ The sum of its digits will also be divisible by 3
⇒ I + 8 + x + 7 + 1 is divisible by 3
⇒ 17 + x is divisible by 3
The sum greater than 17, can be 18, 21, 24, 27
$\therefore$ x can be 1, 4, 7 which are divisible by 3.

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Question 103 Marks

Given that the number $\overline{35\text{a}64}$ is divisible by 3, where a is a digit, what are the possible volues of a?

Answer

The number $\overline{35\text{a}64}$ is divisible by 3
$\because$ The sum of its digits will also be divisible by 3
$\therefore$ 3 + 5 + a + b + 4 is divisible by 3
⇒ 18 + a is divisible by 3
⇒ a is divisible by 3 $(\because$ is divisible by 3$)$
$\therefore$ Values of a can be 0, 3, 6, 9.

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