Playing with Numbers Test

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Playing with Numbers Test - 5

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Weekly Quiz Competition

Question 1
1 / -0

The number (10ⁿ - 1) is divisible by 9 for which of the following values of n?

A
n ∈ odd numbers

B
n ∈ even numbers

C
n ∈ N (where N is a natural number)

D
When n is a multiple of 9

Solution

(10ⁿ - 1) is divisible by 9 for n ∈ N as 10¹ - 1 = 9, 10² -1 = 99, 10³ - 1 = 999 and so on, are divisible by 9.
Question 2
1 / -0

What are the respective values of A and B in the given subtraction?

A
3, 0

B
4, 6

C
3, 2

D
3, 3

Solution

This calculation is possible if the value of A is 3 and that of B is 0.
Question 3
1 / -0

What is the least value of x, so that the number 28345x4 is divisible by 8?

A
8

B
4

C
2

D
0

Solution

28345x4 is divisible by 8, if its last three digits are divisible by 8.
So, we have 5x4 to be divisible by 8.
Therefore, least value of x is 0 (because 504 is divisible by 8).
Question 4
1 / -0

What are the respective values of A and B, where A and B are single-digit numbers and A - B = 3?

A
4, 1

B
7, 5

C
5, 1

D
6, 3

Solution

There are two possible values of A and B.

And

But A - B = 3
So, the only possibility is when A = 4 and B = 1.
Question 5
1 / -0

Suppose, the division leaves a remainder 3 and leaves a remainder 1. What is the units digit of X?

A
9

B
7

C
5

D
3

Solution

It is given that when X is divided by 5, the remainder is 3.
So, X can be 3,8, 13,18, 23, 28, 33, 38, 43, 48, ...

Also, it is given that when X is divided by 2, the remainder is 1.
So, X can be 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, ...

Seeing the common elements in both the cases, we see that numbers common in both the cases have 3 as the units digit.
Question 6
1 / -0

Given that the number 43476p2086 is divisible by 3. Which of the following is the possible value of p, where p is a single digit number?

A
9

B
7

C
6

D
5

Solution

The number 43476p2086 is divisible by 3, if the sum of all the digits is divisible by 3.
So, sum of the digits = 4 + 3 + 4 + 7 + 6 + p + 2 + 0 + 8 + 6 = 40 + p
So, the possible values of p are 2, 5 and 8.
(Because only 42, 45 and 48 are divisible by 3)
Question 7
1 / -0

A prime number N, which is in the range 10-50, remains unchanged when its digits are reversed. The square of N equals

A
289

B
961

C
169

D
121

Solution

If the number is xy, then its reverse is yx. Since the number remains unchanged, 10x + y = 10y + x.
The possible number can be 11, 22, 33, or 44.
The required prime number is 11.
The square of 11 is 121.
Hence, (4) is the correct option.
Question 8
1 / -0

If a number has an even digit at its units place and sum of its digits is divisible by 3, then the number is divisible by which of the following numbers?

A
6

B
7

C
8

D
9

Solution

The prime factors of 6 are 2 and 3. So, if a number is divisible by 2 and 3, it must also be divisible by 6.
As the number has an even digit at its unit place, so the number would be divisible by 2 and also, it is divisible by 3.
So, the number would be divisible by 6 as well.
Question 9
1 / -0

A 5-digit number ab736 is divisible by 3, such that a + b < 4, where a and b are single-digit numbers. What are the possible values of (a, b)?

A
(9, 1) or (1, 1)

B
(2, 0) or (0, 2)

C
(1, 1) only

D
(1, 1) or (2, 0)

Solution

A number is divisible by 3 if the sum of the digits of the number is divisible by 3.
ab736 is divisible by 3 if the (a ＋ b ＋7 ＋ 3 ＋ 6) is divisible by 3.

i.e. 16 + a + b is divisible by 3.
Or a + b = 2, 5, 8
Since a + b < 4;
a + b = 2
So, the only possible values of (a, b) are (1, 1) or (2, 0).
Question 10
1 / -0

What is the value of C, if there is no carry on addition?

A
7

B
6

C
5

D
4

Solution

Given, there is no carry on addition.
The value of B = 5 - 2 = 3
The value of A = 7 - 3 = 4
The value of C = 9 - 4 = 5

=
Question 11
1 / -0

If X divided by 5 leaves a remainder of 4, then the ones digit of X must be either

A
4 or 9

B
4 or 1

C
1 or 9

D
3 or 7

Solution

X leaves a remainder 4, when divided by 5.
(X - 4) is divisible by 5.
The ones digit of X - 4 is either 0 or 5.
So, the ones digit of X is either 4 or 9.
Question 12
1 / -0

The number 383X is divisible by 11, where X is a single digit. What is the least possible value of X?

A
9

B
8

C
7

D
6

Solution

To determine if a number is divisible by 11, we need to find the difference between the sum of the digits at even places and the sum of the digits at odd places.
If that difference is divisible by 11, then the number is also divisible by 11.
So, in 383X:
8 + X - 3 - 3 = 0 or Multiple of 11

(i) 8 + X - 6 = 0
X + 2 = 0
X = -2 (Negative number not possible)

(ii) 8 + X - 6 = 11
X + 2 = 11
X = 9

Putting X = 9, we get
8 + 9 - 3 - 3 = 11
Since 11 is divisible by 11, so the number is 3839.
Question 13
1 / -0

A three-digit number 'bca' is divisible by 3 if

A
a + 2b + 3c is divisible by 3

B
2a + 3b + c is divisible by 3

C
a + b + c is divisible by 3

D
3a + 2b + c is divisible by 3

Solution

The three-digit number is bca.
Sum of its digits = a + b + c
The rule of 3 states that if the sum of the digits of a number is divisible by 3, then the number will also be divisible by 3.
Therefore, if a + b + c is divisible by 3, then 'bca' will be divisible by 3.
Question 14
1 / -0

Find the value of B.

A
0

B
8

C
6

D
5

Solution

Here,
B x 7 = B
So, B = 0, 5
Taking B = 0, we get 7A = 7 and A² = 6 (Both case not possible)
As for A = 1,

Not satisfying the given condition.

So, B = 5.
Now, 5A + 3 + 7A gives unit digit 7.
So, only possibility satisfying the given condition is A = 2.

A = 2 and B = 5
Question 15
1 / -0

The divisor in a question is 10 times the quotient and 6 times the remainder. If the remainder is 45, then what is the dividend?

A
7256

B
7251

C
7335

D
7360

Solution

Divisor = (6 x 45) = 270
10 x Quotient = 270
Quotient = 270 ÷ 10 = 27

Dividend = (Divisor x Quotient) + Remainder
= (270 x 27) + 45
= 7290 + 45
= 7335
Question 16
1 / -0

Which of the following statements is incorrect?

(A) If a number is divisible by 2 and 3, then it is divisible by 6 and 9.
(B) If 5B × B = 399, then value of B = 7.
(C) Number of the form 3y + 2 will leave remainder 2 when divided by 3.
(D) All numbers that are divisible by 10 are divisible by 5.

A
A

B
B

C
C

D
D

Solution

The Rule for 6: The prime factors of 6 are 2 and 3.
So, for a number to be divisible by 6, it must also be divisible by 2 and 3.
Therefore, we need to check if a number is even and then check if the sum of the digits is divisible by 3.
If a number is divisible by 2 and 3, then it is divisible by 6 only.
So, statement A is incorrect.

Question 17

1 / -0

Directions: Study the given information carefully and choose the correct option.

(I) The sum of a two digit number and the number obtained by reversing the digits is always divisible by X.
(II) A three digit number PQR is divisible by 6, if R is an even number and P + Q + R is a multiple of Y.
(III) The difference of a two digit number and the number obtained by reversing the digits is always divisible by Z.

	X	Y	Z
A	9	2	10
B	11	3	9
C	9	2	4
D	11	3	0

A

B

C

D

Solution

(I) Let ab be the required number and number obtained by reversing the digits is ba.
So, ab + ba = (10a + b) + (10b + a) = 11a + 11b = 11(a + b). So, it is divisible by 11.

(II) The Rule for 6: The prime factors of 6 are 2 and 3. So, for a number to be divisible by 6, it must also be divisible by 2 and 3. Therefore, we need to check if a number is even and then check if the sum of the digits is divisible by 3. The number is divisible by both 2 and 3. Since, PQR is divisible by 6, it is also divisible by 2 and 3.

(III) Let ab be the required number and number obtained by reversing the digits is ba.
So, ab - ba = (10a + b) - (10b + a) = 9a - 9b = 9(a - b) it is divisible by 9. Therefore, Z = 9.

Question 18

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Directions: Match the following.

Column I	Column II
(A) If 85235x is divisible by 11, then x =	(i) 8
(B) If 65x92 is divisible by 6, then x =	(ii) 1
(C) If 78430x is divisible by 8, then x =	(iii) 7
(D) If 7869x2 is divisible by both 4 and 3, then x =	(iv) 4

A - (i), B - (ii), C - (iii), D - (iv)

A - (iv), B - (i), C - (iii), D - (ii)

A - (iii), B - (i), C - (iv), D - (ii)

A - (iii), B - (i), C - (ii), D - (iv)

Solution

(A) Given number: 85235x
For divisibility by 11, (8 + 2 + 5) - (5 + 3 + x) = 0 or Multiple of 11
15 - 8 - x = 0, 11
x = 7 (Only possible value)

(B) Given number: 65x92
For divisibility by 6, last digit should be even and the sum of the digits should be divisible by 3.
So, 6 + 5 + x + 9 + 2 = Multiple of 3
22 + x = Multiple of 3
So, x can be 2, 5 or 8.

(C) Given number: 78430x
For divisibility by 8, the number formed by the last three digits of the number should be divisible by 8.

So, the only possible value of 'x' such that 30x is divisible by 8 is 4.

(D) Given number: 7869x2
For divisibility by both 4 and 3, the number formed by the last two digits of the number should be divisible by 4, and the sum of the digits should be divisible by 3.
So, the values of 'x2' in 7869x2, such that it is divisible by 4, can be 12,32,52,72 and 92.

Now, for the number to be also divisible by 3, the sum of the last two digits of the number should also be divisible by 3.(As sum of the remaining digits is divisible by 3)
Among 12, 32, 52, 72 and 92, it only holds true for 12 and 72.
So, x can be either 1 or 7.

Question 19
1 / -0

A five digit number 5XXXX is divisible by 6. What is least value of X?

A
7

B
6

C
5

D
4

Solution

For divisibility by 6, sum of digits has to be a multiple of 3 and the last digit should be even.
A number is divisible by 3, if the sum of the digits is divisible by 3.
5 + X + X + X + X = 4X + 5
4X + 5 should be divisible by 3 and X should be an even number.

Put X = 0.
4X + 5 = 5, which is not divisible by 3.

Put X = 2.
4X + 5 = 13, which is not divisible by 3.

Put X = 4.
4X + 5 = 21, which is divisible by 3.
So, X = 4
Question 20
1 / -0

Find the values of A, B and C in the given operation.

A
A = 5, B = 0, C = 2

B
A = 5, B = 2, C = 2

C
A = 5, B = 5, C = 1

D
A = 3, B = 5, C = 1

Solution

So, AB = 10A + B
CAB = 100 × C + 10 × A + B
(10A + B)5 = 100C + 10A + B
50A + 5B = 100C + 10A + B
40A + 4B = 100C
4(10A + B) = 100C
10A + B = 25C

On putting different values of A, B and C from the options, we find that the equation is only satisfied by the set of values given in option 1, i.e. A = 5, B = 0 and C = 2.
10 × 5 + 0 = 25 × 2
50 = 50
Thus, option 1 is correct.

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Playing with Numbers Test - 5

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Playing with Numbers Test - 5

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SHARING IS CARING

Question 1

n ∈ odd numbers

n ∈ even numbers

n ∈ N (where N is a natural number)

When n is a multiple of 9

Solution

Question 2

3, 0

4, 6

3, 2

3, 3

Solution

Question 3

8

4

2

0

Solution

Question 4

4, 1

7, 5

5, 1

6, 3

Solution

Question 5

9

7

5

3

Solution

Question 6

9

7

6

5

Solution

Question 7

289

961

169

121

Solution

Question 8

6

7

8

9

Solution

Question 9

(9, 1) or (1, 1)

(2, 0) or (0, 2)

(1, 1) only

(1, 1) or (2, 0)

Solution

Question 10

7

6

5

4

Solution

Question 11

4 or 9

4 or 1

1 or 9

3 or 7

Solution

Question 12

9

8

7

6

Solution