Playing with Numbers Test

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Playing with Numbers Test - 6

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Weekly Quiz Competition

Question 1
1 / -0

For what value(s) of N is the number N(10²⁰- 1) divisible by 9?

A
Any natural number

B
Only odd values of N

C
Only even values of N

D
Only if N is multiple of 9

Solution

N(10²⁰ - 1) is divisible by 9 for every natural number (n) as:

10¹ - 1 = 9
10²- 1 = 99
10³ - 1 = 999
...
...
...

10²⁰ - 1 = 999 ..., etc. are divisible by 9.
Question 2
1 / -0

The values of A and B in the given addition respectively are

A
4 and 8

B
3 and 9

C
6 and 6

D
5 and 7

Solution

We need '2' at the units place.
So, A + B = 2 or 12
As (A ≠ B), A + B = 2 is not possible.
A + B = 12 ... (1)

Also,
3 + 3 + 1 (Carried from units place ) = B (tens column)
B = 7

From (1),
A = 12 - 7
A = 5
Question 3
1 / -0

Which of the following numbers is divisible by 8?

A
21,61,316

B
15,41,418

C
13,21,532

D
11,11,640

Solution

According to the divisibility rule of 8, if the number formed by the last three digits is divisible by 8, it means the entire number is divisible by 8.

Option 1: 21,61,316 - 316 is not divisible by 8.
Option 2: 15,41,418 - 418 is not divisible by 8.
Option 3: 13,21,532 - 532 is not divisible by 8.
Option 4: 11,11,640 - 640 is divisible by 8.

Hence, option 4 is the correct answer.
Question 4
1 / -0

If, where X and Y are single-digit numbers, then the values of X and Y respectively are

A
4 and 5

B
4 and 6

C
3 and 5

D
3 and 6

Solution

We have 1 + X = 4, because we will not get any carry from the units place.
X = 3

Now, for the value of Y,
3 × Y = 1Y
There is only one possibility.
3 × 5 = 15
So, Y = 5

So, X = 3 and Y = 5

Hence, option 3 is the correct answer.
Question 5
1 / -0

When an integer k is divided by 6, the remainder is 2. What will be the remainder if 8k is divided by 6?

A
7

B
3

C
9

D
4

Solution

When k is divided by 6, the remainder will be 2.
It can be written as:
k = 6n + 2; where k is the number and n is the quotient.
Multiply all terms by 8.
8k = 8(6n + 2) = 6(8n) + 16
Write 16 as 12 + 4 since 12 = 6 × 2.
8k = 6(8n) + 12 + 4
8k = 6(8n + 2) + 4
The above equation indicates that if 8k is divided by 6, the remainder will be 4.
Question 6
1 / -0

Which of the following can be the values of x and y in 8754x15y4 if it is divisible by 11?

A
x = 3 and y = 2

B
x = 7 and y = 2

C
x = 3 and y = 3

D
x = 4 and y = 2

Solution

If the difference between the sum of the digits at odd places and the sum of the digits at even places is 0 or 11, then the number will be divisible by 11.
Now, (8 + 5 + x + 5 + 4) - (7 + 4 + 1 + y) = 0 or 11
Using option: x = 3 and y = 2;
25 - 14 = 11
Hence, option 1 is the correct answer.
Question 7
1 / -0

The largest number by which the product of any four consecutive odd natural numbers is always divisible is ___.

A
9

B
7

C
5

D
3

Solution

When considering four consecutive odd numbers, i.e. (3, 5, 7, 9) or (23, 25, 27, 29) or (107, 109, 111, 113), one of the four numbers would always be a multiple of 3 as every third odd number is a multiple of 3. If we think about 5, every fifth odd number is a multiple of 5 and hence it is not always possible that the product of four consecutive odd numbers is divisible by 5.
Hence, the product of any four consecutive odd numbers will always be divisible by 3 and 1 only.
Hence, the required number is 3.
Question 8
1 / -0

If the total of all the digits in a number is 9 or any multiple of 9, then the number is definitely divisible by

A
18

B
9

C
6

D
27

Solution

In a number, if the total of all the digits is 9 or any multiple of 9, then the number is definitely divisible by 9. (Divisibility rule of 9)
E.g. 9297
Sum of the digits = 9 + 2 + 9 + 7 = 27 (which is a multiple of 9)
Hence, 9297 is divisible by 9.
Question 9
1 / -0

A 6-digit number ab9698 is divisible by 9 such that a + b < 6, where a and b are single-digit numbers. Find the possible values of a and b.

A
a = 1 and b = 2

B
a = 2 and b = 1

C
a = 2 and b = 2

D
a = 2 and b = 3

Solution

When the total of all the digits is divisible by 9, then the number is divisible by 9.
ab9698
a + b + 9 + 6 + 9 + 8 = a + b + 32
Putting a = 2 and b = 2,
2 + 2 + 32 = 36
2,29,698 is divisible by 9.
Question 10
1 / -0

What can be the value of Z + Y to make the following calculation correct?

A
10

B
11

C
12

D
13

Solution

Ones column: 5 + Y = 2; which means Y = 7 and 5 + 7 = 12.
Tens column: X + 4 + carry(1) = 7
X = 2
Hundreds column: 9 + Z = 3; which means Z = 4 and 9 + 4 = 13.

Y + Z = 7 + 4 = 11
Question 11
1 / -0

Find the numbers between 400 and 550 which when divided by 6, 8 or 9 leave 5 as remainder.

A
421, 493

B
436, 462

C
462, 421

D
509, 437

Solution

LCM of 6, 8 and 9 is 72.
Required number = 72x + 5
Put x = 6 and x = 7 to get the numbers between 400 and 550.
Required numbers = 72 × 6 + 5, 72 × 7 + 5 = 437 and 509
Question 12
1 / -0

Given that the number 57w88 is divisible by 18, where w is a single digit, what is the least possible value of w?

A
4

B
1

C
5

D
8

Solution

57w88 is divisible by 18 if the digit at the units place is even and the sum of the digits is divisible by 9.
5 + 7 + w + 8 + 8 = 28 + w is divisible by 9.
The multiples of 9 are 9, 18, 27, 36 ...
Next multiple of 9 after 27 is 36, so we need to add 8 to make it 36.
So, least possible value of w = 8
Question 13
1 / -0

A 3-digit number 'dfg' is divisible by 6 if

A
dfg is even and d + 2f + g is divisible by 3

B
dfg is even and d + f + g is divisible by 3

C
dfg is odd and 2d + 3f + g is divisible by 3

D
dfg is odd and 2d + 2f + g is divisible by 3

Solution

A number is said to be divisible by 6 if it is divisible by 2 and 3.
Option 2 is correct here because dfg is an even number or it is divisible by 2. Also, the sum of the digits in the number is divisible 3.
E.g. 756
756 is an even number.
Also, 7 + 5 + 6 = 18, which is divisible by 3.
So, 756 is divisible by 6.
Question 14
1 / -0

If, where A, B and C are single-digit numbers, then the values of A, B and C are:

A
A = 4, B = 2, C = 5

B
A = 4, B = 2, C = 6

C
A = 4, B = 3, C = 6

D
A = 4, B = 3, C = 8

Solution

We have B + B = A, because we will not get any carry from the units place
Now, in hundreds column = 8 + 1 + C = 5 , so the only possibility here is 15 (Not 5 or 25)
8 + 1 + C = 15, So C = 6.
1 + A = 5
A = 4

Now, for value of B,
In the first step, 4 × 6 = 24
4 is at the units place and 2 will be carried forward to the tens place.
Then 4 × 5 = 20, then 20 + 2 = 22
So, B = 2
Checking calculation after putting the values of A, B and C:

So, A = 4
B = 2
C = 6
Question 15
1 / -0

In a division, if the dividend is 12 times the quotient and 6 times the remainder, then what will be the divisor?

A
11

B
10

C
9

D
8

Solution

According to the question,
Dividend = Divisor × Quotient + Remainder

Let the quotient be Q and divisor be D.
So, dividend is 12Q.
Also, remainder will (1/6) times the dividend i.e. 2Q.

Therefore,
Dividend = Divisor × Quotient + Remainder
12Q = D × Q + 2Q
10Q = DQ
D = 10
Question 16
1 / -0

Which of the following statements is INCORRECT?

A. If the lasttwo digits of a whole number are divisible by 4, then the entire number is divisible by 4.
B. If the sum of the digits of a number is divisibleby 9, then the actual number is also divisibleby 9.
C. If thelast three digits of a whole number are divisibleby 7, then theentire number is divisibleby 7.
D. Take the difference between alternating sumof the digits in the number. If that is divisible by 11, so is the original number.

A
A

B
B

C
C

D
D

Solution

(A) If the lasttwo digits of a whole number are divisible by 4, then the entire number is divisible by 4.
E.g. 4316; 16 is divisible by 4, which means 4316 is divisible by 4.
So, this statement is true.

(B) If the sum of the digits is divisibleby 9, then the actual number is also divisibleby 9.
E.g. 1233; sum of the digits = 1 + 2 + 3 + 3 = 9 (which is divisible by 9). So, 1233 is also divisible by 9.
So, this statement is true.

(C) If thelast three digits of a whole number are divisibleby 7, then theentire number is divisibleby 7.
E.g. 1133; 133 is divisible by 7, but 1133 is not divisible by 7.
So, this statement is not true.

(D) Take the difference between alternating sumof the digits in the number. If that is divisible by 11, so is the original number.
For instance, 7282 has the alternating sum of digits as 7 - 2 + 8 - 2 = 11. Since 11 is divisible by 11, 7282 is divisible by 11.
So, this statement is true.

Question 17

1 / -0

Fill in the blanks with the help of the table given below these statements:

(i) If 42a5 is a multiple of 9 and 'a' is a single digit number, then the value of 'a' is ___P____.
(ii) The usual form of the number 9 × 100 + 7 × 1 is ____Q____.
(iii) If M is a number such that M ÷ 5 gives a remainder of 1, then the number at the ones place of M is ____R____.

	P	Q	R
(A)	7	907	6
(B)	7	907	1 or 6
(C)	16	907	1
(D)	16	970	1 or 6

(A)

(B)

(C)

(D)

Solution

(i) If 42a5 is a multiple of 9 and 'a' is a digit, then the sum of the digits is: 4 + 2 + a + 5 = 11 + a, which should be divisible by 9.
So, the value of 'a' will be 7, as 11 + 7 = 18, which is divisible by 9.
(ii) The usual form of the number 9 × 100 + 7 × 1 is: 900 + 7 = 907.
(iii) It is given that M is a number such that M ÷ 5 gives a remainder of 1.
Since every multiple of 5 ends at 0 or 5, to get the remainder of 1, the ones digit should be one more than 0 or 5. So, the ones digit must be 1 or 6.
Hence, option 2 is the correct answer.

Question 18

1 / -0

Match the columns:

Column I	Column II
(A) If 3a74 is a number divisible by 9, then the least value of 'a' is	(i) 3
(B) The number 28,221 is divisible by	(ii) 4
(C) If a number is divisible by 5, then the digit at the ones place can be	(iii) 14
(D) If 2X28 is a number divisible by 11, then the least value of 2X is	(iv) 5

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

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

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

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

Solution

For a number to be divisible by 9, the sum of the digits should be divisible by 9.
That is, 3 + 7 + 4 + a = 14 + a = 14 + 4 = 18, which is divisible by 9.
(A) 3474 ÷ 9 = 386
(B) Sum of the digits = 2 + 8 + 2 + 2 + 1 = 15, which is divisible by 3.
Hence, 28,221 ÷ 3 = 9407
(C) If a number is divisible by 5, then the digit on the ones place can be either 0 or 5.
Hence, the correct answer is 5.
(D) For a number to be divisible by 11, take the alternating sum of the digits in the number, read from left to right. If that is divisible by 11, so is the original number. So, 2X28 has the alternating sum of digits as (2 + 2) - (X + 8) = a number that is divisible by 11.
Therefore, the least number can by 7 as (2 + 2) - (7 + 8) = 4 - 15 = -11, which is divisible by 11.
Hence, 2 X = 2 × 7 = 14.

Question 19
1 / -0

How many five-digit numbers of the form AABAA are divisible by 9?

A
9

B
7

C
6

D
3

Solution

If the sum of the digits is a multiple of 9 then, the number is divisible by 9.
Hence, if the number is AABAA, then 4A + B must be a multiple of 9.
For any value of A, B can be adjusted to make 4A + B a multiple of 9.
Therefore, there will be 9 such numbers as the value of A will vary from 1 to 9.

If A = 1, then B = 5
If A = 2, then B = 1
If A = 3, then B = 6
If A = 4, then B = 2
If A = 5, then B = 7
If A = 6, then B = 3
If A = 7, then B = 8
If A = 8, then B = 4
If A = 9, then B = 0

Question 20

1 / -0

Find the respective values of A, B and C with the help of the table given below.

(i) (A, B, C)	(ii) (A, B, C)
I. 3, 2, 4	I. 2, 3, 1
II. 1, 5, 6	II. 2, 7, 3
III. 2, 6, 5	III. 3, 7, 3
IV. 1, 2, 3	IV. 4, 3, 1

I

II

III

IV

Solution

(i)
We can derive the values from the answers
The ones place digit is 4 only. Hence, there is nothing to be exchanged to the tens column.
The tens place digit is 3. Now, the sum of the digits can be 3 or 13. It will be 13 because the number cannot be smaller than 8.
If the number is 13, then the value of B is 5 (B + 8 = 13). 1 will be exchanged to the hundreds column.
Now, A + 3 = 4
So, A = 1
A = 1, B = 5, C = ?
As we know, 9 X C = 4 at the ones place. The only possibility for C will be 6 because this will give us 54 (4 at ones place).

Checking the answer after putting the values:

(ii)
The values can be derived from the answer (AA8645) itself here.
As we know that there is nothing to be exchanged from the ones place;
Tens place = C + 1 = 4; so C = 3 (It cannot be 14 because the value should be 9 or less than 9).
Hundreds place = A + 4 = 6 or 16. It cannot be 16, so A = 2.
Thousands place = 1 + B = 8, B = 7

Checking the answer after putting the values:

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Re-Attempt Weekly Quiz Competition

Playing with Numbers Test - 6

Result

Playing with Numbers Test - 6

Score

-

Rank

-

SHARING IS CARING

Question 1

Any natural number

Only odd values of N

Only even values of N

Only if N is multiple of 9

Solution

Question 2

4 and 8

3 and 9

6 and 6

5 and 7

Solution

Question 3

21,61,316

15,41,418

13,21,532

11,11,640

Solution

Question 4

4 and 5

4 and 6

3 and 5

3 and 6

Solution

Question 5

7

3

9

4

Solution

Question 6

x = 3 and y = 2

x = 7 and y = 2

x = 3 and y = 3

x = 4 and y = 2

Solution

Question 7

9

7

5

3

Solution

Question 8

18

9

6

27

Solution

Question 9

a = 1 and b = 2

a = 2 and b = 1

a = 2 and b = 2

a = 2 and b = 3

Solution

Question 10

10

11

12

13

Solution

Question 11

421, 493

436, 462

462, 421

509, 437

Solution

Question 12

4

1

5

8

Solution