Selina Concise Solutions for ICSE Class 8 Mathematics Chapter 5 Playing with Number

Exercise 5(A)

Question 1. Write the quotient when the sum of 73 and 37 is divided by
(i) 11
(ii) 10
Answer:
Let us take the two-digit number \( ab = 73 \). Reversing the digits gives \( ba = 37 \).
Here, we have \( a = 7 \) and \( b = 3 \).
(i) The rule for adding reversed two-digit numbers says that when \( ab + ba \) is divided by 11, the quotient is \( a + b \).
So, quotient = \( a + b = 7 + 3 = 10 \).
(ii) When \( ab + ba \) is divided by 10 (which is \( a + b \)), the rule shows that the quotient is 11.
So, quotient = 11.
In simple words: When you add a two-digit number to its reversed form, the sum can always be divided evenly by 11 and also by the sum of its digits. If you divide by 11, you get the sum of the digits. If you divide by the sum of the digits, you get 11.

Exam Tip: Remember the general formula \( \frac{ab + ba}{11} = a + b \) to quickly find the answer without doing long addition and division.

Question 2. Write the quotient when the sum of 94 and 49 is divided by
(i) 11
(ii) 13
Answer:
Let the original number be \( ab = 94 \). Its reversed counterpart is \( ba = 49 \).
From these numbers, we get \( a = 9 \) and \( b = 4 \).
(i) Dividing the sum \( ab + ba \) by 11 gives a quotient equal to \( a + b \).
So, quotient = \( a + b = 9 + 4 = 13 \).
(ii) Dividing the sum by 13 (which is \( a + b \)) gives a quotient of 11.
So, quotient = 11.
In simple words: Adding 94 and 49 gives 143. If we divide 143 by 11, we get 13. If we divide 143 by 13, we get 11.

Exam Tip: Always make sure to write down the values of \( a \) and \( b \) clearly before applying the divisibility rule.

Question 3. Find the quotient when 73 - 37 is divided by
(i) 9
(ii) 4
Answer:
Let the two-digit number be \( ab = 73 \) and its reversed form be \( ba = 37 \).
Here, the digits are \( a = 7 \) and \( b = 3 \).
(i) When the difference of a reversed two-digit number \( ab - ba \) is divided by 9, the quotient is \( a - b \).
So, quotient = \( a - b = 7 - 3 = 4 \).
(ii) When the difference \( ab - ba \) is divided by 4 (which is \( a - b \)), the quotient is 9.
So, quotient = 9.
In simple words: Subtracting 37 from 73 gives 36. If we divide 36 by 9, we get 4. If we divide 36 by 4, we get 9.

Exam Tip: For subtracting reversed numbers, the difference is always divisible by 9 and by \( a - b \).

Question 4. Find the quotient when 94 - 49 is divided by
(i) 9
(ii) 5
Answer:
Let us take \( ab = 94 \) and \( ba = 49 \).
Here, we have \( a = 9 \) and \( b = 4 \).
(i) The quotient when \( ab - ba \) is divided by 9 is \( a - b \).
So, quotient = \( 9 - 4 = 5 \).
(ii) The quotient when \( ab - ba \) is divided by 5 (which is \( a - b \)) is 9.
So, quotient = 9.
In simple words: The difference between 94 and 49 is 45. Dividing 45 by 9 gives 5, and dividing 45 by 5 gives 9.

Exam Tip: Keep the formula \( \frac{ab - ba}{9} = a - b \) in mind to easily solve subtraction-based number games.

Question 5. Show that 527 + 752 + 275 is exactly divisible by 14.
Answer:
Let us write any three-digit number \( abc \) and its cyclic permutations \( bca \) and \( cab \) in their expanded forms:
\( abc = 100a + 10b + c \) ...(i)
\( bca = 100b + 10c + a \) ...(ii)
\( cab = 100c + 10a + b \) ...(iii)
Adding equations (i), (ii), and (iii), we get:
\( abc + bca + cab = 111a + 111b + 111c \)
\( = 111(a + b + c) \)
\( = 3 \times 37(a + b + c) \)

Now, let us use this property for the given sum \( 527 + 752 + 275 \).
Comparing this with our general form, we have \( a = 5 \), \( b = 2 \), and \( c = 7 \).
Thus, the sum can be written as:
\( 527 + 752 + 275 = 3 \times 37(5 + 2 + 7) \)
\( = 3 \times 37 \times 14 \)
Since 14 is one of the factors of the resulting expression, the sum \( 527 + 752 + 275 \) is exactly divisible by 14.
In simple words: When we add three-digit numbers made by cycling the same digits, the total sum is always a multiple of the sum of those three digits. Here, the sum of the digits is 14, so the total must be divisible by 14.

Exam Tip: For any cyclic sum \( abc + bca + cab \), remember that \( 111(a + b + c) \) is the key algebraic identity to prove divisibility.

Question 6. If a = b, show that abc = bac.
Answer:
Given that \( a = b \).
We need to show that \( abc = bac \).
Using the place value expansion of three-digit numbers, we have:
\( abc = 100a + 10b + c \) ...(i)
\( bac = 100b + 10a + c \) ...(ii)
Since we are given \( a = b \), we can substitute \( a \) with \( b \) in both equations (i) and (ii):
\( abc = 100b + 10b + c \) ...(iii)
\( bac = 100b + 10b + c \) ...(iv)
Subtracting equation (iv) from equation (iii), we obtain:
\( abc - bac = 0 \)

\( \implies abc = bac \)
Hence, we have proved the statement.
In simple words: Since \( a \) is equal to \( b \), swapping their positions in the hundreds and tens places does not change the value of the number at all.

Exam Tip: Expand the numbers using place values first, then substitute the given equality to show that both expressions are identical.

Question 7. If a > c; show that abc - cba = 99(a - c).
Answer:
We are given that \( a > c \).
We want to prove that \( abc - cba = 99(a - c) \).
Let us write the place value expansions for both three-digit numbers:
\( abc = 100a + 10b + c \) ...(i)
\( cba = 100c + 10b + a \) ...(ii)
Now, subtract equation (ii) from equation (i):
\( abc - cba = (100a + 10b + c) - (100c + 10b + a) \)
\( = 100a + 10b + c - 100c - 10b - a \)
The \( 10b \) terms cancel each other out, leaving us with:
\( abc - cba = 99a - 99c \)
Factoring out 99 gives:
\( abc - cba = 99(a - c) \)
This completes the proof.
In simple words: When you subtract a reversed three-digit number from the original, the tens digit cancels out. The difference is always 99 times the difference between the first and last digits.

Exam Tip: Remember that the middle digit \( b \) always cancels out during subtraction of reversed three-digit numbers.

Question 8. If c > a; show that cba - abc = 99(c - a).
Answer:
Given that \( c > a \).
We need to show that \( cba - abc = 99(c - a) \).
Expanding both numbers in terms of their place values:
\( cba = 100c + 10b + a \) ...(i)
\( abc = 100a + 10b + c \) ...(ii)
Subtracting equation (ii) from equation (i):
\( cba - abc = (100c + 10b + a) - (100a + 10b + c) \)
\( = 100c + 10b + a - 100a - 10b - c \)
Since the \( 10b \) terms subtract to zero, we are left with:
\( cba - abc = 99c - 99a \)
Taking out the common factor of 99:
\( cba - abc = 99(c - a) \)
Hence, the statement is proved.
In simple words: If the last digit of a three-digit number is larger than the first digit, subtracting the original number from its reversed version gives 99 times the difference of those two digits.

Exam Tip: Ensure you subtract the smaller place-value expansion from the larger one to keep the resulting difference positive.

Question 9. If a = c, show that cba - abc = 0.
Answer:
Given that \( a = c \).
We need to show that \( cba - abc = 0 \).
Using place value expansion, we write:
\( cba = 100c + 10b + a \) ...(i)
\( abc = 100a + 10b + c \) ...(ii)
Since \( a = c \), we substitute \( a \) with \( c \) in both equations:
\( cba = 100c + 10b + c \) ...(iii)
\( abc = 100c + 10b + c \) ...(iv)
Subtracting equation (iv) from equation (iii), we get:
\( cba - abc = (100c + 10b + c) - (100c + 10b + c) = 0 \)

\( \implies cba = abc \)
Hence proved.
In simple words: If the first and last digits of a three-digit number are equal, reversing the digits doesn't change the number. So, subtracting them gives zero.

Exam Tip: Substituting the given condition \( a = c \) directly shows that both expanded forms are exactly the same, making their difference zero.

Question 10. Show that 954 - 459 is exactly divisible by 99.
Answer:
Let us express \( 954 \) and \( 459 \) using their place values, with \( a = 9 \), \( b = 5 \), and \( c = 4 \):
\( abc = 100a + 10b + c \)

\( \implies 954 = 100 \times 9 + 10 \times 5 + 4 = 900 + 50 + 4 \) ...(i)
And its reversed form is:
\( 459 = 100 \times 4 + 10 \times 5 + 9 = 400 + 50 + 9 \) ...(ii)
Now, let us subtract equation (ii) from equation (i):
\( 954 - 459 = (900 + 50 + 4) - (400 + 50 + 9) \)
\( = 900 + 50 + 4 - 400 - 50 - 9 \)
Simplifying the terms, we get:
\( 954 - 459 = 500 - 5 \)

\( \implies 954 - 459 = 495 \)
Since \( 495 \) can be factored as \( 99 \times 5 \), it is a multiple of 99.
Hence, \( 954 - 459 \) is exactly divisible by 99.
In simple words: The difference between 954 and 459 is 495. Since 495 divided by 99 is exactly 5 with no remainder, the difference is perfectly divisible by 99.

Exam Tip: Expressing the numbers in expanded form first is a systematic way to prove divisibility by 99 without just doing standard subtraction.

Exercise 5(B)

Question 1. Solve the following addition:
  3 A
+ 2 5
------
  B 2
Answer:
Looking at the units place, we have \( A + 5 = 2 \) or a number ending in 2 (which must be 12).
So, \( A + 5 = 12 \)
\( \implies A = 12 - 5 = 7 \)
With \( A = 7 \), there is a carryover of 1 to the tens place.
Now, looking at the tens place, we have:
\( 3 + 2 + 1 \text{ (carryover)} = B \)
\( \implies B = 6 \)
Therefore, we find \( A = 7 \) and \( B = 6 \).
The completed sum is:
  3 7
+ 2 5
------
  6 2
In simple words: To get a 2 at the end of the sum, \( A \) must be 7 because \( 7 + 5 = 12 \). Carrying over the 1 to the next column, we add \( 1 + 3 + 2 \) to get \( B = 6 \).

Exam Tip: Always check if your sum has a carryover, as failing to add it to the next column is a very common error.

Question 2. Solve the following addition:
  9 8
+ 4 A
------
C B 3
Answer:
From the units column, we have \( 8 + A = 13 \) (since the result ends in 3 and must be greater than 8).
\( \implies A = 13 - 8 = 5 \)
This gives a carryover of 1 to the tens column.
Adding the tens column:
\( 9 + 4 + 1 \text{ (carryover)} = 14 \)
Comparing this with \( C B \), we see that \( B = 4 \) and \( C = 1 \).
Thus, the values are \( A = 5 \), \( B = 4 \), and \( C = 1 \).
The completed sum is:
  9 8
+ 4 5
------
1 4 3
In simple words: We need a digit \( A \) that when added to 8 gives a number ending in 3. That digit is 5, making the sum 13. We carry 1 to the tens place and add \( 1 + 9 + 4 = 14 \), so \( B = 4 \) and \( C = 1 \).

Exam Tip: Work column by column from right to left, and write down the carryover explicitly to keep track of it.

Question 3. Solve the following addition:
  A 1
+ 1 B
------
  B 0
Answer:
Looking at the units place, we have \( 1 + B = 10 \) (since the sum ends in 0).
\( \implies B = 9 \)
This addition results in a carryover of 1 to the tens place.
Now, considering the tens column, we have:
\( A + 1 + 1 \text{ (carryover)} = B \)
Substitute \( B = 9 \):
\( A + 2 = 9 \)
\( \implies A = 7 \)
Thus, \( A = 7 \) and \( B = 9 \).
The completed sum is:
  7 1
+ 1 9
------
  9 0
In simple words: Adding 1 to \( B \) gives 10, so \( B \) must be 9. This gives a carry of 1. In the tens column, \( A + 1 \) plus the carry of 1 must equal \( B \) (which is 9). So \( A \) must be 7.

Exam Tip: Always substitute the value found for the first letter back into the rest of the problem to verify it works for the whole sum.

Question 4. Solve the following addition:
  2 A B
+ A B 1
-------
  B 1 8
Answer:
From the units column, we see that:
\( B + 1 = 8 \)
\( \implies B = 7 \)
In the tens column, the sum of \( A + B \) ends in 1, which means:
\( A + B = 11 \)
Substituting \( B = 7 \):
\( A + 7 = 11 \)
\( \implies A = 4 \)
Let us verify this in the hundreds column by adding the digits and the carryover of 1 from the tens place:
\( 2 + A + 1 \text{ (carryover)} = 2 + 4 + 1 = 7 \)
Since this matches \( B = 7 \), our solution is correct.
Hence, \( A = 4 \) and \( B = 7 \).
The completed sum is:
  2 4 7
+ 4 7 1
-------
  7 1 8
In simple words: In the first column, \( B + 1 = 8 \), so \( B \) is 7. In the middle column, \( A + 7 = 11 \), which gives \( A = 4 \) and carries 1. In the last column, \( 2 + 4 + 1 = 7 \), which matches our \( B \) value.

Exam Tip: Verify all columns, especially the leftmost one, to ensure no carryover was missed.

Question 5. Solve the following addition:
  1 2 A
+ 6 A B
-------
  A 0 9
Answer:
In the units column, we have:
\( A + B = 9 \)
In the tens column, the sum is 0, which means \( 2 + A \) must equal 10:
\( 2 + A = 10 \)
\( \implies A = 8 \)
Now, substitute \( A = 8 \) back into the units column equation:
\( 8 + B = 9 \)
\( \implies B = 1 \)
Let us check the hundreds column, including the carryover of 1 from the tens place:
\( 1 + 6 + 1 \text{ (carryover)} = 8 \)
This perfectly matches \( A = 8 \).
Thus, the values are \( A = 8 \) and \( B = 1 \).
The completed sum is:
  1 2 8
+ 6 8 1
-------
  8 0 9
In simple words: The tens column shows \( 2 + A = 10 \), so \( A \) is 8, and we carry over 1. In the units column, \( 8 + B = 9 \), which means \( B \) is 1.

Exam Tip: When the sum in a column is 0, it usually indicates a sum of 10, which generates a carryover for the next column.

Question 6. Solve the following multiplication:
  1 A
x   A
------
  9 A
Answer:
We need to find a digit \( A \) such that the product \( 1A \times A \) equals \( 9A \).
Looking at the units digit, \( A \times A \) must end in the digit \( A \). The possible single digits for \( A \) are 0, 1, 5, and 6.
Let us test these possibilities:
- If \( A = 1 \), then \( 11 \times 1 = 11 \) (which does not match \( 91 \)).
- If \( A = 5 \), then \( 15 \times 5 = 75 \) (which does not match \( 95 \)).
- If \( A = 6 \), then \( 16 \times 6 = 96 \). This perfectly matches \( 9A \) with \( A = 6 \).
Therefore, the value of \( A \) is 6.
The completed multiplication is:
  1 6
x   6
------
  9 6
In simple words: We want a number of the form \( 1A \) multiplied by \( A \) to give \( 9A \). By trying different digits, we find that \( 16 \times 6 = 96 \), which works perfectly.

Exam Tip: For problems like \( A \times A \) ending in \( A \), limit your trials to the set of numbers {0, 1, 5, 6}.

Question 7. Solve the following multiplication:
  A B
x   6
------
B B B
Answer:
We are looking for digits \( A \) and \( B \) such that \( AB \times 6 = BBB \).
From the units column, \( B \times 6 \) must result in a number ending in \( B \). The possible even digits for \( B \) are 0, 2, 4, 6, and 8.
Let us test these values:
- If \( B = 4 \), then \( 4 \times 6 = 24 \). The units digit is 4, and the carryover is 2.
Now, for the tens and hundreds places, we must have \( A \times 6 + 2 = 44 \) (since the final product is \( 444 \)).
\( \implies A \times 6 = 42 \)
\( \implies A = 7 \)
This gives \( 74 \times 6 = 444 \), which is correct.
Thus, the solution is \( A = 7 \) and \( B = 4 \).
The completed multiplication is:
  7 4
x   6
------
4 4 4
In simple words: Multiplying a two-digit number by 6 gives a three-digit number with all identical digits. By testing different options for \( B \), we find that \( 74 \times 6 \) gives exactly 444.

Exam Tip: Since the final answer \( BBB \) has all digits same, use the units digit rule of multiplication by 6 to quickly narrow down the value of \( B \).

Question 8. Solve the following multiplication:
  A B
x   3
------
C A B
Answer:
We want to find \( A \), \( B \), and \( C \) such that \( AB \times 3 = CAB \).
Starting with the units place, the product \( B \times 3 \) must end in \( B \). The only digits that satisfy this are 0 and 5.
Let us try \( B = 0 \):
- If \( B = 0 \), then \( 0 \times 3 = 0 \). There is no carryover.
Now, for the tens place, we need \( A \times 3 \) to end in \( A \).
Using the same logic, \( A \) must be 0 or 5. Since \( A \) is the first digit of a two-digit number, \( A \neq 0 \). Thus, \( A = 5 \).
Now we have \( 50 \times 3 = 150 \).
Comparing this with \( CAB \), we get \( C = 1 \), \( A = 5 \), and \( B = 0 \). This is a perfect match.
Therefore, \( A = 5 \), \( B = 0 \), and \( C = 1 \).
The completed multiplication is:
  5 0
x   3
------
1 5 0
In simple words: The only way to multiply a number ending in \( B \) by 3 and still get a number ending in \( B \) is if \( B \) is 0 (or 5, which fails later). If \( B = 0 \), then to get \( A \) in the tens place of the answer, \( A \) must be 5 because \( 5 \times 3 = 15 \).

Exam Tip: When \( B \times 3 \) ends in \( B \), check both 0 and 5 as possible values, then test which one makes the tens column work.

Question 9. Solve the following multiplication:
  A B
x   5
------
C A B
Answer:
We need to find digits \( A \), \( B \), and \( C \) such that \( AB \times 5 = CAB \).
For the units digit, \( B \times 5 \) must end in \( B \). The possible values are \( B = 0 \) or \( B = 5 \).
Let us test \( B = 0 \):
If \( B = 0 \), there is no carryover from the units column.
For the tens column, we require \( A \times 5 \) to end in \( A \). Since \( A \) is the leading digit of a two-digit number, \( A \) cannot be 0. Thus, we have \( A = 5 \).
Now, let us calculate the product:
\( 50 \times 5 = 250 \).
Comparing this with \( CAB \), we get \( C = 2 \), \( A = 5 \), and \( B = 0 \). This is a consistent solution.
Therefore, \( A = 5 \), \( B = 0 \), and \( C = 2 \).
The completed multiplication is:
  5 0
x   5
------
2 5 0
In simple words: Since any number multiplied by 5 ends in either 0 or 5, \( B \) must be 0 or 5. Choosing \( B = 0 \) means we need \( A \times 5 \) to end in \( A \). This leads us to \( A = 5 \), giving a total product of 250.

Exam Tip: Multiplying by 5 simplifies the search space since units digits can only be 0 or 5. Try the 0 case first as it simplifies carryovers.

Question 10. Solve the following addition:
  8 A 5
+ 9 4 A
-------
1 A 3 3
Answer:
From the units column, we have:
\( 5 + A = 13 \)
\( \implies A = 13 - 5 = 8 \)
With \( A = 8 \), there is a carryover of 1 to the tens column.
Let us verify if \( A = 8 \) is consistent with the tens column:
\( A + 4 + 1 \text{ (carryover)} = 8 + 4 + 1 = 13 \)
The sum in the tens column is 13, which ends in 3 and gives a carryover of 1 to the hundreds column. This is consistent with the sum shown.
Now, let us verify the hundreds column:
\( 8 + 9 + 1 \text{ (carryover)} = 18 \)
Since the hundreds and thousands digits are \( 1A \), this matches \( 18 \) perfectly with \( A = 8 \).
Hence, the value of \( A \) is 8.
The completed sum is:
  8 8 5
+ 9 4 8
-------
1 8 3 3
In simple words: Adding 5 and \( A \) must give 13, which means \( A \) is 8. Putting 8 in place of \( A \) in the rest of the sum works out perfectly for all columns.

Exam Tip: When a single letter \( A \) appears in multiple columns, find its value from the simplest column first and then substitute it into the other columns to verify.

Question 11. Solve the following addition:
  6 A B 5
+ D 5 8 C
---------
  9 3 5 1
Answer:
Let us solve the addition column by column, starting from the units place:
1. Units column:
\( 5 + C = 11 \)
\( \implies C = 11 - 5 = 6 \)
We have a carryover of 1 to the tens column.

2. Tens column:
\( B + 8 + 1 \text{ (carryover)} = 15 \)
\( \implies B + 9 = 15 \)
\( \implies B = 6 \)
We have a carryover of 1 to the hundreds column.

3. Hundreds column:
\( A + 5 + 1 \text{ (carryover)} = 13 \)
\( \implies A + 6 = 13 \)
\( \implies A = 7 \)
We have a carryover of 1 to the thousands column.

4. Thousands column:
\( 6 + D + 1 \text{ (carryover)} = 9 \)
\( \implies D + 7 = 9 \)
\( \implies D = 2 \)
Thus, the values are \( A = 7 \), \( B = 6 \), \( C = 6 \), and \( D = 2 \).
The completed sum is:
  6 7 6 5
+ 2 5 8 6
---------
  9 3 5 1
In simple words: We find \( C = 6 \) by looking at the first column. This gives a carryover, which we use in the next column to find \( B = 6 \). Repeating this column by column gives \( A = 7 \) and \( D = 2 \).

Exam Tip: Be very careful to add the carryover at each step, especially in multi-variable additions where one mistake cascades through the rest.

Exercise 5(C)

Question 1. Find which of the following numbers are divisible by 2:
(i) 192
(ii) 1660
(iii) 1101
(iv) 2079
Answer:
A number is divisible by 2 if its units digit is even (i.e., 0, 2, 4, 6, or 8).
(i) 192: The last digit is 2, which is even. So, it is divisible by 2.
(ii) 1660: The last digit is 0, which is even. So, it is divisible by 2.
(iii) 1101: The last digit is 1, which is odd. So, it is not divisible by 2.
(iv) 2079: The last digit is 9, which is odd. So, it is not divisible by 2.
Therefore, the numbers 192 and 1660 are divisible by 2.
In simple words: Any number that ends with 0, 2, 4, 6, or 8 can be divided by 2. Out of the given numbers, only 192 and 1660 end with these digits.

Exam Tip: To test divisibility by 2, you only need to look at the very last digit, regardless of how large the rest of the number is.

Question 2. Find which of the following numbers are divisible by 3:
(i) 261
(ii) 111
(iii) 6657
(iv) 2574
Answer:
A number is divisible by 3 if the sum of its digits is a multiple of 3.
(i) 261: Sum of digits = \( 2 + 6 + 1 = 9 \). Since 9 is divisible by 3, 261 is divisible by 3.
(ii) 111: Sum of digits = \( 1 + 1 + 1 = 3 \). Since 3 is divisible by 3, 111 is divisible by 3.
(iii) 6657: Sum of digits = \( 6 + 6 + 5 + 7 = 24 \). Since 24 is divisible by 3, 6657 is divisible by 3.
(iv) 2574: Sum of digits = \( 2 + 5 + 7 + 4 = 18 \). Since 18 is divisible by 3, 2574 is divisible by 3.
Therefore, all the given numbers (261, 111, 6657, and 2574) are divisible by 3.
In simple words: Add up the digits of each number. If that sum can be divided by 3, then the original number is also divisible by 3. For all of these numbers, their digit sums are divisible by 3.

Exam Tip: Don't divide the whole number by 3. Just sum the digits to quickly check for divisibility by 3.

Question 3. Find which of the following numbers are divisible by 4:
(i) 360
(ii) 3180
(iii) 5348
(iv) 7756
Answer:
A number is divisible by 4 if the number formed by its last two digits is divisible by 4.
(i) 360: The last two digits are 60. Since \( 60 \) is divisible by 4, 360 is divisible by 4.
(ii) 3180: The last two digits are 80. Since \( 80 \) is divisible by 4, 3180 is divisible by 4.
(iii) 5348: The last two digits are 48. Since \( 48 \) is divisible by 4, 5348 is divisible by 4.
(iv) 7756: The last two digits are 56. Since \( 56 \) is divisible by 4, 7756 is divisible by 4.
Therefore, all the given numbers (360, 3180, 5348, and 7756) are divisible by 4.
In simple words: To see if a number can be divided by 4, just look at its last two digits. If that two-digit number is divisible by 4, then the whole number is too.

Exam Tip: For divisibility by 4, ignore everything except the tens and units digits of the number.

Question 4. Find which of the following numbers are divisible by 5:
(i) 3250
(ii) 5557
(iii) 39255
(iv) 8204
Answer:
A number is divisible by 5 if its last digit is either 0 or 5.
(i) 3250: The units digit is 0, so it is divisible by 5.
(ii) 5557: The units digit is 7, so it is not divisible by 5.
(iii) 39255: The units digit is 5, so it is divisible by 5.
(iv) 8204: The units digit is 4, so it is not divisible by 5.
Therefore, the numbers 3250 and 39255 are divisible by 5.
In simple words: Only numbers that end in 0 or 5 are divisible by 5. Here, only 3250 and 39255 meet this condition.

Exam Tip: Just check the units digit. If it is 0 or 5, the entire number is divisible by 5.

Question 5. Find which of the following numbers are divisible by 10:
(i) 5100
(ii) 4612
(iii) 3400
(iv) 8399
Answer:
A number is divisible by 10 if its last digit is 0.
(i) 5100: Ends in 0, so it is divisible by 10.
(ii) 4612: Ends in 2, so it is not divisible by 10.
(iii) 3400: Ends in 0, so it is divisible by 10.
(iv) 8399: Ends in 9, so it is not divisible by 10.
Therefore, 5100 and 3400 are divisible by 10.
In simple words: Any number that has a 0 at the very end can be divided by 10.

Exam Tip: Divisibility by 10 only requires checking if the unit digit is exactly 0.

Question 6. Which of the following numbers are divisible by 11:
(i) 2563
(ii) 8307
(iii) 95635
Answer:
A number is divisible by 11 if the difference between the sum of the digits at odd positions and the sum of the digits at even positions is either 0 or a multiple of 11.
(i) 2563:
- Sum of digits at odd places (from right): \( 3 + 5 = 8 \)
- Sum of digits at even places (from right): \( 6 + 2 = 8 \)
- Difference = \( 8 - 8 = 0 \)
Since the difference is 0, 2563 is divisible by 11.

(ii) 8307:
- Sum of digits at odd places: \( 7 + 3 = 10 \)
- Sum of digits at even places: \( 0 + 8 = 8 \)
- Difference = \( 10 - 8 = 2 \)
Since 2 is not a multiple of 11, 8307 is not divisible by 11.

(iii) 95635:
- Sum of digits at odd places: \( 5 + 6 + 9 = 20 \)
- Sum of digits at even places: \( 3 + 5 = 8 \)
- Difference = \( 20 - 8 = 12 \)
Since 12 is not a multiple of 11, 95635 is not divisible by 11.
Therefore, only 2563 is divisible by 11.
In simple words: Add up every second digit starting from the right, and then add up the remaining digits. Subtract the smaller total from the larger one. If the result is 0, 11, 22, etc., the number is divisible by 11.

Exam Tip: Be consistent when identifying odd and even places; always count from the rightmost digit (units place) to avoid errors.

Exercise 5(D)

Question 1. For what value of digit x, is \( 1x5 \) divisible by 3?
Answer:
For the three-digit number \( 1x5 \) to be divisible by 3, the sum of its digits must be a multiple of 3.
Sum of digits = \( 1 + x + 5 = 6 + x \)
Since \( x \) is a single digit (0 to 9), the possible multiples of 3 for \( 6 + x \) are 6, 9, 12, and 15.
- If \( 6 + x = 6 \implies x = 0 \)
- If \( 6 + x = 9 \implies x = 3 \)
- If \( 6 + x = 12 \implies x = 6 \)
- If \( 6 + x = 15 \implies x = 9 \)
Therefore, the possible values for the digit \( x \) are 0, 3, 6, or 9.
In simple words: Add the known digits: \( 1 + 5 = 6 \). To make the total sum divisible by 3, the missing middle digit \( x \) must be 0, 3, 6, or 9.

Exam Tip: Remember that \( x \) can be 0, and since it is in the tens place, 0 is a valid digit.

Question 2. For what value of digit x, is \( 31x5 \) divisible by 3?
Answer:
The four-digit number \( 31x5 \) is divisible by 3 if the sum of its digits is a multiple of 3.
Sum of digits = \( 3 + 1 + x + 5 = 9 + x \)
For \( 9 + x \) to be a multiple of 3, where \( x \) is a single digit:
- If \( 9 + x = 9 \implies x = 0 \)
- If \( 9 + x = 12 \implies x = 3 \)
- If \( 9 + x = 15 \implies x = 6 \)
- If \( 9 + x = 18 \implies x = 9 \)
Therefore, the possible values for \( x \) are 0, 3, 6, or 9.
In simple words: The sum of the given digits is 9. Since 9 is already divisible by 3, the digit \( x \) can be 0, 3, 6, or 9.

Exam Tip: If the sum of the known digits is already a multiple of 3, the unknown digit can be 0 or any multiple of 3 (3, 6, 9).

Question 3. For what value of digit x, is \( 28x6 \) a multiple of 3?
Answer:
For \( 28x6 \) to be a multiple of 3, the sum of its digits must be divisible by 3.
Sum of digits = \( 2 + 8 + x + 6 = 16 + x \)
Since \( x \) is a digit, \( 16 + x \) must be a multiple of 3 greater than or equal to 16:
- If \( 16 + x = 18 \implies x = 2 \)
- If \( 16 + x = 21 \implies x = 5 \)
- If \( 16 + x = 24 \implies x = 8 \)
Thus, the possible values for \( x \) are 2, 5, or 8.
In simple words: The sum of the known digits is 16. The next multiples of 3 are 18, 21, and 24. Subtracting 16 from these gives the possible values of \( x \) as 2, 5, or 8.

Exam Tip: Stop checking once the value of \( x \) exceeds 9, as \( x \) must be a single digit.

Question 4. For what value of digit x, is \( 24x \) divisible by 6?
Answer:
A number is divisible by 6 if it is divisible by both 2 and 3.
1. For divisibility by 2, the units digit \( x \) must be even. So, \( x \) can be 0, 2, 4, 6, or 8.
2. For divisibility by 3, the sum of the digits must be a multiple of 3:
Sum of digits = \( 2 + 4 + x = 6 + x \)
This means \( 6 + x \) must be 6, 9, 12, or 15, which gives \( x = 0, 3, 6, \text{ or } 9 \).

By finding the common values between both conditions, we see that \( x \) must be both even and a multiple of 3.
Thus, \( x \) can be 0 or 6.
In simple words: To be divisible by 6, a number must be even and its digit sum must be divisible by 3. Comparing the options, only 0 and 6 satisfy both conditions.

Exam Tip: Always check both divisibility conditions (by 2 and by 3) and find their intersection to solve for divisibility by 6.

Question 5. For what value of digit x, is \( 3x26 \) a multiple of 6?
Answer:
A number is a multiple of 6 if it is divisible by both 2 and 3.
The units digit of \( 3x26 \) is 6, which is an even number, so the number is already divisible by 2 for any digit value of \( x \).
Now, we only need to ensure divisibility by 3:
Sum of digits = \( 3 + x + 2 + 6 = 11 + x \)
For \( 11 + x \) to be a multiple of 3, the possible single-digit values are:
- If \( 11 + x = 12 \implies x = 1 \)
- If \( 11 + x = 15 \implies x = 4 \)
- If \( 11 + x = 18 \implies x = 7 \)
Therefore, the possible values for \( x \) are 1, 4, or 7.
In simple words: Since the number already ends in 6, it is even. We just need the sum of the digits, which is \( 11 + x \), to be a multiple of 3. This gives us \( x = 1, 4, \text{ or } 7 \).

Exam Tip: When the units digit is already even, the question of divisibility by 6 simplifies to just checking the divisibility rule for 3.

Question 6. For what value of digit x, is \( 42x8 \) divisible by 4?
Answer:
A number is divisible by 4 if the last two digits form a number that is divisible by 4.
The number formed by the last two digits of \( 42x8 \) is \( x8 \).
For the two-digit number \( x8 \) to be divisible by 4, we can test all single-digit values of \( x \) from 0 to 9:
- For \( x = 0 \), the number is 08, which is divisible by 4.
- For \( x = 2 \), the number is 28, which is divisible by 4.
- For \( x = 4 \), the number is 48, which is divisible by 4.
- For \( x = 6 \), the number is 68, which is divisible by 4.
- For \( x = 8 \), the number is 88, which is divisible by 4.
Other digits (1, 3, 5, 7, 9) result in numbers (18, 38, 58, 78, 98) that are not divisible by 4.
Thus, the possible values for \( x \) are 0, 2, 4, 6, or 8.
In simple words: A number is divisible by 4 if its last two digits can be divided by 4. Testing different digits for \( x \) in \( x8 \) shows that any even digit (0, 2, 4, 6, or 8) works.

Exam Tip: Always test all possible single digits from 0 to 9 for the unknown digit to ensure you do not miss any valid solution like 0.

Question 7. For what value of digit x, is \( 9142x \) a multiple of 4?
Answer:
For \( 9142x \) to be a multiple of 4, the number formed by its last two digits, which is \( 2x \), must be divisible by 4.
Let us test the values of \( x \) (from 0 to 9) to see which ones make \( 2x \) a multiple of 4:
- If \( x = 0 \), the last two digits are 20, which is divisible by 4.
- If \( x = 4 \), the last two digits are 24, which is divisible by 4.
- If \( x = 8 \), the last two digits are 28, which is divisible by 4.
For all other digits, \( 2x \) is not divisible by 4.
Thus, the possible values for the digit \( x \) are 0, 4, or 8.
In simple words: The last two digits are \( 2x \). To make this two-digit number divisible by 4, \( x \) must be 0, 4, or 8, giving us 20, 24, or 28.

Exam Tip: In divisibility by 4, always check if 0 is a valid solution, as 20 is divisible by 4.

Question 8. For what value of digit x, is \( 7x34 \) divisible by 9?
Answer:
A number is divisible by 9 if the sum of its digits is a multiple of 9.
Sum of digits = \( 7 + x + 3 + 4 = 14 + x \)
For \( 14 + x \) to be a multiple of 9, and since \( x \) is a single digit:
\( 14 + x = 18 \)
\( \implies x = 18 - 14 = 4 \)
Therefore, the value of \( x \) is 4.
In simple words: Add the known digits: \( 7 + 3 + 4 = 14 \). The next multiple of 9 is 18. So we need \( x \) to be 4 to reach a total sum of 18.

Exam Tip: Divisibility by 9 is similar to divisibility by 3 - just find the sum of all digits and ensure it equals a multiple of 9.

Question 9. For what value of digit x, is \( 5x555 \) a multiple of 9?
Answer:
For \( 5x555 \) to be a multiple of 9, the sum of all its digits must be divisible by 9.
Sum of digits = \( 5 + x + 5 + 5 + 5 = 20 + x \)
Since \( x \) is a single digit, the nearest multiple of 9 greater than or equal to 20 is 27.
\( 20 + x = 27 \)
\( \implies x = 7 \)
Thus, the value of \( x \) is 7.
In simple words: The sum of the known digits is 20. The next number divisible by 9 is 27. So, \( x \) must be 7.

Exam Tip: Double check your addition of the known digits to make sure the base sum is correct before finding the missing digit.

Question 10. For what value of digit x, is \( 3x2 \) divisible by 11?
Answer:
A three-digit number \( 3x2 \) is divisible by 11 if the difference between the sum of the digits at odd positions and even positions is either 0 or a multiple of 11.
- Sum of the digits at the odd positions (units and hundreds): \( 2 + 3 = 5 \)
- Sum of the digits at the even position (tens): \( x \)
The difference between these sums is \( 5 - x \).
For divisibility by 11, this difference must be 0:
\( 5 - x = 0 \)
\( \implies x = 5 \)
Therefore, the value of \( x \) is 5.
In simple words: The sum of the outer digits is \( 3 + 2 = 5 \). For the number to be divisible by 11, the middle digit \( x \) must be 5 so that the difference is 0.

Exam Tip: For any three-digit number to be divisible by 11, the middle digit must equal the sum of the outer digits (unless that sum is 10 or more).

Question 11. For what value of digit x, is \( 5x2 \) a multiple of 11?
Answer:
For \( 5x2 \) to be a multiple of 11, the difference between the sum of the digits at the odd and even places must be 0 or divisible by 11.
- Sum of the digits at the odd places (first and third digits): \( 5 + 2 = 7 \)
- Sum of the digits at the even place (second digit): \( x \)
The difference is \( 7 - x \).
Setting the difference to 0:
\( 7 - x = 0 \)
\( \implies x = 7 \)
Therefore, the value of \( x \) is 7.
In simple words: The first and last digits add up to 7. To make this number divisible by 11, the middle digit \( x \) must also be 7.

Exam Tip: Just like the previous question, the middle digit \( x \) is simply the sum of the outer digits since it is less than 10.