GSEB Class 8 Maths Solutions Chapter 16 Playing with Numbers Exercise 16.1

Question 1. Find the values of the letters in each of the following and give reasons for the steps involved?
1.
  3 A
+ 2 5
----
  B 2
Answer:
For the units column, \( A + 5 \) ends in 2. This means \( A + 5 = 12 \), so \( A = 7 \). (Since \( A \) cannot be 17)
Now, in the tens column, we have \( 1 \) (carry-over from units) \( + 3 + 2 = 6 \). This value is \( B \). So \( B = 6 \).
Therefore, the values are \( A = 7 \) and \( B = 6 \).
Check:
  3 7
+ 2 5
----
  6 2
In simple words: Start with the last column. \( A \) plus 5 must end in 2, so \( A \) is 7 with a carry-over of 1. Then, add the numbers in the middle column along with the carry-over to find \( B \).

Exam Tip: Always work from the rightmost column (units place) to the leftmost column, carefully tracking any carry-overs to the next column.

Question 1. Find the values of the letters in each of the following and give reasons for the steps involved?
2.
  4 A
+ 9 8
----
 C B 3
Answer:
For the units column, \( A + 8 \) ends in 3. This means \( A + 8 = 13 \), so \( A = 5 \).
Now, for the tens column, we have \( 1 \) (carry-over from units) \( + 4 + 9 \). This sum is \( 1 + 4 + 9 = 14 \). So, the tens digit \( B \) is 4, and there is a carry-over of 1 to the hundreds column.
For the hundreds column, we have \( 1 \) (carry-over from tens) and no other digits, so \( C \) is 1.
Therefore, the values are \( A = 5 \), \( B = 4 \), and \( C = 1 \).
Check:
  4 5
+ 9 8
----
1 4 3
In simple words: First, find \( A \) by looking at the units digits. Then, find \( B \) using the tens digits and the carry-over from the units. Finally, find \( C \) using the hundreds digits and any carry-overs.

Exam Tip: Be sure to manage carry-overs carefully in each column as you move from right to left, as they are crucial for solving these puzzles accurately.

Question 1. Find the values of the letters in each of the following and give reasons for the steps involved?
3.
  1 A
× A
----
  9 A
Answer:
For the units column, \( A \times A \) must result in a number whose units digit is \( A \). Possible digits for \( A \) are 0, 1, 5, or 6:
- If \( A = 0 \), then \( 10 \times 0 = 0 \). The product would be 00, but we need 9A. So \( A \ne 0 \).
- If \( A = 1 \), then \( 11 \times 1 = 11 \). The product should be 9A, which is 91. Since \( 11 \ne 91 \), \( A \ne 1 \).
- If \( A = 5 \), then \( 15 \times 5 = 75 \). The product should be 9A, which is 95. Since \( 75 \ne 95 \), \( A \ne 5 \).
- If \( A = 6 \), then \( 16 \times 6 = 96 \). The product 96 matches 9A (where \( A = 6 \)). This works.
Therefore, the required value for \( A \) is 6.
Check:
  1 6
× 6
----
  9 6
In simple words: Look for a number \( A \) where \( A \) times \( A \) ends in \( A \). Then, try those possible \( A \) values in the full multiplication to see which one makes the entire sum correct.

Exam Tip: When dealing with digit puzzles involving multiplication, consider the units digit first to narrow down the possibilities for the unknown letter. Then, test these possibilities to find the correct solution.

Question 1. Find the values of the letters in each of the following and give reasons for the steps involved?
4.
  A B
+ 3 7
----
  6 A
Answer:
For the units column, \( B + 7 \) ends in \( A \). This suggests a carry-over is likely. If \( B + 7 = 10 + A \), then \( B + 7 \ge 10 \).
For the tens column, \( 1 \) (carry-over from units) \( + A + 3 = 6 \).
So, \( A + 4 = 6 \), which gives \( A = 2 \).
Now that we know \( A = 2 \), let's use the units column: \( B + 7 \) ends in 2. This means \( B + 7 = 12 \), so \( B = 5 \).
Therefore, the values are \( A = 2 \) and \( B = 5 \).
Check:
  2 5
+ 3 7
----
  6 2
In simple words: Start with the tens column, remembering a possible carry-over. Find \( A \). Once you have \( A \), use it in the units column to find \( B \).

Exam Tip: Sometimes, starting with the tens or hundreds column first (if it involves fewer unknowns or a clear carry-over) can simplify the problem. Always cross-check the solution by performing the original operation.

Question 1. Find the values of the letters in each of the following and give reasons for the steps involved?
5.
  A B
× 3
----
C A B
Answer:
For the units column, \( B \times 3 \) must end in \( B \). This means \( B \) can be 0 (since \( 0 \times 3 = 0 \)) or 5 (since \( 5 \times 3 = 15 \), which ends in 5).
Let's consider these two possibilities for \( B \):
**Case 1: \( B = 5 \)**
If \( B = 5 \), there is a carry-over of 1 to the tens column. For the tens column, \( A \times 3 + 1 \) must end in \( A \). Let's test values for \( A \):
- If \( A = 0 \), \( 0 \times 3 + 1 = 1 \) (does not end in 0).
- If \( A = 1 \), \( 1 \times 3 + 1 = 4 \) (does not end in 1).
- If \( A = 2 \), \( 2 \times 3 + 1 = 7 \) (does not end in 2).
- If \( A = 3 \), \( 3 \times 3 + 1 = 10 \) (ends in 0, not 3).
- If \( A = 4 \), \( 4 \times 3 + 1 = 13 \) (ends in 3, not 4).
- If \( A = 5 \), \( 5 \times 3 + 1 = 16 \) (ends in 6, not 5).
- If \( A = 6 \), \( 6 \times 3 + 1 = 19 \) (ends in 9, not 6).
- If \( A = 7 \), \( 7 \times 3 + 1 = 22 \) (ends in 2, not 7).
- If \( A = 8 \), \( 8 \times 3 + 1 = 25 \) (ends in 5, not 8).
- If \( A = 9 \), \( 9 \times 3 + 1 = 28 \) (ends in 8, not 9).
Since no value for \( A \) works when \( B = 5 \), we conclude \( B \ne 5 \).
**Case 2: \( B = 0 \)**
If \( B = 0 \), there is no carry-over to the tens column. For the tens column, \( A \times 3 \) must end in \( A \). This means \( A \) can be 0 or 5.
- If \( A = 0 \), then \( 00 \times 3 = 000 \). But \( CAB \) implies \( C \ne 0 \), so \( A \ne 0 \).
- If \( A = 5 \), then \( 50 \times 3 = 150 \). Here, \( A = 5 \), \( B = 0 \), and \( C = 1 \). This solution works!
Therefore, the values are \( A = 5 \), \( B = 0 \), and \( C = 1 \).
Check:
  5 0
× 3
----
1 5 0
In simple words: First, figure out what \( B \) could be by checking its units digit. Then, test each possible \( B \) value. For each \( B \), check what \( A \) must be. Remember to use carry-overs and ensure all parts of the answer match.

Exam Tip: In multiplication puzzles, analyze the units digit of the product and the multiplier to deduce possible values for the unknown. This systematic trial-and-error approach often leads to the correct answer.

Question 1. Find the values of the letters in each of the following and give reasons for the steps involved?
6.
  A B
× 5
----
C A B
Answer:
For the units column, \( B \times 5 \) must end in \( B \). This means \( B \) can be 0 (since \( 0 \times 5 = 0 \)) or 5 (since \( 5 \times 5 = 25 \), which ends in 5).
Let's consider these two possibilities for \( B \):
**Case 1: \( B = 5 \)**
If \( B = 5 \), there is a carry-over of 2 to the tens column. For the tens column, \( A \times 5 + 2 \) must end in \( A \). Let's test values for \( A \):
- If \( A = 2 \), \( 2 \times 5 + 2 = 12 \). This ends in 2, which matches \( A \). So, \( A = 2 \) is a possible value.
If \( A = 2 \) and \( B = 5 \), then \( 25 \times 5 = 125 \). Here, \( A = 2 \), \( B = 5 \), and \( C = 1 \). This works!
**Case 2: \( B = 0 \)**
If \( B = 0 \), there is no carry-over to the tens column. For the tens column, \( A \times 5 \) must end in \( A \). This means \( A \) can be 0 or 5.
- If \( A = 0 \), then \( 00 \times 5 = 000 \). But \( CAB \) implies \( C \ne 0 \), so \( A \ne 0 \).
- If \( A = 5 \), then \( 50 \times 5 = 250 \). Here, \( A = 5 \), \( B = 0 \), and \( C = 2 \). This solution works!
Both cases provide a valid solution. We will follow the solution provided, which is \( A = 5 \), \( B = 0 \), and \( C = 2 \).
Check:
  5 0
× 5
----
2 5 0
In simple words: First, find possible values for \( B \) by checking the units digit. Then, for each possible \( B \), find \( A \) by checking the tens digit, including any numbers carried over. Confirm the result by checking if it makes sense for the hundreds digit.

Exam Tip: Be aware that some puzzles might have more than one mathematical solution. If the problem implies a unique answer, ensure your chosen solution fits all conditions and constraints provided.

Question 1. Find the values of the letters in each of the following and give reasons for the steps involved?
7.
  A B
× 6
----
B B B
Answer:
For the units column, \( B \times 6 \) must end in \( B \). Let's test digits for \( B \):
- If \( B = 0 \), then \( 0 \times 6 = 0 \). But in \( BBB \), \( B \) cannot be 0 (as \( BBB \) would be 000, implying \( A=0 \), which is trivial). So \( B \ne 0 \).
- If \( B = 2 \), then \( 2 \times 6 = 12 \) (ends in 2, carry 1).
- If \( B = 4 \), then \( 4 \times 6 = 24 \) (ends in 4, carry 2).
- If \( B = 6 \), then \( 6 \times 6 = 36 \) (ends in 6, carry 3).
- If \( B = 8 \), then \( 8 \times 6 = 48 \) (ends in 8, carry 4).
So, \( B \) can be 2, 4, 6, or 8.
The product \( BBB \) can be written as \( B \times 111 \). We have \( AB \times 6 = B \times 111 \).
Dividing both sides by \( B \) (since \( B \ne 0 \)), we get \( A \times 6 = 111 \) which is not right, or rather we should think about it as \( AB = (B \times 111) / 6 \).
Since \( 111 \) is divisible by 3 (\( 1+1+1=3 \)), but not by 2, \( B \times 111 \) must be an even number divisible by 6. This means \( B \) must be an even digit.
Possible products \( BBB \) are 222, 444, 666, 888. However, 111 and 333 are rejected since they are not divisible by 6.
- If \( BBB = 222 \), then \( AB = 222 \div 6 = 37 \). Here, \( A = 3 \), \( B = 7 \). This doesn't match our possible \( B \) values (2, 4, 6, 8). So \( B \ne 7 \).
- If \( BBB = 444 \), then \( AB = 444 \div 6 = 74 \). Here, \( A = 7 \), \( B = 4 \). This matches our possible \( B \) values! So \( A = 7 \) and \( B = 4 \).
- If \( BBB = 666 \), then \( AB = 666 \div 6 = 111 \). This is a 3-digit number, but \( AB \) must be a 2-digit number. So this is not possible.
- If \( BBB = 888 \), then \( AB = 888 \div 6 = 148 \). This is also a 3-digit number, which is not possible.
Therefore, the values are \( A = 7 \) and \( B = 4 \).
Check:
  7 4
× 6
----
4 4 4
In simple words: First, find possible digits for \( B \) by checking the units place. Then, use the fact that \( AB \times 6 = BBB \) (meaning \( B \times 111 \)) to test which possible \( B \) gives a valid two-digit \( AB \).

Exam Tip: For multiplication puzzles, consider the algebraic relationship between the terms (\( AB \times X = YYY \)) to simplify the problem. Remember that digits must satisfy the constraints of both value and position.

Question 1. Find the values of the letters in each of the following and give reasons for the steps involved?
8.
  A 1
+ 1 B
----
  B 0
Answer:
For the units column, \( 1 + B \) must end in 0. This means \( 1 + B = 10 \), so \( B = 9 \). There is a carry-over of 1 to the tens column.
For the tens column, we have \( 1 \) (carry-over from units) \( + A + 1 \). This sum must equal \( B \).
So, \( A + 2 = B \). Since we found \( B = 9 \), we have \( A + 2 = 9 \), which means \( A = 7 \).
Therefore, the values are \( A = 7 \) and \( B = 9 \).
Check:
  7 1
+ 1 9
----
  9 0
In simple words: Start by adding the units column to find \( B \) and any number carried over. Then, use that carry-over in the tens column's sum to discover \( A \).

Exam Tip: Always solve for the unknown letters systematically, starting with the column that offers the most direct information (often the units column). Work through carry-overs for subsequent columns.

Question 1. Find the values of the letters in each of the following and give reasons for the steps involved?
9.
 2 A B
+ A B 1
----
 B 1 8
Answer:
For the units column, \( B + 1 \) ends in 8. This means \( B + 1 = 8 \), so \( B = 7 \).
For the tens column, \( A + B \) (which is \( A + 7 \)) ends in 1. This means \( A + 7 = 11 \), so \( A = 4 \). There is a carry-over of 1 to the hundreds column.
For the hundreds column, we have \( 1 \) (carry-over from tens) \( + 2 + A \). This sum must equal \( B \).
So, \( 1 + 2 + 4 = 7 \). This matches our value for \( B = 7 \).
Therefore, the values are \( A = 4 \) and \( B = 7 \).
Check:
 2 4 7
+ 4 7 1
----
 7 1 8
In simple words: Determine \( B \) from the units column. Then, use that \( B \) and the tens column to find \( A \), remembering any carry-overs. Finally, check if these values fit in the hundreds column too.

Exam Tip: When the same letter appears in multiple places in the puzzle, its value must be consistent throughout the entire problem. Use one column to find a letter, then apply that value to solve other columns.

Question 1. Find the values of the letters in each of the following and give reasons for the steps involved?
10.
 1 2 A
+ 6 A B
----
 A 0 9
Answer:
For the tens column, \( 2 + A \) ends in 0. This means \( 2 + A = 10 \), so \( A = 8 \). There is a carry-over of 1 to the hundreds column.
For the hundreds column, we have \( 1 \) (carry-over from tens) \( + 1 + 6 \). This sum must equal \( A \).
So, \( 1 + 1 + 6 = 8 \). This matches our value for \( A = 8 \).
Now, for the units column, \( A + B \) ends in 9. Since \( A = 8 \), we have \( 8 + B \) ends in 9. This means \( 8 + B = 9 \), so \( B = 1 \). (There is no carry-over from units to tens in this case).
Therefore, the values are \( A = 8 \) and \( B = 1 \).
Check:
 1 2 8
+ 6 8 1
----
 8 0 9
In simple words: Solve for \( A \) using the tens column first, considering a carry-over to the hundreds. Confirm \( A \) with the hundreds column. Then, use the value of \( A \) to find \( B \) from the units column.

Exam Tip: When an unknown letter is present in multiple columns, use the column that provides the most definitive information first. Often, a unique solution for one letter can unlock the rest of the puzzle.