GSEB Class 8 Maths Solutions Chapter 16 Playing with Numbers InText Questions

Try These (Page 250)

Question 1. Write the following numbers in generalised form?
1. 25
2. 73
3. 129
4. 302
Answer:
1. \( 25 = 20 + 5 = (10 \times 2) + (5 \times 1) = (10 \times 2) + 5 \)
2. \( 73 = 70 + 3 = (10 \times 7) + (3 \times 1) = (10 \times 7) + 3 \)
3. \( 129 = 100 + 20 + 9 = (100 \times 1) + (10 \times 2) + (1 \times 9) = (100 \times 1) + (10 \times 2) + 9 \)
4. \( 302 = 100 \times 3 + 10 \times 0 + 1 \times 2 = 300 + 0 + 2 \)
In simple words: To write a number in generalised form, you break it down by its place value. For example, a two-digit number \( AB \) is \( (10 \times A) + B \), and a three-digit number \( ABC \) is \( (100 \times A) + (10 \times B) + C \).

Exam Tip: Always remember that the generalised form shows a number as the sum of the products of its digits and their respective place values (powers of 10).

Question 2. Write the following in the usual form?
1. \( 10 \times 5 + 6 \)
2. \( 100 \times 7 + 10 \times 1 + 8 \)
3. \( 100 \times a + 10 \times c + b \)
Answer:
1. \( 10 \times 5 + 6 = 50 + 6 = 56 \)
2. \( 100 \times 7 + 10 \times 1 + 8 = 700 + 10 + 8 = 718 \)
3. \( 100 \times a + 10 \times c + b = 100a + 10c + b \)
This can be represented as \( abc \), where \( b \) is the ones digit, \( c \) is the tens digit, and \( a \) is the hundreds digit.
In simple words: When you have a number in its expanded, generalised form (like \( 10 \times 5 + 6 \)), you just do the math to get the normal number. If it has letters, you show the number by writing the letters in the correct order for their place values.

Exam Tip: The usual form is simply the standard way we write numbers. Generalized form helps us understand the place value of each digit within a number.

Try These (Page 251)

Adding the number with reversed digits to the chosen numbers.

Question 1. Check what the result would have been if Sundaram had chosen the numbers shown below?
1. 27
2. 39
3. 64
4. 17
Answer:
1. Chosen number = 27
Number with reversed digits = 72
Sum of the two numbers = \( 27 + 72 = 99 \)
Now, \( 99 = 11 \times 9 = 11 \times (2 + 7) \)
\( \implies 11 \times (\text{Sum of the digits of the chosen number}) \)
2. Chosen number = 39
Number with reversed digits = 93
Sum of the two numbers = \( 39 + 93 = 132 \)
Now, \( 132 \div 11 = 12 \)
i.e., \( 132 = 11 \times 12 = 11 \times (3 + 9) \)
\( \implies 11 \times (\text{Sum of the digits of the chosen number}) \)
3. Chosen number = 64
Number with reversed digits = 46
Sum = \( 64 + 46 = 110 \)
Now, \( 110 = 11 \times 10 = 11 \times (6 + 4) \)
\( \implies 11 \times (\text{Sum of the digits of the chosen number}) \)
4. Chosen number = 17
Number with reversed digits = 71
Sum = \( 17 + 71 = 88 \)
Now, \( 88 = 11 \times 8 = 11 \times (1 + 7) \)
\( \implies 11 \times (\text{Sum of the digits of the chosen number}) \)
In simple words: When you pick a two-digit number, reverse its digits, and then add the original number and the reversed number, the total sum you get will always be 11 times the sum of the original two digits.

Exam Tip: This is a fun number trick. The key idea is that any two-digit number \( (10a+b) \) added to its reverse \( (10b+a) \) results in \( 11a+11b = 11(a+b) \), which explains the pattern.

Try These (Page 251)

Finding the difference of the chosen number and the number obtained by reversing the digits.

Question 1. Check what result would have been if Sundaram had chosen the numbers shown?
1. 17
2. 21
3. 96
4. 37
Answer:
1. Chosen number = 17
Number with reversed digits = 71
Difference = \( 71 - 17 = 54 = 9 \times [6] \)
\( \implies 9 \times (\text{Difference of the digits of the chosen number } (7 - 1 = 6)) \)
2. Chosen number = 21
Number with reversed digits = 12
Difference = \( 21 - 12 = 9 = 9 \times [1] \)
\( \implies 9 \times (\text{Difference between the digits of the chosen number } (2 - 1 = 1)) \)
3. Chosen number = 96
Number with reversed digits = 69
Difference = \( 96 - 69 = 27 = 9 \times [3] \)
\( \implies 9 \times (\text{Difference between the digits of the chosen number } (9 - 6 = 3)) \)
4. Chosen number = 37
Number with reversed digits = 73
Difference = \( 73 - 37 = 36 = 9 \times [4] \)
\( \implies 9 \times (\text{Difference between the digits of the chosen number } (7 - 3 = 4)) \)
In simple words: When you take a two-digit number, reverse its digits, and then subtract the smaller number from the larger one, the result will always be 9 times the difference between the two digits of the original number.

Exam Tip: This number pattern works because the difference between a two-digit number \( (10a+b) \) and its reverse \( (10b+a) \) is \( |(10a+b) - (10b+a)| = |9a - 9b| = 9|a-b| \).

Try These (Page 252)

Question 1. Check what the result would have been if Minakshi had chosen the numbers shown below. In each case keep a record of the quotient obtained at the end?
1. 132
2. 469
3. 737
4. 901
Answer:
1. Chosen number = 132
Reversed number = 231
Difference = \( 231 - 132 = 99 \)
We have \( 99 \div 99 = 1 \), remainder = 0
2. Chosen number = 469
Reversed number = 964
Difference = \( 964 - 469 = 495 \)
We have \( 495 \div 99 = 5 \), remainder = 0
3. Chosen number = 737
Reversed number = 737
We have Difference = \( 737 - 737 = 0 \)
\( 0 \div 99 = 0 \), remainder = 0
4. Chosen number = 901
Reversed number = 109
Difference = \( 901 - 109 = 792 \)
We have \( 792 \div 99 = 8 \), remainder = 0
In simple words: When you pick a three-digit number, reverse its digits, find the difference between them, and then divide that difference by 99, you will always get a whole number as the quotient, usually with no remainder.

Exam Tip: This pattern stems from the fact that a three-digit number \( (100a+10b+c) \) minus its reverse \( (100c+10b+a) \) simplifies to \( 99a - 99c = 99(a-c) \), making it always divisible by 99.

Note: Forming three-digit number with given three digits
Generalised form of \( abc = 100a + 10b + c \)
Generalised form of \( cab = 100c + 10a + b \)
Generalised form of \( bca = 100b + 10c + a \)
Adding \( abc + cab + bca \)
\( = (100a + 10b + c) + (100c + 10a + b) + (100b + 10c + a) \)
\( = 111a + 111b + 111c \)
\( = 111(a + b + c) \)
\( = (37 \times 3)(a + b + c) \) which is divisible by 37

Try These (Page 253)

Question 1. Check what the result would have been if Sundaram had chosen the numbers shown below?
1. 417
2. 632
3. 117
4. 937
Answer:
1. Chosen number = 417
Two other numbers with the same digits are 741 and 174
Sum of the three numbers = \( 417 + 741 + 174 = 1332 \)
We have \( 1332 \div 37 = 36 \), remainder = 0
2. Chosen number = 632
Two other numbers are 263 and 326
Sum of the three numbers = \( 632 + 263 + 326 = 1221 \)
We have \( 1221 \div 37 = 33 \), remainder = 0
3. Chosen number = 117
Other numbers are 711 and 171
Sum of the three numbers = \( 117 + 711 + 171 = 999 \)
We have \( 999 \div 37 = 27 \), remainder = 0
4. Chosen number = 937
Other two numbers are 793 and 379
Sum of the three numbers = \( 937 + 793 + 379 = 2109 \)
We have \( 2109 \div 37 = 57 \), remainder = 0
In simple words: If you take a three-digit number, then form all possible unique numbers using its digits (usually two more numbers if all digits are different), and add them up, the total sum will always be perfectly divisible by 37.

Exam Tip: This property is true because the sum of all permutations of a three-digit number \( abc \) (where digits can be repeated) is \( 222(a+b+c) = (37 \times 6)(a+b+c) \), which is clearly divisible by 37.

Try These (Page 257)

Question 1. If the division N ÷ 5 leaves a remainder of 1, what might be the ones digit of N?
Answer: If a number \( N \) is divided by 5 and leaves a remainder of 1, its ones digit must be either 1 or 6. This occurs because numbers ending in 1 or 6 are the only ones that leave a remainder of 1 when divided by 5.
In simple words: When you divide a number by 5 and get a remainder of 1, the last digit of that number must be either 1 or 6.

Exam Tip: For divisibility by 5, the ones digit determines the remainder. If the ones digit is 0 or 5, the remainder is 0. Otherwise, the remainder is the remainder of the ones digit divided by 5.

Question 2. If the division N ÷ 5 leaves a remainder of 1, what might be the ones digit of N?
Answer: If the remainder is 1, then the ones digit of 'N' must be either 1 or 6.
In simple words: For a number to have a remainder of 1 when divided by 5, its last digit has to be a 1 or a 6.

Exam Tip: The ones digit rule for division by 5 is simple: if the ones digit is 0 or 5, it's divisible. For other ones digits, the remainder will be that digit's remainder when divided by 5.

Question 3. If the division N ÷ 5 leaves a remainder of 4, what might be the ones digit of N?
Answer: If the remainder is 4, then the ones digit of 'N' must be either 9 or 4.
In simple words: If a number leaves a remainder of 4 when divided by 5, its last digit must be 4 or 9.

Exam Tip: Quickly check these rules by testing a few numbers: 4, 9, 14, 19, 24, 29, etc. All these numbers end in 4 or 9 and leave a remainder of 4 when divided by 5.

Try These (Page 257)

Question 1. If the division N ÷ 2 leaves a remainder of 1, what might be the ones digit of N?
Answer: \( N \) is an odd number; so its ones digit must also be odd. Therefore, the ones digit can be 1, 3, 5, 7 or 9.
In simple words: If you divide a number by 2 and get a remainder of 1, that number is odd. So, its last digit must be an odd number, like 1, 3, 5, 7, or 9.

Exam Tip: Remember that all odd numbers leave a remainder of 1 when divided by 2, and all even numbers leave a remainder of 0.

Question 2. If the division N ÷ 2 leaves no remainder (i.e. zero remainder), what might be the one digit of N?
Answer: Since the remainder is 0, the ones digit can be 0, 2, 4, 6 or 8.
In simple words: If a number is perfectly divisible by 2 (meaning no remainder), it's an even number. So, its last digit must be an even number like 0, 2, 4, 6, or 8.

Exam Tip: This is the definition of an even number. Knowing the ones digit quickly tells you if a number is even or odd.

Question 3. Suppose that the division N ÷ 5 leaves a remainder of 4 and the division N ÷ 2 leaves a remainder of I. What must be the ones digit of N?
Answer: For \( N \div 5 \) to leave a remainder of 4, the ones digit of \( N \) can be 4 or 9.
For \( N \div 2 \) to leave a remainder of 1, \( N \) must be an odd number. This means its ones digit must be odd.
Combining both conditions, the only digit that is both 4 or 9 AND odd is 9.
Thus, the ones digit of \( N \) must be 9 only.
In simple words: If a number leaves a remainder of 4 when divided by 5 (so it ends in 4 or 9) AND it leaves a remainder of 1 when divided by 2 (so it's an odd number), then its last digit must be 9.

Exam Tip: When given multiple conditions, list the possibilities for each condition separately, then find the common elements that satisfy all conditions simultaneously.

Try These (Page 259)

Question 1. Check the divisibility of the following numbers by 9?
1. 108
2. 616
3. 294
4. 432
5. 927
Answer:
1. 108
Sum of digits = \( 1 + 0 + 8 = 9 \)
Since 9 is divisible by 9, then 108 is divisible by 9.
2. 616
Sum of digits = \( 6 + 1 + 6 = 13 \)
Since 13 is not divisible by 9 (it leaves a remainder of 4), then 616 is also not divisible by 9.
3. 294
Sum of digits = \( 2 + 9 + 4 = 15 \)
Since 15 is not divisible by 9 (it leaves a remainder of 6), then 294 is also not divisible by 9.
4. 432
Sum of digits = \( 4 + 3 + 2 = 9 \)
Since 9 is divisible by 9, then 432 is divisible by 9.
5. 927
Sum of digits = \( 9 + 2 + 7 = 18 \)
Since 18 is divisible by 9, then 927 is also divisible by 9.
In simple words: To check if a number can be divided evenly by 9, just add up all its digits. If that sum can be divided by 9 without any remainder, then the original number can also be divided by 9.

Exam Tip: The divisibility rule for 9 is very similar to the rule for 3. The sum of the digits must be a multiple of 9 for the number to be divisible by 9.

Try These (Page 260)

Question 1. Check the divisibility of the following numbers by 3?
1. 108
2. 616
3. 294
4. 432
5. 927
Answer:
1. 108
Sum of digits = \( 1 + 0 + 8 = 9 \)
Since 9 is divisible by 3, then 108 is divisible by 3.
2. 616
Sum of digits = \( 6 + 1 + 6 = 13 \)
Since 13 is not divisible by 3 (it leaves a remainder of 1), then 616 is also not divisible by 3.
3. 294
Sum of digits = \( 2 + 9 + 4 = 15 \)
Since 15 is divisible by 3, then 294 is also divisible by 3.
4. 432
Sum of digits = \( 4 + 3 + 2 = 9 \)
Since 9 is divisible by 3, then 432 is divisible by 3.
5. 927
Sum of digits = \( 9 + 2 + 7 = 18 \)
Since 18 is divisible by 3, then 927 is also divisible by 3.
In simple words: To find out if a number can be evenly divided by 3, just add up all its digits. If the sum you get can be divided by 3 without any remainder, then the original number is also divisible by 3.

Exam Tip: The divisibility rule for 3 is a quick way to determine if a number is a multiple of 3. It's often helpful to quickly sum digits for larger numbers.