RBSE Solutions Class 8 Maths Chapter 4 Mental Exercise 4.1

Question 1. If 3-digit number 24x is divisible by 9 then find the value of x. Where x is a digit.
Answer: For a number to be divisible by 9, the sum of its digits must also be divisible by 9. The given number is \( 24x \). Let's find the sum of its digits: \( 2 + 4 + x = 6 + x \). Since \( x \) is a single digit, the possible values for \( 6 + x \) that are divisible by 9 are 9 or 18. Case 1: If \( 6 + x = 9 \) \( x = 9 - 6 \) \( x = 3 \) Case 2: If \( 6 + x = 18 \) \( x = 18 - 6 \) \( x = 12 \) However, \( x \) must be a single digit (0-9). So, \( x \neq 12 \). Thus, the only valid single digit value for \( x \) is 3. In simple words: To check if a number can be divided by 9, add up all its digits. If this sum can be divided by 9, then the number can too. For "24x", the sum is 6 + x. Since x must be a single digit, x has to be 3 for the total sum (9) to be divisible by 9.

🎯 Exam Tip: Remember that "x is a digit" means x must be a whole number from 0 to 9. This constraint is crucial in finding the correct answer.

Question 2. If 3-digit number 89y is divisible by 9 then find the value of y?
Answer: For a number to be divisible by 9, the sum of its digits must also be divisible by 9. The given number is \( 89y \). Let's find the sum of its digits: \( 8 + 9 + y = 17 + y \). Since \( y \) is a single digit, the possible values for \( 17 + y \) that are divisible by 9 are 18, 27, etc. Case 1: If \( 17 + y = 18 \) \( y = 18 - 17 \) \( y = 1 \) Case 2: If \( 17 + y = 27 \) \( y = 27 - 17 \) \( y = 10 \) However, \( y \) must be a single digit (0-9). So, \( y \neq 10 \). Thus, the only valid single digit value for \( y \) is 1. In simple words: Just like with the number 24x, for 89y to be divisible by 9, the sum of its digits (17 + y) must be a multiple of 9. With y being a single digit, y must be 1 to make the sum 18.

🎯 Exam Tip: Always list out the possible multiples of 9 (or 3, or 11, etc.) after finding the sum of digits, and then check which ones allow the variable to be a valid digit.

Question 3. 31M5 is a multiple of 9 and two values are obtained by M. Why? Where M is a digit?
Answer: For the number \( 31M5 \) to be a multiple of 9, the sum of its digits must be divisible by 9. Let's find the sum of its digits: \( 3 + 1 + M + 5 = 9 + M \). Since \( M \) is a single digit (0-9), the possible values for \( 9 + M \) that are multiples of 9 are 9 or 18. Case 1: If \( 9 + M = 9 \) \( M = 9 - 9 \) \( M = 0 \) Case 2: If \( 9 + M = 18 \) \( M = 18 - 9 \) \( M = 9 \) If \( 9 + M \) were 27, then \( M \) would be 18, which is not a single digit. Therefore, \( M \) can be either 0 or 9. This is why two values are obtained for \( M \). In simple words: For the number 31M5 to be divisible by 9, the sum of its digits (which is 9 + M) must also be divisible by 9. Since M is a single digit, M can be 0 (making the sum 9) or M can be 9 (making the sum 18). Both 9 and 18 are divisible by 9, so M has two possible values.

🎯 Exam Tip: When a question asks "Why?", always provide the step-by-step reasoning that leads to the conclusion, explaining each condition and outcome clearly.

Question 4. If 3-digit number 24y is a multiple of 3, then find the value of y?
Answer: For a number to be a multiple of 3, the sum of its digits must be divisible by 3. The given number is \( 24y \). Let's find the sum of its digits: \( 2 + 4 + y = 6 + y \). Since \( y \) is a single digit (0-9), the possible values for \( 6 + y \) that are multiples of 3 are 6, 9, 12, 15. (The next multiple 18 would make y=12, which is not a digit). Case 1: If \( 6 + y = 6 \) \( y = 6 - 6 \) \( y = 0 \) Case 2: If \( 6 + y = 9 \) \( y = 9 - 6 \) \( y = 3 \) Case 3: If \( 6 + y = 12 \) \( y = 12 - 6 \) \( y = 6 \) Case 4: If \( 6 + y = 15 \) \( y = 15 - 6 \) \( y = 9 \) If \( 6 + y \) were 18, then \( y \) would be 12, which is not a single digit. Therefore, the possible values for \( y \) are 0, 3, 6, and 9. In simple words: For 24y to be a multiple of 3, the sum of its digits (6 + y) must also be a multiple of 3. Since y is a single digit, y can be 0, 3, 6, or 9. Each of these values makes the sum (6, 9, 12, or 15) perfectly divisible by 3.

🎯 Exam Tip: Remember that "multiple of 3" implies the sum of digits can be 0, 3, 6, 9, 12, 15, and so on. Always check all possibilities within the range of a single digit (0-9).

Question 5. Test the divisibility of following numbers by 3, 9 and 11
(i) 294
(ii) 4455
(iii) 1041966
Answer:
(i) For the number 294:
Divisibility by 3: Sum of digits \( = 2 + 9 + 4 = 15 \). Since 15 is divisible by 3, 294 is also divisible by 3.
Divisibility by 9: Sum of digits \( = 2 + 9 + 4 = 15 \). Since 15 is not divisible by 9, 294 is not divisible by 9.
Divisibility by 11: Alternating sum of digits \( = (4 + 2) - 9 = 6 - 9 = -3 \). Since -3 is not divisible by 11, 294 is not divisible by 11.

(ii) For the number 4455:
Divisibility by 3: Sum of digits \( = 4 + 4 + 5 + 5 = 18 \). Since 18 is divisible by 3, 4455 is also divisible by 3.
Divisibility by 9: Sum of digits \( = 4 + 4 + 5 + 5 = 18 \). Since 18 is divisible by 9, 4455 is also divisible by 9.
Divisibility by 11: Alternating sum of digits \( = (5 - 5) + (4 - 4) = 0 + 0 = 0 \). Since 0 is divisible by 11, 4455 is also divisible by 11.

(iii) For the number 1041966:
Divisibility by 3: Sum of digits \( = 1 + 0 + 4 + 1 + 9 + 6 + 6 = 27 \). Since 27 is divisible by 3, 1041966 is also divisible by 3.
Divisibility by 9: Sum of digits \( = 1 + 0 + 4 + 1 + 9 + 6 + 6 = 27 \). Since 27 is divisible by 9, 1041966 is also divisible by 9.
Divisibility by 11: Alternating sum of digits (starting from the rightmost digit, with alternating signs) \( = 6 - 6 + 9 - 1 + 4 - 0 + 1 = 13 \). Since 13 is not divisible by 11, 1041966 is not divisible by 11. In simple words: For divisibility by 3 and 9, add up all the digits. If the sum can be divided by 3 (or 9), the number can too. For divisibility by 11, take the sum of digits at odd places and subtract the sum of digits at even places. If this result is 0 or a multiple of 11, the number is divisible by 11.

🎯 Exam Tip: Always show the sum of digits for 3 and 9 divisibility and the alternating sum for 11 divisibility. This demonstrates your understanding of the rules.

Question 6. If R = 4 in number 31R1 then by the rule of divisibility find that this number is divisible by 11.
Answer: If \( R = 4 \), the number becomes 3141. To check divisibility by 11, we find the alternating sum of its digits, starting from the rightmost digit. Alternating sum \( = 1 - 4 + 1 - 3 \) \( = 2 - 7 \) \( = -5 \) Since -5 is not divisible by 11, the number 3141 (or 31R1 when \( R=4 \)) is not divisible by 11. In simple words: We replace R with 4, making the number 3141. Then we check the rule for 11: subtract digits in an alternating pattern (1 - 4 + 1 - 3). The result is -5. Since -5 cannot be divided by 11, the number 3141 is not divisible by 11.

🎯 Exam Tip: When given a variable in a number, first substitute the value to form the complete number, then apply the divisibility rule. Clearly state your final conclusion based on the test result.

Question 7. If 31P5 is a multiple of 3 then find the value of P, where P is a digit?
Answer: For the number \( 31P5 \) to be a multiple of 3, the sum of its digits must be divisible by 3. Let's find the sum of its digits: \( 3 + 1 + P + 5 = 9 + P \). Since \( P \) is a single digit (0-9), the possible values for \( 9 + P \) that are multiples of 3 are 9, 12, 15, 18. (The next multiple 21 would make P=12, which is not a digit). Case 1: If \( 9 + P = 9 \) \( P = 9 - 9 \) \( P = 0 \) Case 2: If \( 9 + P = 12 \) \( P = 12 - 9 \) \( P = 3 \) Case 3: If \( 9 + P = 15 \) \( P = 15 - 9 \) \( P = 6 \) Case 4: If \( 9 + P = 18 \) \( P = 18 - 9 \) \( P = 9 \) If \( 9 + P \) were 21, then \( P \) would be 12, which is not a single digit. Therefore, the possible values for \( P \) are 0, 3, 6, and 9. In simple words: For the number 31P5 to be divisible by 3, we add its digits (3 + 1 + P + 5 = 9 + P). This sum must be a multiple of 3. Since P is a single digit, P can be 0, 3, 6, or 9, as these values make the sum (9, 12, 15, or 18) divisible by 3.

🎯 Exam Tip: List all possible multiples of 3 that the sum of digits (9+P) could be. Then, for each multiple, calculate the value of P and check if it falls within the single-digit range (0-9).