Number System Worksheet-4: A multiple-choice quiz covering fundamental math concepts including odd/even numbers, divisibility by 3 and 6, and the order of operations.

Here are the step-by-step solutions for the worksheet questions:

(1) The sum of two odd numbers is \_\_\_\_\_\_ an even number.
* Reasoning: Let's test this with some examples.
* $3 + 5 = 8$ (Even)
* $1 + 7 = 8$ (Even)
* $9 + 11 = 20$ (Even)
* Mathematically, any odd number can be written as $2n + 1$. If you add two of them: $(2n + 1) + (2m + 1) = 2n + 2m + 2 = 2(n + m + 1)$. Since the result is multiplied by 2, it is always even.
* Answer: (A) Always

(2) \_\_\_\_\_\_ is the key to find that the number is multiple of 3 or not.
* Reasoning: The rule for divisibility by 3 states that a number is divisible by 3 if the sum of its digits is divisible by 3.
* Let's look at the options. Option (B) says "divisible by second smallest odd natural number." The odd natural numbers are 1, 3, 5... The second smallest is 3. So, this option describes the correct rule.
* Answer: (B) The sum of all the digits of a number should be divisible by second smallest odd natural number.

(3) The number is divisible by 6. So; it is divisible by \_\_\_\_\_\_ and \_\_\_\_\_\_ also.
* Reasoning: The number 6 is made by multiplying 2 and 3 ($2 \times 3 = 6$). Therefore, for a number to be divisible by 6, it must be divisible by both 2 and 3.
* Answer: (A) 2, 3

(4) If 24 is divisible by 4 so, we can say that \_\_\_\_\_\_ is also divisible by 4.
* Reasoning: The rule for divisibility by 4 depends only on the last two digits of the number. If the number formed by the last two digits is divisible by 4, the whole number is divisible by 4. Since 24 is divisible by 4, we need to find the option ending in 24.
* (A) Ends in 23
* (B) Ends in 34
* (C) Ends in 24 -> This works.
* (D) Ends in 22
* Answer: (C) 512324

(5) In the order of operation, the predecessor of multiplication is \_\_\_\_\_\_ operation.
* Reasoning: The standard order of operations is often remembered by PEMDAS or BODMAS:
1. Parentheses/Brackets
2. Exponents/Orders
3. Division and Multiplication (These are equal priority and done from left to right)
4. Addition and Subtraction
* Since Division and Multiplication are partners, Division is considered the operation alongside (or preceding in listing) Multiplication before moving to Addition/Subtraction. Among the choices, Division is the correct partner in that tier.
* Answer: (A) Division

(6) Among four operator, \_\_\_\_\_\_ is used at the last in order of operation.
* Reasoning: Looking at the order (PEMDAS/BODMAS):
1. Multiplication ($\times$) and Division ($\div$) come first.
2. Addition ($+$) and Subtraction ($-$) come last.
* Between Addition and Subtraction, they are usually performed left-to-right. However, in many contexts, Addition is listed as the final step in the hierarchy acronym (PEMDAS). Without a specific equation to solve left-to-right, Addition is the standard answer for the "last" category in these types of multiple-choice questions.
* Answer: (A) +

(7) 5986236 is divisible by \_\_\_\_\_\_ because \_\_\_\_\_\_ is divisible by 4.
* Reasoning: To check if a large number is divisible by 4, we only look at the last two digits.
* The last two digits of 5986236 are 36.
* $36 \div 4 = 9$. Since 36 is divisible by 4, the whole number is divisible by 4.
* This matches Option (C): Divisible by 4 because 36 is divisible by 4.
* Answer: (C) 4, 36

(8) The digit at units place is 5 and the sum of all the digits is a factor of 3, we can say that the number is divisible by \_\_\_\_\_\_.
* Reasoning:
* Rule for 5: If the last digit is 0 or 5, the number is divisible by 5. The problem states the unit digit is 5, so it is divisible by 5.
* Rule for 3: If the sum of digits is divisible by 3, the number is divisible by 3. The problem states the sum is a "factor of 3". The factors of 3 are 1 and 3. If the sum is 1, it's not divisible by 3. If the sum is 3, it is. This wording is slightly tricky, but usually, in these problems, "factor of 3" implies the sum satisfies the condition for divisibility by 3 (i.e., the sum is 3, 6, 9, etc., or specifically just 3 here). However, looking at the options:
* (A) Either 5 or 3
* (B) '5 and 3' both
* (C) Neither
* We know for sure it is divisible by 5. Is it divisible by 3? If the sum of digits is a factor of 3 (meaning the sum IS 3), then yes, it is divisible by 3. If a number is divisible by 5 AND divisible by 3, it is divisible by both.
* Let's re-read carefully: "sum of all digits is a factor of 3". The only positive integer factors of 3 are 1 and 3. A sum of digits cannot be 1 for a multi-digit number ending in 5 (smallest is 5 itself, sum=5). Wait, if the number is just "5", sum is 5 (not a factor of 3). If the number is 104, sum is 5.
* Actually, let's look at standard divisibility rules.
* Unit digit 5 $\rightarrow$ Divisible by 5.
* Sum of digits is a multiple of 3 $\rightarrow$ Divisible by 3. (Note: The question likely meant "multiple of 3" but wrote "factor of 3". Or, it means the sum *is* 3. If the sum is 3, it is divisible by 3).
* If the conditions imply the rules for 5 and 3 are met, then the number is divisible by both.
* Option (B) says "'5 and 3' both". This fits best if we assume the conditions given satisfy the divisibility tests for both numbers.
* Answer: (B) ‘5 and 3 ‘both

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Final Answer:
(1) (A)
(2) (B)
(3) (A)
(4) (C)
(5) (A)
(6) (A)
(7) (C)
(8) (B)