This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

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Here are the solutions to the problems:
18. A number Z is such that when it is divided by 30, 36 and 47 the remainder is always 7. Find the smallest possible value of Z.
Step 1: Understand the problem. If a number Z leaves a remainder of 7 when divided by 30, 36, and 47, it means that is perfectly divisible by 30, 36, and 47. Therefore, is a common multiple of 30, 36, and 47. To find the smallest possible value of Z, we need to find the Least Common Multiple (LCM) of 30, 36, and 47, and then add 7 to it.
Step 2: Find the prime factorization of each number.
Step 3: Calculate the LCM of 30, 36, and 47. The LCM is found by taking the highest power of all prime factors present in any of the numbers.
Step 4: Find the value of Z. Since , we have: The smallest possible value of Z is .
19. Which of the following number is divisible by both 3 and 4: 36, 744, 101, 302, 805
To be divisible by both 3 and 4, a number must satisfy the divisibility rules for both 3 and 4. • Divisibility Rule for 3: The sum of the digits of the number must be divisible by 3. • Divisibility Rule for 4: The number formed by the last two digits must be divisible by 4.
Let's check each number:
36:
744:
101:
302:
805:
The numbers divisible by both 3 and 4 are .
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18. A number Z is such that when it is divided by 30, 36 and 47 the remainder is always 7.
This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.