All questions of Playing with Numbers for Class 6 Exam

The maximum even multiple of 25 between 500 and 1000 is 975.

LCM = hcf ÷ a ×b
2175 =145÷725×b is farmula se anwer aa ye ga

Solution:
The largest number that will divide 76, 113 and 186 leaving remainder 4, 5, 6 respectively is also known as the highest common factor (HCF) of (76-4), (113-5) and (186-6).

Step-by-step solution:
• Subtract the remainder from each number to get the actual number.
• So, the actual numbers are 72, 108, and 180.
• Now, find the common factors of these numbers.
• The common factors of 72, 108, and 180 are 1, 2, 3, 4, 6, 9, 12, 18, 36.
• Out of these common factors, select the highest one which leaves the same remainder when divided by 4, 5, and 6 respectively.
• The highest common factor (HCF) of (76-4), (113-5) and (186-6) is 36.
• Hence, the correct option is (c) 36.

Therefore, the largest number that will divide 76, 113 and 186 leaving remainder 4, 5, 6 respectively is 36.

To determine the two numbers nearest to 1870 that are each divisible by 22, we can use the concept of divisibility and apply it to the given options.

Divisibility Rule of 22: A number is divisible by 22 if it is divisible by both 2 and 11.

Let's analyze the given options one by one to see if they satisfy the divisibility rule of 22.

a) 1848, 1892:
- 1848: This number is divisible by 2 because the last digit is even (8). However, to determine if it is divisible by 11, we can use the divisibility rule of 11, which states that the difference between the sum of the digits in the odd place and the sum of the digits in the even place should be either 0 or a multiple of 11.
- In this case, 1 - 8 + 4 - 8 = -11, which is a multiple of 11. Therefore, 1848 is divisible by 11 and consequently divisible by 22.
- 1892: This number is divisible by 2 because the last digit is even (2). However, to determine if it is divisible by 11, we can use the divisibility rule of 11.
- In this case, 1 - 8 + 9 - 2 = 0, which is a multiple of 11. Therefore, 1892 is divisible by 11 and consequently divisible by 22.

Since both 1848 and 1892 satisfy the divisibility rule of 22, option A is correct.

b) 1893, 1914:
- 1893: This number is not divisible by 2 because the last digit is odd (3). Therefore, it is not divisible by 22.
- 1914: This number is divisible by 2 because the last digit is even (4). However, it is not divisible by 11 since 1 - 9 + 1 - 4 = -11, which is not a multiple of 11. Therefore, it is not divisible by 22.

Option B is incorrect.

c) 1826, 1914:
- 1826: This number is divisible by 2 because the last digit is even (6). However, it is not divisible by 11 since 1 - 8 + 2 - 6 = -11, which is not a multiple of 11. Therefore, it is not divisible by 22.
- 1914: This number is divisible by 2 because the last digit is even (4). However, it is not divisible by 11 since 1 - 9 + 1 - 4 = -11, which is not a multiple of 11. Therefore, it is not divisible by 22.

Option C is incorrect.

d) None of these:
Since we have found that option A is correct, option D is incorrect.

Therefore, the two numbers nearest to 1870 that are each divisible by 22 are 1848 and 1892, as stated in option A.

To find the smallest possible number that is divisible by 28, 36, and 45, we need to find the least common multiple (LCM) of these three numbers.

Step 1: Find the prime factors of each number.
- Prime factors of 28: 2, 2, 7 (2 x 2 x 7)
- Prime factors of 36: 2, 2, 3, 3 (2 x 2 x 3 x 3)
- Prime factors of 45: 3, 3, 5 (3 x 3 x 5)

Step 2: Identify the largest exponent for each prime factor.
- Largest exponent of 2: 2 (from 2 x 2)
- Largest exponent of 3: 3 (from 3 x 3)
- Largest exponent of 7: 1 (from 7)
- Largest exponent of 5: 1 (from 5)

Step 3: Multiply the prime factors with their largest exponents.
2^2 x 3^3 x 7^1 x 5^1 = 4 x 27 x 7 x 5 = 3780

Step 4: Add 19 to the result.
3780 + 19 = 3799

Hence, the smallest possible number that on adding 19 becomes exactly divisible by 28, 36, and 45 is 3799.

Therefore, option B (1241) is incorrect.

Explanation:

Divisibility by 8:
- To determine if a number is divisible by 8, we need to check if the last three digits of the number are divisible by 8.
- In the number 1509344, the last three digits are 344, which is divisible by 8 because 344 ÷ 8 = 43.
- Therefore, 1509344 is divisible by 8.

Divisibility by 11:
- To determine if a number is divisible by 11, we can use the rule that states that the difference between the sum of the digits at odd places and the sum of the digits at even places should be a multiple of 11.
- For the number 72493, the sum of the digits at odd places (7 + 4 + 3 = 14) is 14, and the sum of the digits at even places (2 + 9 = 11) is 11.
- The difference between 14 and 11 is 3, which is not a multiple of 11.
- Therefore, 72493 is not divisible by 11.

Conclusion:
- Based on the explanations above, the statement "1509344 is divisible by 8" is true.