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Find the smallest number exactly divisible by 15, 25, 30, and 45.
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Solution: Step 1: Calculate the LCM of 15, 25, 30, and 45. Step 2: Prime factorizations: 15 = 3 * 5, 25 = 5^2, 30 = 2 * 3 * 5, 45 = 3^2 * 5. Step 3: LCM = 2 * 3^2 * 5^2 = 2 * 9 * 25 = 450. Step 4: Therefore, the least number exactly divisible by 15, 25, 30, and 45 is 450.
2
Find the smallest amount that when divided by Rs. 16, Rs. 18, Rs. 20, or Rs. 25 leaves a remainder of Rs. 4 each time.
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Solution: Step 1: Calculate the LCM of 16, 18, 20, and 25. LCM = 3600. Step 2: Since the amount leaves a remainder of Rs. 4 when distributed into these groups, the required amount = LCM + 4 = 3600 + 4 = Rs. 3604. Step 3: Verify that 3604 is indeed divisible by each of the given amounts with a remainder of 4. Hence, the smallest amount is Rs. 3604.
3
Find the smallest 4-digit number that, when divided by 7, 9, and 11, leaves remainders of 3, 5, and 7 respectively.
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Solution: Step 1: Observe the differences between divisors and remainders: 7-3 = 4, 9-5 = 4, 11-7 = 4. Step 2: The number is of the form LCM(7, 9, 11)k - 4. Step 3: Calculate LCM(7, 9, 11) = 7 * 9 * 11 = 693. Step 4: Find the smallest 4-digit multiple of 693: 693 * 2 = 1386. Step 5: The required number = 1386 - 4 = 1382.
4
Three individuals have step lengths of 252 cm, 280 cm, and 308 cm. What is the minimum distance each should cover so that all can complete the distance in whole steps?
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Solution: Step 1: Find the LCM of 252, 280, and 308. Step 2: Prime factorization: - 252 = 2^2 * 3^2 * 7 - 280 = 2^3 * 5 * 7 - 308 = 2^2 * 7 * 11 Step 3: LCM = 2^3 * 3^2 * 5 * 7 * 11 = 27,720 cm Step 4: The minimum distance is 27,720 cm
5
Four bells start tolling together at intervals of 8, 10, 12, and 16 seconds. After how many seconds will they toll together again?
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Solution: Step 1: Find the prime factorization of each interval: 8 = 2^3 10 = 2 * 5 12 = 2^2 * 3 16 = 2^4 Step 2: Calculate the LCM using the highest powers of prime factors: LCM = 2^4 * 3 * 5 = 16 * 3 * 5 = 240 Step 3: Therefore, the bells will toll together again after 240 seconds.
6
What is the smallest number that is exactly divisible by 7, 18, 56, and 36, leaving a remainder of zero?
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Solution: Step 1: Find the prime factorization of each number. 7 = 7 18 = 2 × 3^2 56 = 2^3 × 7 36 = 2^2 × 3^2 Step 2: Identify the maximum frequency of each prime factor. - 2 occurs maximum 3 times - 3 occurs maximum 2 times - 7 occurs maximum 1 time Step 3: Calculate the LCM = 2^3 × 3^2 × 7 = 8 × 9 × 7 = 504. Hence, the smallest number that leaves a remainder of zero when divided by 7, 18, 56, and 36 is 504.
7
What is the smallest 4-digit number divisible by 12, 13, and 15?
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Solution: Step 1: Find the LCM of 12, 13, and 15. Step 2: The prime factorizations are: 12 = 2^2 * 3, 13 = 13, 15 = 3 * 5. Step 3: LCM = 2^2 * 3 * 5 * 13 = 780. Step 4: Since 780 is a 3-digit number, find the smallest 4-digit multiple of 780. Step 5: The smallest 4-digit number is 1000, so find the next multiple of 780 after 1000. Step 6: 780 * 2 = 1560, which is the smallest 4-digit number divisible by 12, 13, and 15. Step 7: Therefore, the answer is 1560.
8
The difference between two numbers is 14. Their Least Common Multiple (LCM) and Highest Common Factor (HCF) are 441 and 7, respectively. What are the two numbers?
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Solution: Step 1: Let the numbers be 7x and 7y (since HCF is 7) Step 2: Difference = 14 => 7x - 7y = 14 => x - y = 2 Step 3: Product of numbers = LCM * HCF => 7x * 7y = 441 * 7 Step 4: xy = 63 (from step 3) Step 5: Solve x - y = 2 and xy = 63 => x = 9, y = 7 Step 6: Numbers are 7x and 7y => 63 and 49
9
Find the smallest number that when doubled is exactly divisible by 12, 18, 21, and 30.
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Solution: Step 1: Find the LCM of 12, 18, 21, and 30. Step 2: Prime factorize each number: 12 = 2^2 * 3, 18 = 2 * 3^2, 21 = 3 * 7, 30 = 2 * 3 * 5. Step 3: Calculate LCM = 2^2 * 3^2 * 5 * 7 = 1260. Step 4: The number that when doubled gives this LCM is 1260 / 2 = 630. Step 5: Confirm 1260 is the correct number as it satisfies all divisibility conditions.
10
A ticket collector has a certain number of tickets. When arranged in pairs, one ticket remains. The same happens when arranged in groups of 3, 4, 5, or 6. What is the remainder when these tickets are arranged in groups of 8?
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Solution: Step 1: Understand that the number of tickets = LCM(2, 3, 4, 5, 6) + 1 Step 2: Calculate LCM(2, 3, 4, 5, 6) = 60 Step 3: So, the number of tickets = 60 + 1 = 61 Step 4: Find the remainder when 61 is divided by 8 Step 5: 61 / 8 gives a remainder of 5
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