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Determine the number nearest to 100000 but greater than 100000 which is exactly divisible by each of 8, 15 and 21?
- a)100200
- b)100800
- c)240800
- d)260040
Correct answer is option 'B'. Can you explain this answer?
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Determine the number nearest to 100000 but greater than 100000 which i...
Reminder= 40 84040=800 100000+800=100800
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Determine the number nearest to 100000 but greater than 100000 which i...
To find the number nearest to 100,000 but greater than 100,000 which is exactly divisible by each of 8, 15, and 21, we need to find the least common multiple (LCM) of these three numbers.
The LCM is the smallest number that is divisible by all the given numbers.
Here's how we can find the LCM of 8, 15, and 21:
1. Prime factorize each number:
- 8 = 2^3
- 15 = 3 * 5
- 21 = 3 * 7
2. Identify the highest power of each prime factor:
- The highest power of 2 is 3.
- The highest power of 3 is 1.
- The highest power of 5 is 1.
- The highest power of 7 is 1.
3. Multiply the prime factors with their highest powers:
- LCM = 2^3 * 3^1 * 5^1 * 7^1
= 8 * 3 * 5 * 7
= 840
So, the LCM of 8, 15, and 21 is 840.
To find the number nearest to 100,000 but greater than 100,000 which is exactly divisible by 840, we need to find the smallest multiple of 840 that is greater than 100,000.
Dividing 100,000 by 840 gives us a quotient of 119 and a remainder of 40.
To find the next multiple of 840 after 100,000, we add the remainder (40) to the product of the quotient (119) and the divisor (840):
Next multiple = 119 * 840 + 40
= 100,360 + 40
= 100,400
So, the number nearest to 100,000 but greater than 100,000 which is exactly divisible by each of 8, 15, and 21 is 100,400, which is option B.
The LCM is the smallest number that is divisible by all the given numbers.
Here's how we can find the LCM of 8, 15, and 21:
1. Prime factorize each number:
- 8 = 2^3
- 15 = 3 * 5
- 21 = 3 * 7
2. Identify the highest power of each prime factor:
- The highest power of 2 is 3.
- The highest power of 3 is 1.
- The highest power of 5 is 1.
- The highest power of 7 is 1.
3. Multiply the prime factors with their highest powers:
- LCM = 2^3 * 3^1 * 5^1 * 7^1
= 8 * 3 * 5 * 7
= 840
So, the LCM of 8, 15, and 21 is 840.
To find the number nearest to 100,000 but greater than 100,000 which is exactly divisible by 840, we need to find the smallest multiple of 840 that is greater than 100,000.
Dividing 100,000 by 840 gives us a quotient of 119 and a remainder of 40.
To find the next multiple of 840 after 100,000, we add the remainder (40) to the product of the quotient (119) and the divisor (840):
Next multiple = 119 * 840 + 40
= 100,360 + 40
= 100,400
So, the number nearest to 100,000 but greater than 100,000 which is exactly divisible by each of 8, 15, and 21 is 100,400, which is option B.
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Determine the number nearest to 100000 but greater than 100000 which is exactly divisible by each of 8, 15 and 21?a)100200 b)100800 c)240800d)260040Correct answer is option 'B'. Can you explain this answer?
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