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What is the highest integer power of 6 that can divide 73!–72! ?
- a)14
- b)16
- c)34
- d)36
- e)68
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
What is the highest integer power of 6 that can divide73!–72! ?a...
We want to find the highest power of 6 that can divide 73!. Since 6 = 2 * 3, we need to count the number of factors of 2 and 3 in 73!.
Let's count the number of factors of 2 first. We know that every even number contributes at least one factor of 2. There are 36 even numbers between 1 and 73, so there are at least 36 factors of 2 in 73!.
But some numbers contribute more than one factor of 2. Every multiple of 4 contributes an additional factor of 2, and there are 18 multiples of 4 between 1 and 73. Similarly, every multiple of 8 contributes an additional factor of 2, and there are 9 multiples of 8 between 1 and 73. Continuing this pattern, we can count the total number of factors of 2 as:
36 + 18 + 9 + 4 + 2 + 1 = 70
Now let's count the number of factors of 3. There are 24 multiples of 3 between 1 and 73, but some of these multiples contribute more than one factor of 3. Every multiple of 9 contributes an additional factor of 3, and there are 8 multiples of 9 between 1 and 73. Continuing this pattern, we can count the total number of factors of 3 as:
24 + 8 + 2 = 34
Since we need a power of 6, which is 2 * 3, the highest power of 6 that can divide 73! is min(70, 34) = 34.
Let's count the number of factors of 2 first. We know that every even number contributes at least one factor of 2. There are 36 even numbers between 1 and 73, so there are at least 36 factors of 2 in 73!.
But some numbers contribute more than one factor of 2. Every multiple of 4 contributes an additional factor of 2, and there are 18 multiples of 4 between 1 and 73. Similarly, every multiple of 8 contributes an additional factor of 2, and there are 9 multiples of 8 between 1 and 73. Continuing this pattern, we can count the total number of factors of 2 as:
36 + 18 + 9 + 4 + 2 + 1 = 70
Now let's count the number of factors of 3. There are 24 multiples of 3 between 1 and 73, but some of these multiples contribute more than one factor of 3. Every multiple of 9 contributes an additional factor of 3, and there are 8 multiples of 9 between 1 and 73. Continuing this pattern, we can count the total number of factors of 3 as:
24 + 8 + 2 = 34
Since we need a power of 6, which is 2 * 3, the highest power of 6 that can divide 73! is min(70, 34) = 34.
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Community Answer
What is the highest integer power of 6 that can divide73!–72! ?a...
We can simplify the expression (73! - 72!) as follows:
73! - 72! = 72! * (73 - 1) = 72! * 72
73! - 72! = 72! * (73 - 1) = 72! * 72
Since we are looking for the highest power of 6 that can divide the expression, we need to determine how many factors of 6 are present in 72!.
Let's find the highest power of 6 in 72!. We can do this by finding the highest power of 6 in the prime factorization of 72.
Prime factorization of 72: 23 * 32
To find the highest power of 6, we need to consider the highest power of 2 and the highest power of 3.
Highest power of 2: 3 (23)
Highest power of 3: 2 (32)
Highest power of 3: 2 (32)
Therefore, the highest power of 6 that can divide 72! is 62, which is 36.
Thus, the highest integer power of 6 that can divide (73! - 72!) is 36.
The correct answer is D: 36.
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What is the highest integer power of 6 that can divide73!–72! ?a)14b)16c)34d)36e)68Correct answer is option 'D'. Can you explain this answer?
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