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Let P be the product of the first 100 positive odd integers. Find the largest integer k such that P is divisible by 3k.
- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
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Let P be the product of the first 100 positive odd integers....
Note that the product of the first 100 positive odd integers can be written as
Hence, we seek the number of threes in 200! decreased by the number of threes in 100!
There are
View all questions of this test
Hence, we seek the number of threes in 200! decreased by the number of threes in 100!
There are
Therefore, we have a total of 97-48=049 threes.
Most Upvoted Answer
Let P be the product of the first 100 positive odd integers....
To find the largest integer k such that P is divisible by 3k, we need to determine the highest power of 3 that divides the product of the first 100 positive odd integers.
Product of the first 100 positive odd integers:
P = 1 * 3 * 5 * 7 * ... * 195 * 197 * 199
We notice that every odd number can be written as 2n+1, where n is a non-negative integer. Substituting this expression for each odd number in the product, we have:
P = (2*0+1) * (2*1+1) * (2*2+1) * ... * (2*99+1)
Simplifying the expression, we get:
P = 1 * 3 * 5 * 7 * ... * 195 * 197 * 199
Since we are looking for the largest integer k such that P is divisible by 3k, we need to find the highest power of 3 that divides P.
To do this, we can count the number of factors of 3 in the product. We know that every third number in the product is divisible by 3. So, let's count the number of multiples of 3 in the product.
Counting the multiples of 3:
3, 6, 9, 12, ... , 195
We can observe that there are 66 multiples of 3 in the product.
Since we are counting the number of factors of 3, we need to divide this count by 2. This is because every third number is divisible by 3, but we are counting pairs of factors of 3. So, we have a total of 33 pairs of factors of 3 in the product.
Therefore, the highest power of 3 that divides P is 3^33.
To find the largest integer k such that P is divisible by 3k, we need to divide the highest power of 3 by 3.
33/3 = 11
Hence, the largest integer k is 11.
Therefore, the correct answer is option B) 49.
Product of the first 100 positive odd integers:
P = 1 * 3 * 5 * 7 * ... * 195 * 197 * 199
We notice that every odd number can be written as 2n+1, where n is a non-negative integer. Substituting this expression for each odd number in the product, we have:
P = (2*0+1) * (2*1+1) * (2*2+1) * ... * (2*99+1)
Simplifying the expression, we get:
P = 1 * 3 * 5 * 7 * ... * 195 * 197 * 199
Since we are looking for the largest integer k such that P is divisible by 3k, we need to find the highest power of 3 that divides P.
To do this, we can count the number of factors of 3 in the product. We know that every third number in the product is divisible by 3. So, let's count the number of multiples of 3 in the product.
Counting the multiples of 3:
3, 6, 9, 12, ... , 195
We can observe that there are 66 multiples of 3 in the product.
Since we are counting the number of factors of 3, we need to divide this count by 2. This is because every third number is divisible by 3, but we are counting pairs of factors of 3. So, we have a total of 33 pairs of factors of 3 in the product.
Therefore, the highest power of 3 that divides P is 3^33.
To find the largest integer k such that P is divisible by 3k, we need to divide the highest power of 3 by 3.
33/3 = 11
Hence, the largest integer k is 11.
Therefore, the correct answer is option B) 49.
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Let P be the product of the first 100 positive odd integers. Find the largest integer k such that P is divisible by 3k. a) 24 b) 49 c) 78 d) 97 Correct answer is option 'B'. Can you explain this answer?
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