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How many trailing zeros will be there after the rightmost non-zero digit in the value of 25!?
- a)25
- b)8
- c)6
- d)5
- e)2
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
How many trailing zeros will be there after the rightmost non-zero dig...
5! means factorial 25 whose value = 25 * 24 * 23 * 22 *....* 1
When a number that is a multiple of 5 is multiplied with an even number, it results in a trailing zero.
(Product of 5 and 2 is 10 and any number when multiplied with 10 or a power of 10 will have one or as many zeroes as the power of 10 with which it has been multiplied)
(Product of 5 and 2 is 10 and any number when multiplied with 10 or a power of 10 will have one or as many zeroes as the power of 10 with which it has been multiplied)
In 25!, the following numbers have 5 as their factor: 5, 10, 15, 20, and 25.
25 is the square of 5 and hence it has two 5s in it.
In toto, it is equivalent of having six 5s.
25 is the square of 5 and hence it has two 5s in it.
In toto, it is equivalent of having six 5s.
There are at least 6 even numbers in 25!
Hence, the number 25! will have 6 trailing zeroes in it.
Hence, the number 25! will have 6 trailing zeroes in it.
Choice C is the correct answer.
Most Upvoted Answer
How many trailing zeros will be there after the rightmost non-zero dig...
Solution:
When we multiply any number by 10, we add one zero to the end of that number. For example, 10, 20, 30, and so on.
Similarly, when we multiply any number by 100, we add two zeros to the end of that number. For example, 100, 200, 300, and so on.
Therefore, the number of trailing zeros in a factorial is equal to the number of times the factorial can be divided by 10.
To find the number of trailing zeros in 25!, we need to find the number of times 25! can be divided by 10.
Step 1: Prime factorize 10
10 = 2 × 5
Step 2: Count the number of factors of 5 in 25!
We can count the number of factors of 5 in 25! by dividing 25 by 5, then dividing the result by 5 again, and so on until we get a quotient less than 5.
25 ÷ 5 = 5
5 ÷ 5 = 1
Therefore, there are 5 + 1 = 6 factors of 5 in 25!
Step 3: Count the number of factors of 2 in 25!
We can count the number of factors of 2 in 25! by dividing 25 by 2, then dividing the result by 2 again, and so on until we get a quotient less than 2.
25 ÷ 2 = 12
12 ÷ 2 = 6
6 ÷ 2 = 3
3 ÷ 2 = 1
Therefore, there are 12 + 6 + 3 + 1 = 22 factors of 2 in 25!
Step 4: Determine the number of trailing zeros
Since there are 6 factors of 5 and 22 factors of 2 in 25!, we can only create 6 pairs of 2 and 5, which will result in 6 trailing zeros.
Therefore, the answer is option C, 6.
When we multiply any number by 10, we add one zero to the end of that number. For example, 10, 20, 30, and so on.
Similarly, when we multiply any number by 100, we add two zeros to the end of that number. For example, 100, 200, 300, and so on.
Therefore, the number of trailing zeros in a factorial is equal to the number of times the factorial can be divided by 10.
To find the number of trailing zeros in 25!, we need to find the number of times 25! can be divided by 10.
Step 1: Prime factorize 10
10 = 2 × 5
Step 2: Count the number of factors of 5 in 25!
We can count the number of factors of 5 in 25! by dividing 25 by 5, then dividing the result by 5 again, and so on until we get a quotient less than 5.
25 ÷ 5 = 5
5 ÷ 5 = 1
Therefore, there are 5 + 1 = 6 factors of 5 in 25!
Step 3: Count the number of factors of 2 in 25!
We can count the number of factors of 2 in 25! by dividing 25 by 2, then dividing the result by 2 again, and so on until we get a quotient less than 2.
25 ÷ 2 = 12
12 ÷ 2 = 6
6 ÷ 2 = 3
3 ÷ 2 = 1
Therefore, there are 12 + 6 + 3 + 1 = 22 factors of 2 in 25!
Step 4: Determine the number of trailing zeros
Since there are 6 factors of 5 and 22 factors of 2 in 25!, we can only create 6 pairs of 2 and 5, which will result in 6 trailing zeros.
Therefore, the answer is option C, 6.
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