For how many positive integers n does 1/n yield a terminating decimal with a non-zero hundredths digit?
Note that the numbers whose decimal representations begin $0.00\ldots$ are the positive real numbers less than $1/100$. Therefore, the hundredths digit of $1/n$ is zero for all $n > 100$. Also, recall that $1/n$ is a terminating decimal if and only if $n$ is divisible by no primes other than 2 and 5. The fifteen integers up to 100 whose prime factorizations contain only twos and fives are 1, 2, 4, 5, 8, 10, 16, 20, 25, 32, 40, 50, 64, 80, and 100. One way to generate this list systematically is to consider integers of the form $2^m5^n$, starting with $n=0$ and $m=0,1,2,3,4,5,6$, then $n=1$ and $m=0,1,2,3,4$, etc. Not all of these 15 integers have nonzero hundredths digits, however. For $n\leq 10$, it is possible that the tenths digit is nonzero while the hundredths digit is zero. Checking the values of $n$ up to 10, we find that the hundredths digits of 1, 1/2, 1/5, and 1/10 are zero. Therefore, there are $15 - 4 = \boxed{11}$ fractions that yield terminating decimals with a nonzero digit two places to the right of the decimal point.
@above you just copied the alcumus answer
Why did the decimal go to therapy?
Because it had too many issues!
In this case, we need to find the value of n for which 1/n yields a terminating decimal with a non-zero hundredths digit.
To do this, we need to look at the prime factorization of n. A terminating decimal occurs when the denominator consists only of 2s and/or 5s, and any additional factors will cause the decimal to be non-terminating.
Let's break it down:
- If n is a multiple of 5, then it will have a 5 in its prime factorization. However, having a 5 as a factor will not give us a non-zero hundredths digit, since any fractional number divided by 5 will only have a non-zero tenths digit.
- If n is a multiple of 2 but not 5, then it will have a 2 in its prime factorization. Dividing 1 by a power of 2 will give us a terminating decimal, but it will have a zero in the hundredths place.
- If n is a multiple of both 2 and 5, then it will have both a 2 and a 5 in its prime factorization. Dividing 1 by a number with both a power of 2 and a power of 5 will give us a terminating decimal with a non-zero hundredths digit.
Therefore, the positive integers n for which 1/n yields a terminating decimal with a non-zero hundredths digit are the positive integers that have both a power of 2 and a power of 5 in their prime factorization.
So, how many positive integers n satisfy this condition?
Drumroll, please...
The answer is infinite! There are an infinite number of positive integers that have both a power of 2 and a power of 5 in their prime factorization.
I hope that explanation wasn't too "divisive"!
To find the number of positive integers n for which 1/n yields a terminating decimal with a non-zero hundredths digit, we need to understand the conditions under which a fraction will terminate.
A fraction in its simplest form will terminate if its denominator has prime factors only of 2 and/or 5. This is because other prime factors in the denominator would result in a recurring decimal.
Let's analyze the prime factors of n:
- If n has only prime factors of 2 and/or 5, then the decimal representation of 1/n will terminate.
- If n has any prime factors other than 2 and 5, then the decimal representation of 1/n will be recurring.
Now, let's consider the condition of having a non-zero hundredths digit. A terminating decimal with a non-zero hundredths digit means that the denominator n should have at least one prime factor other than 2 and 5.
From this analysis, we can conclude that the number of positive integers n where 1/n yields a terminating decimal with a non-zero hundredths digit is equal to the number of positive integers that have prime factors other than 2 and 5.
To determine this count, we can use the concept of complementary counting. We can find the total count of positive integers that only have prime factors of 2 and 5 and subtract it from the total count of positive integers.
Since there are infinitely many positive integers, the total count of positive integers is infinite. However, for practical purposes, we can define a range or limit based on the context of the problem.
For example, if we want to find the count of positive integers n up to a certain value such as 100, we can iterate through the numbers from 1 to 100 and check if each number has prime factors other than 2 and 5.
Alternatively, if we want to find the count of positive integers up to a certain power of 10, such as up to 10^6, we can use number theory techniques to calculate it more efficiently, such as counting the numbers in a range that have prime factors other than 2 and 5.
It is important to note that even though the total count of positive integers is infinite, the count of positive integers n with 1/n yielding a terminating decimal with a non-zero hundredths digit is finite within the given range or limit.
Overall, to determine the exact number of positive integers n, you would need to define the range or limit for the analysis and employ number theory techniques to calculate it accurately.
just check a few. All the denominators must have only factors which are powers of 2 and 5. 1/n will have a zero hundredths digit of n > 100
So, how many numbers are there less than 100, which contain only powers of 2 and 5?
Watch out for 0.10, 0.20, 0.40, 0.50 which also have a zero hundredths digit.