The double factorial, denoted by $n!!$, returns the product of all of the odd integers that are less than or equal to $n$. For example, $7!! = 7 * 5 *3 * 1$. What is the units digit of $1!! + 3!! + 5!! + 7!! + ... + 49!!$?
To find the units digit of the given expression, we first need to compute the double factorial of each odd number from 1 to 49, and then sum them up. Let's break down the problem into smaller steps:
Step 1: Compute the double factorial of each odd number from 1 to 49.
The double factorial of an odd number $n$ is calculated by multiplying all the odd numbers less than or equal to $n$. For example, $7!! = 7 \times 5 \times 3 \times 1$. We can compute the double factorial using a loop or by multiplying the odd numbers manually.
Step 2: Find the units digit of each double factorial.
To find the units digit of a number, we only need to consider the units digit of each digit multiplication. For instance, when computing $7!! = 7 \times 5 \times 3 \times 1$, we only need to consider the units digits of each product: $7 \times 5 \times 3 \times 1 = 105$. Therefore, the units digit of $7!!$ is 5.
Step 3: Sum up all the units digits.
Once we have calculated the units digit for each double factorial, we can simply sum them up to obtain the units digit of the expression $1!! + 3!! + 5!! + 7!! + ... + 49!!$.
Let's calculate the units digit for each double factorial:
$1!! = 1$
$3!! = 5$
$5!! = 5$
$7!! = 5$
...
$49!! = 5$
Notice that for each odd number, the units digit of its double factorial is always 5. Since we have 25 odd numbers from 1 to 49, and the units digit of each double factorial is 5, the sum of all these units digits will be $25 \times 5 = 125$. The units digit of this sum is 5.
Therefore, the units digit of $1!! + 3!! + 5!! + 7!! + ... + 49!!$ is 5.
well, you know that n!! ends in 5 for all n >= 5
There are (49+1)/2 = 25 terms in your sum. The last 23 of them end in 5.
So, what do you think?