The answer would be True.
Sorry for this but to be honest we do t get taught this in the school I go to sorry
Answer:
Replace /* Your solution goes here */ with:
cin>>matchValue;
numMatches = 0;
for (i = 0; i < userValues.size(); ++i) {
if(matchValue == userValues.at(i))
{
numMatches++;
}
}
Explanation:
This line gets input for matchValue
<em>cin>>matchValue;
</em>
This line initializes numMatches to 0
<em>numMatches = 0;
</em>
The following iteration checks for the number of matches (numMatches) of the matchValue
<em>for (i = 0; i < userValues.size(); ++i) {
</em>
<em>if(matchValue == userValues.at(i))
</em>
<em>{
</em>
<em> numMatches++;
</em>
<em>}
</em>
<em>}
</em>
<em>See Attachment for full source code</em>
Answer:
- common = []
- num1 = 8
- num2 = 24
- for i in range(1, num1 + 1):
- if(num1 % i == 0 and num2 % i == 0):
- common.append(i)
- print(common)
Explanation:
The solution is written in Python 3.
Firstly create a common list to hold a list of the common factor between 8 and 24 (Line 1).
Create two variables num1, and num2 and set 8 and 24 as their values, respectively (Line 3 - 4).
Create a for loop to traverse through the number from 1 to 8 and use modulus operator to check if num1 and num2 are divisible by current i value. If so the remainder of both num1%i and num2%i will be zero and the if block will run to append the current i value to common list (Line 6-8).
After the loop, print the common list and we shall get [1, 2, 4, 8]
