Introduction
In a car, a car racer knows that with every rev of the engine, the odometer needle climbs steadily 7 times. If he revs it seven times without the odometer needle falling at all, if that is all possible, how many times will he rev his engine to reach 35 miles per hour?
Consider the factors of 35 to figure out how to determine the answer to this problem. There are multiple ways to solve this problem. However, using the factors will always help.
Factors of 35
A factorization tree can show you the factors of 35.
With the number 35, you may already know that 1 and 35 are already factors of 35. You can rule out 2 because it is not an even number. Now, you need to try 3, but 35 is not a multiple of 3. Now, check 4. Nope, four isn’t a factor because when you divide 4 into 35, there is a remainder. Five is a factor because five times 7 is 35. Let’s try number 6, but this is not a factor. There would be a remainder. The only thing left to do is to try more numbers. Eight, nine, ten, eleven, twelve, and thirteen are not factors. If you tried, you would see that the rest of the numbers up to 35 are also not factors.
With the car racer, we know that there are only 4 factors for 35. These are 1, 5, 7, and 35. We know that the engine was revved 7 times and it reached 35. So, 7’s factor partner is 5. This means that seven times five is 35. The odometer needle went up 5 miles per hour on the odometer with each rev of the engine.
Challenging Questions
Question 1List the factors of 35
Which positive whole numbers divide 35 without leaving a remainder?
The factors of 35 are 1, 5, 7, 35.
Each listed number divides the original number evenly.
Question 2Find the factor pairs of 35
Which multiplication pairs have a product of 35?
35 has these positive factor pairs: 1 × 35; 5 × 7.
Multiplying the two numbers in any pair gives the original number.
Question 3Prime-factorize 35
What is the prime factorization of 35?
35 = 5 × 7.
Every number in the product is prime.
