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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.
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.
You are baking a chocolate souflee for a competition. You don’t want the souflee to fall, so you figure you will check on the dessert every so many minutes. You are being strict because your souflee could fall. Instead of opening the oven door, you plan on clicking the oven light on to see how the dessert is doing. The dessert takes 35 minutes to bake. What possibilities could there be to check on the cake?
There are really only two ways to check since you don’t want to be checking every single minute. So, you could check seven times every five minutes, or you could check five times every seven minutes. Those two ways seem to be the most logical.
A handyman places a 35-foot extension ladder against the wall of the house to paint the side of the house. He wants to paint the wall in sections. He can usually paint about 5 feet with one cup of his paint before he needs to go back down the ladder for more paint. How many times will need to go up and down the ladder for paint? In other words, how sections of painting will the handyman do?
You can use the factors of 35 to find the answer to this word problem. Since there are 5 feet in a 35 foot area of the wall, this means that you need to find 5’s factor partner to 35. This would be 7 since 5 times 7 equals 35.
A seamstress wants to gather the 35-centimeter waist of a skirt. She will need to gather it in sections of 3. Would she be able to do that so that there would be an equal number of centimeters gathered in each section? Explain why or why not.
In this case, the seamstress would not be able to do that. Instead, she would have 11 sections that would be the same amount of centimeters. However, one section would have only 2 centimeters of gathering instead of the 3 centimeters of gathering in the other groups.
There are only 4 factors for the number 35. They are 1, 5, 7, and 35.
There is a difference between divisibility rules and factoring. Factoring has to do with finding the numbers that equally divide into a number without having a remainder. Divisibility rules are the rules that are associated with how to divide certain numbers. In this case, the rules of divisibility will apply to factoring. You will use these divisibility rules to find the factors. Factoring can also have to do with another math skill.
If a number ends in 5 or 0, that number is evenly divisible by 5. So, the number 35, 45, 55, 60, 70 and 80 are just a few of the numbers that can be divided by 5 without having a remainder. This will influence someone finding the factors of that number because you would be trying to find those numbers that can be divided by a number without having a remainder. This would make it easy since you would automatically know that 5 would be a factor of this number like 35. You would then have to find its factor partner. For 35, 5’s factor partner is 7 because 5 times 7 is 35.
To find the factors of 35, first write down 1 and the number itself. Then see if it is an even number. Since 35 isn’t, let’s see if 3 will evenly divide into 35 without having a remainder. There would be a remainder, so 3 is not a factor of 35. Then, see if 4 can divide into 35 without a remainder. It doesn’t either. The next step is to try each number after that. In this case, there are only two more numbers that would be factors for 35. Those two numbers are 5 and 7.
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