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"Are there any tricks that can help you easily factor three digit numbers (without using a calculator)?" ~Jay
Hi Jay, I assume you're talking about tricks besides the normal divisibility rules (for example, if the digits add to three, the number is a multiple of three, if it ends in 0 or 5 the number is divisble by 5, etc). If you're not familiar with those rules, you might want to take a look at this unit here: Divisibility Rules.
Beyond that, there are some tricks that sometimes help. Here's my favorite. Let's say you wanted to factor the number 483. Here's what I would do:
- Multiply the first and last digit: 4 x 3 = 12
- Find two numbers that multiply to 12 and add to the middle digit (8). The numbers are 6 and 2 (6 + 2 = 8 and 6 x 2 = 12).
- Now rewrite the number using those two numbers we just found: 483 = 460 + 23 (the tens place got split into two pieces using our numbers, and the entire number was rewritten as a sum of two numbers).
- Now factor the result: 460 + 23 = 23(20 + 1) = 23 x 21.
- Finish factoring: 23 x 7 x 3
Unfortunately, this doesn't always work. For example, it won't work for 648, because you can't find two numbers that add to 4 and multiply to 48. But maybe if we can find a way of regrouping this number, we might get around that. My first thought is to pull out one of the "hundreds" and put it into the tens place. So we're thinking of 648 as being rewritten 5(14)8. Now we do 5 x 8 = 40, and realize that our two numbers must be 4 and 10 (4 + 10 = 14 and 4 x 10 = 40). So we rewrite the number: (600 + 48 = 24(25 + 2) = 24 x 27. Then we just finish the prime factorization from there.
If the number is one of those special numbers (like 483) that can be factored without regrouping, it's a straightforward, foolproof process. But if the number has to be regrouped, it requires a bit of intuition to work it out. However, if you don't have a calculator, it might be worth doing!
Thabang from Lesotho writes, "how do we rationalize a denominator consisting of a cube root with another constant added to it or subtracted from it?"
Good morning, Thabang, and thank you for your question. This is actually something I don't remember ever seeing before, so I had to give it some thought before answering.
What you're looking for is, how do we rationalize the denominator, if the denominator is something like "The cube root of three, plus two" or "the cube root of three, minus two"?
In order to solve this, it's important to remember two factoring rules you may have learned in an Algebra class:
x3 + y3 = (x + y)(x2 - xy + y2)
x3 - y3 = (x - y)(x2 + xy + y2)
Let's say your denominator is the cube root of three, plus two. Then I'm going to do the following substitutions:
Let x = the cube root of three, let y = 2.
Now your denominator is x + y, and if you multiply the numerator and denominator of the fraction by (x2 - xy + y2), you will have turned the denominator into x3 + y3 = 3 + 8 = 11, which is rational.
That was using the first factoring rule shown above. If the denominator had a subtraction (the cube root of three, minus two), we'd just use the second factoring rule, and multiply by (x2 + xy + y2).
Thanks again for asking, Thabang.
Navya asks: "Why do we have names for numbers?"
There are two answers to this question. The first answer is: because it's impossible to talk about numbers verbally unless you have names for them. If we didn't have the names "one", "two", "three" and so forth, how would we ever say "I have five apples"?
That explanation is sufficient for why we have names for the numbers from zero to nine, but it's not sufficient for numbers like eleven, twelve, and so on. After all, if we didn't have the name "eleven", we could still say the number by saying "one one."
Thus, we would count like this (starting at ten): "one zero, one one, one two, one three..." and so forth.
And in some cases, that would be quicker. The name "eleven" has three syllables, while "one one" just has two. Even worse would be a number like "three thousand, four hundred, sixty three" which takes nine syllables instead of the four syllables required for "three four six three".
So, since the number names aren't always quicker to say than just reciting off the digits, why do we bother? The answer is that using number names allows us to get an immediate order-of-magnitude sense for how big the number is. Look at it this way - if I say "seven million, two hundred twenty three thousand, four hundred twelve," the moment I said "seven million" you had a very good sense for how big that number is. But if instead I had said "seven two two three four one two" you would not have any way of determining the number's magnitude until I was all done reciting the digits, and you knew how many digits there were. And if you lost track of how many digits there were, you still wouldn't have a good sense for how big the number is!
So number names are very helpful for order-of-magnitude sense of the size of a number.
Laina asks for an explanation of the puzzle shown below.
The text reads: "THE BATTLE OF ENGLISH AND MATHEMATICS.
1 rabbit saw 6 elephants while going to the river.
Every elephant saw 2 monkeys going towards the river.
Every monkey holds 1 parrot in their hands.
How many animals are going towards the river???
Lets see who went to shcool..."
First, I doubt I'll be able to solve this, since I never went to shcool...
However, my answer is, simply, there is no solution, because there is not enough information given.
This is like a poor man's version of the "St. Ives" riddle. In case you don't know that riddle, it goes like this: "As I was going to St. Ives, I met a man with seven wives. Every wife had seven sacks, every sack had seven cats, every cat had seven kitts. Kitts, cats, sacks, wives, how many were going to St. Ives?"
With the St. Ives riddle, the narrator meets a man with seven wives. The word "meets" strongly implies* that the man with seven wives was going in the opposite direction.
Notice, though, that in our question here, it does not say that the rabbit "met" the elephants, only that he saw them, which could certainly mean that he overtook them as they were also on their way to the river. We simply don't know what the elephants are doing, and therefore there is not enough information to definitively answer the question.
I could end there, because the answer is indeterminate. However, I think it's worth exploring the question further.
The next question that arises is, how many monkeys are there? The two most common answers to this are two and twelve.
The first answer comes from assuming that each elephant saw the same two monkeys, and the second answer comes from assuming that each elephant saw two monkeys that none of the others saw. The most compelling argument for "two" as the answer is that the problem uses the word "every" instead of "each." The word "every" is a group word, implying the group as a whole did something. Thus, the group, as a whole, saw two monkeys. It's an interesting argument, but not all that convincing; grammarly states that each and every can be used interchangeable in some circumstances, and Cambridge Dictionary gives us such an example:
Each person has a bicycle.
Every person has a bicycle.
Both of these statements mean that there is a bicycle for each person. If we aren't going to assume that every person has the SAME bicycle (meaning, of course, that there's only one bicycle), then we can't assume that every elephant saw the same two monkeys.
Nor can we assume that every elephant saw two distinct monkeys that none of the others saw.
Therefore, there must be somewhere between 2 and 12 monkeys, inclusive, but the exact quantity of monkeys is unknown. Giving us, once again, an indeterminate answer.
There was another argument I saw for two monkeys - some people were saying, "since the problem doesn't state explicitly, implicit rules apply, making it two." Implicit rules? What is that? Is it just a fancy way of saying, "Since they didn't specify, I get to pick whichever makes the most sense to me, and call it the 'default' meaning."?
Regarding the parrots, it's interesting to me that the people who present the "each vs. every" argument are inconsistent; If "every elephant saw two monkeys" means that the group as a whole saw two monkeys, then "every monkey holds one parrot" means that the group as a whole is holding a parrot, and there is one parrot, instead of two parrots. It seems to me that you can't have it both ways.
Finally, there really is no battle between English and Mathematics. I've been hired to write competition math problems in at least three different states, and I can tell you that English and Mathematics work side-by-side to make a good problem. Ambiguity in the use of English makes the problem unsolvable. If this problem showed up in any math competition anywhere, the appeals committee for the competition would be inundated with appeals written by students arguing for different answers. The appeals committee would have to accept multiple answers or throw the problem out entirely (I saw this happen just a couple weeks ago at a state math competition; because of one missing word in a problem, the problem was thrown out, and no one received any points for it).
I asked my problem proofreader what he would do if I sent this problem to him to proofread. He assured me that he would throw it out. He also said:
"I think this is a wonderful problem for a class discussion, but not for any type of grade or for a competition. There is far too much ambiguity for it to be a problem with a well defined answer."
* Even the St. Ives problem is ambiguous; there isn't actually a solid reason to conclude that the man with seven wives is going in the opposite direction; the narrator could have simply caught up with the man (all things considered, he's probably going fairly slowly!) and struck up a conversation with him. That is an appropriate usage of the word "meet," and the St. Ives problem is therefore ambigious.
Martha asks, "In a straight line equation, why is the slope change in y over change in x, instead of change in x over change in y?"
That's a great question, Martha. One simple answer would be to just say, "because that's the way it is." But there are better answers than that. Let me give you one. Consider the equation for a line, when written in slope-intercept form. It is:
y = mx + b, and we say that m is Δy/Δx. If you're not familiar with the Greek letter delta used here (Δ), it simply means "the change in", so we would read that fraction as "the change in y over the change in x." So the equation now is:
y = (Δy/Δx)x + b.
Now that we've established that, let's talk about units. In real-world situations, the axes would both have units. For example, if the linear equation was part of a word problem about a car traveling from place to place, the horizontal axis might have a time unit, such as hours, and the vertical axis might have a distance unit like miles.
So now let's look at how those units would fit into our equation:
y is a vertical value, so its units would be miles.
Δy is a vertical change, so its units would also be miles.
Δx is a horizontal change, so its units would be hours.
x is a horizontal value, so its units would also be hours.
b represents the y-intercept (which is a y-value when x = 0) so it also has the vertical unit, miles.
Therefore, the equation, written only in terms of its units, would look like this:
miles = (miles/hour)hour + miles
Notice what happens when you multiply out (miles/hour)hour - you get just miles, leading to:
miles = miles + miles.
This is exactly what we need; when you add a two mileages, you get another mileage.
But consider what this would look like if m was Δx/Δy instead of the other way around. We would have:
miles = (hour/miles)hour + miles
miles = hour2/miles + miles, which is about the most absurd conglomeration of nonsensical units I've ever seen.
So that's one way of looking at the question, "Why is the slope Δy/Δx?" Because if it was the other way around, it would produce absurd results.