Let’s go over the answers to those questions in the SAT number properties quiz. Each question is repeated below so you don’t have to keep going back and forth between posts.
In the three-digit integer , in which X, Y, and Z each represent a different integer, . If is even, how many different integers can it represent?
Explanation: First, set up three spaces on your paper for that three digit integer.
__ __ __
We’re going to have to plug in a couple of numbers to these digits and look for a pattern. Never hesitate to plug in. Let’s start with 1.
1 __ __
But that would make Y=1 also, and the question told us that X and Y are different integers. No go.
So let’s try X=2.
2 __ __
That means Y=4
2 4 __
So two-hundred and forty-something works. But because it’s even, we can only put five numbers into the last blank: 0,2,4,6, or 8.
BUT keep in mind that the digits have to be distinct! That means neither 2 nor 4 can be Z, because they’ve been used for X and Y already.
So we have three possibilities. (A) is out. Can we pick other numbers for X? Try 3.
3 __ __
3 9 __
This time, we can use all five even digits numbers for Z. So we’re up to eight possibilities. (B) is out.
And let’s try another…
4 __ __
4 16 __
16 can’t be a digit. That means 4 is too big to be the value of X, which means our only options were 2 and 3.So we have a total of eight numbers to be XYZ.
Shortcut: Since we know that Y can’t be greater than 9, and Y is X squared, we know that X can’t be greater than 3. Then just check 2 and 1 quickly.
If , which of the following must be true?
(A) I only
(B) II only
(C) I and II only
(D) II and III only
(E) I, II, and III
Explanation: Again, plug in some numbers! The question gives us some hints as to what to try in I, II, and III. Since I is a bit hard to test, start with II.
If , then and , so that doesn’t work. And we can rule out any other positive integer for the same reason; increasing the exponent increases the value. And 1 is out, since it neither increases nor decreases with an exponent. So II is true. That eliminates (A).
Next, let’s check III.
If , then and . That means is the lowest of the three, so that doesn’t make sense. And, come to think of it, all negative numbers will follow the same pattern, going from negative to positive to negative as the exponent increases. So we can eliminate III, which means (D) and (E) are gone.
Now we know that X must be between -1 and 1. Plug in and to check III, and you’ll see that works. So (C) it is.
Shortcut: Since , we know that must be true. Both have a larger, odd exponent creating a lesser value than a smaller, even exponent. There’s no real number that would work in one of those inequalities and not the other.
Also, knowing that natural numbers always increase with larger exponents would prove fact II right away. No number greater than 1 would follow the strange pattern provided. Then it’s just a matter of plugging in negative less than -1 to check III.
The sum of 4 distinct positive integers is 29. If the greatest integer is less than 10, what is the least of the four?
Explanation: Use the numbers in the answer choices if you’re not sure how to tackle this. Since we’re looking for the least, we’ll start by checking (A).
If the lowest number is 0, then the other three together must total 29. But if they’re all less than 10, that’s not possible. The highest they can be is 9+8+7=24. So (A) is out.
(B) gives us a similar problem. The other three would have to total 26, but 24 is as high as we can get.
(C), on the other hand, works. 29 – 5 = 24, and we already know that’s possible.
Shortcut: In order to get a very low number in a set that adds up to something specific, we want to use the highest possible numbers for the others in the set.
For example, if we were trying to get a sum of 10 from two numbers, and we wanted to use a low number like 1, we’d need to pair it with a high number, 9.
In this case, we use the highest possible numbers—9, 8, and 7—for the other three, and then just check what the difference is. 29 – 24 = 5.
When the prime number k is multiplied by 2, the result is 1 less than another prime number. Which of the following is NOT a possible value of k?
Explanation: Once again, we’re going to want to use these answer choices to our advantage. In fact, we have to. Any question that asks “which of the following is NOT” has to be solved by process of elimination.
So let’s take (A). If k is 2, then multiply that by 2 to get 4. Is 4 one less than a prime number? Yep. 5 is prime.
For (B), 3 multiplied by 2 gives us 6, which is one less then 7. Since 7 is prime, this also works. Cross it off.
(C) has the same problem; 2 times 5 is 10, which is one less than 11.
(D) is the only one that doesn’t make another prime. 2 times 7 is 14, which is one less than 15, and 15 isn’t prime.
Shortcut: Since you have to go by process of elimination, there isn’t a faster way to do this, really. Writing out 2k + 1 = prime might help you speed up your math, though.
If a and b are distinct positive integers and , then what is one possible value of a?
Answer: 8, 27, 64, 125, or any other perfect cube.
Explanation: No answer choices to work with here… and it’s not immediately obvious if plugging in any old number for the value of a or b will help us.
We need to simplify the equation. Get b on only one side.
Then get rid of that root 3 by cubing each side.
Get the square root of each side to make it simpler.
And you’re pretty much done. We know that b is an integer, so plug one in—say, 2—and the value of a is 8. Or whatever else, depending on what you choose for b.
Shortcut: Once we get to , we can cancel out the 2s in each exponent to get
And since we know b is an integer, a must be a perfect cube; just pick one.
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