broccoli fractal (source) |

Pattern questions on the SAT aren’t super common, but they tend to give people all sorts of difficulty when they do appear. Let’s take one apart.

- A farmer is planting a row of plants. He first plants 2 broccoli plants, then 3 cabbage plants, then 1 apple tree, then 2 orange trees, then 1 dill weed plant. He repeats this pattern over and over again until he’s filled up all the land on his (very unorthodox) farm. What kind of plant is the 782
^{nd}one he plants?

(A) cabbage

(B) apple

(C) broccoli

(D) dill weed

(E) orange

Obviously I’m taking some liberties with the writing style of the test makers, but take away the goofiness and this question could totally appear on your SAT. How do you solve it?

Start by writing a few iterations of the pattern on top of each other:

B, B, C, C, C, A, O, O, M,

B, B, C, C, C, A, O, O, M,

B, B, C, C, C, A, O, O, M, …

Now number the ends of the lines (in green here):

(1)B, B, C, C, C, A, O, O, M (9),

(10)B, B, C, C, C, A, O, O, M (18),

(18)B, B, C, C, C, A, O, O, M(27), …

Note that down the RIGHT side of your lines you’re basically counting up by multiples of 9. On the LEFT side, you’re also counting up by 9’s, but in a slightly more confusing way. The mistake a lot of people make on pattern questions is that they start counting from the beginning of the pattern, and they get all screwed up. Count from the *end* of the pattern to make this easy on yourself!

Every even multiple of 9 will be a dill weed plant. We want the 782^{nd} plant. 782 ÷ 9 = 86 *remainder 8*, which means the 782^{nd} plant will be the 8^{th} in our pattern. In other words, it’ll be an orange tree. (E) is our answer. Another way to think of this: the 783^{rd} plant will be a dill weed plant, so the 782^{nd} will be an orange tree.

No sweat, right? Sit down, SAT. You can’t slow me down with that weak sauce.

Of course, repeating patterns like the one above are only one kind of pattern that might get thrown at you, just the kind that seem to give students the most trouble, in my experience. The drill below contains more pattern-type questions.

##### Try these examples:

You need to be registered and logged in to take this quiz. Log in or Register

I’m still not clear on what to do with number 18

For #18, try to identify the pattern in the units digits of the powers of 3. They repeat. 🙂

can you explain 18 one more time? Still unclear. Thanks !

Sure. If you look at the first few terms, you’ll see a pattern: 3, 9, 27, 81, 243, 729, 2187,…

The units digits repeat the following pattern: 3, 9, 7, 1, 3, 9, 7, 1, …

3, 9, 7, 1,

3, 9, 7, 1,

3, 9, 7, 1, …

Every 4th term will have a units digit of 1. Since 1,000,000,000 is a multiple of 4, the units digit of that term will be 1.

Is not 2 a multiple of 1,000,000,000? Please respond. I came here because I have the book and I can’t wrap my head around the question. Thank you very much.

2 is a factor of 1,000,000,000, yes. But since the pattern repeats every four terms, you don’t want to worry about multiples of 2. For example, 6 and 8 are both multiples of 2, but only the 8th term will be a 1. The 6th term will be a 9. Likewise, 1,000,000,002 is divisible by 2, but the 1,000,000,002nd term will be 9, while the 1,000,000,000th term will be 1.

Is that helpful? If not, please let me know.

Yes that was helpful, but is there a better way of answering that problem because the first time I tried it I picked 9 since I looked at the tenth term and assumed that the billionth term would be the same, which of course is wrong. Thanks.

There’s not really a faster way. For questions of this type (which, admittedly, are rare) you need to figure out how often the pattern repeats, and work from there. Your first solution didn’t work because 10 isn’t a multiple of 4. In the future, if you’re faced with one of these beasts, make it your mission to discover where the pattern repeats. Once you know that, these questions aren’t so scary. 🙂

Alright thank you.

There’s also another pattern i discovered. the 10th term has a ones digit of 9, hundreds digit has 1, thousands digit has 9, and so on… but in the end, the pattern proves to say that 1,000,000,000 has a ones digit of 9…..?

I don’t fully follow. Why would the digits of the 10th term (especially the hundreds and thousands digits) help you find the units digit of the 1,000,000,000th term?

can you explain 14? Thanks!

Sure! Every 5th night, it’s going to be Elsie deciding what’s on TV. So on the 40th night, Elsie is in charge. Back up 2 nights from there: on the 39th night, it’s David, and on the 38th night, it’s Charice.

sorry, i meant 10. Sorry for the confusion!

Oh, OK. The trick to #10 is that it’s totally not important what the individual terms are, since we’re only concerned with the difference between two terms. For example, if you were asked about the difference between the 100th and 101st term, it’d be easy, right? Each term is 3 greater than the one before it, so the difference between the 100th and 101st would just be 3!

29th term –> START

30th term –> +3

31st term –> +3

32nd term –> +3

33rd term –> +3

34th term –> +3

35th term –> +3

DIFFERENCE = 18

so I solved it in another way:

2nd term =>we added +3 to the first term

3rd term=> 6 we added 6 to the first term

4th term => 9 ..

I noticed that 9/3= 3

so does 6/2 = 3

so the answer : (3 times 35) – (3 times 29) = 18

it just happened I didn’t know how I solved it after I did :

Why can’t #20 be C? It is also divisible by 3.

For the same reason it can’t be the 6th day. The answer needs to be a number divisible by 3, but only every other number divisible by 3 works. Does that help?

oooooooh, because then it could also be Shayla; I get it now, thanks!

That’s right. Another way to think about it is that the right answer needs to be

n+ 3, wherenis a multiple of 6.I’m sorry for asking so many questions but I am so confused. I think I get the part where it has to be divisible by 3 because it is the third term, but like said above, 279 is divisible by 3, but not divisible by 6. 84 is divisible by 3 and 6, but not 9. I thought the reasoning that 279 does not work was because the answer must be divisible by all multiples of 3?

It does have to be divisible by 3, but NOT because it’s the third term. If the pattern repeated every 5 terms instead of every 6, for example, then the correct answer would NOT have to be divisible by 3. It WOULD still have to be 2 less than a multiple of 5.

In other words,

what the number is divisible by isn’t important for questions like this. What is important is how close the number is to a multiple of the number of repeating terms. So in this question, the only thing that matters is the relative position of the answer choices to multiples of 6.I know what I just wrote is super confusing, but it’s worth reading a few times to try to wrap your head around.

Can you please explain #4 on the diagnostic? 1, 7, 49, 343, …

4. Each term in the sequence above after the

first is determined by multiplying the

previous term by 7. What will be the units

(ones) digit of the 96th term?

I see that it will be 7^95 but I can’t follow the pattern of the one’s digit.

Thank you!

Hi Angelica,

Sure can! The trick is to look for the pattern in the ones digits of powers of 7.

74

934

3240

11680

711764

9The ones digits go like this: 7, 9, 3, 1, 7, 9, 3, 1, …

You’re right that the 96th term of the original sequence will be 7^95. What is the 95th term in the ones digits pattern above?

Note that the 4th term is 1. So is the 8th term. If you keep going, every 4th term will be 1. 96 is a multiple of 4, so the 96th term will be 1. The 95th term, then, will be whatever comes before 1 in the pattern every time. In this pattern, 3 always comes before 1, so 7^95 will have a ones digit of 3.

[Link to the diagnostic drill this comes from]

can you please solve section 6 #8? 🙁

http://www.collegeboard.com/prod_downloads/sat/sat-practice-test.pdf

Yeah. This one sucks, right?

The key to this one is to lay out a 14-day calendar on your paper, then write in letters for each family that could be staying in the hotel that day. Note that the Liu and Benton family do not overlap, and stayed a total of 14 days, which guarantees that there was always SOMEONE at the hotel.

Try stacking all the families other than Liu and Benton all on one side. If you put them all in the first part of the 14 day period, then the Jackson family would cover all the answer choice days.

If you put them all at the end of the 14 day period, then the Jackson family shows up on the 4th day, and stays until the end. They don’t need to be there the 3rd day, nor does anyone else except either Benton or Liu. That’s why it’s (A).

(FYI: The reason your comment didn’t appear right away is that it contained a link. Comments with links get automatically set aside for approval before they go live, so they basically have to wait until I get home.)

Oh my god! I can’t believe how easy these are c’:

Can you please explain #3 in the diagnostic drill #3? (In the sequence above, there is one 1,followed by two 2s, three 3s, four 4s, and so on, etc.)

After the first occurrence of 80, there will be 79 more 80’s. Then there will be 81 81s, 82 82s, 83 83s, and 84 84’s. So the answer is 79 + 81 + 82 + 83 + 84 = 409.

Can you please explain #20?

It rather frightens me

Did you read the thread starting here?

Yes, I did. But, actually, I don’t understand how to do it from the very beginning. Can you tell me what I should do first?

OK. First, recognize that the pattern repeats every 6 days, so the 6th day, the 12th day, the 18th day, etc. will all be Shayla days.

Now recognize that 3 days before each of those days will always be Fantasia days. So the 6-3=3rd day, 12-3=9th day, 18-3=15th day, etc.

What you’re looking for in the answer choices is a number that’s 3 less than a multiple of 6. Only (E) is.

Oh! Thank you!

Thank you also for the awesome blog! You are the most inspiring writer I’ve ever seen!

Hi, how are you so I ran into an sat question and I’m not sure if it’s a pattern question, it was a grid-in.. The question is n=100110021003…1020

In the sequence n above the sequence starts from 1001 and goes up to 2020 so how much is the sum of all the digits in the sequence from 1001 to 2020?

I’m having a hard time figuring out exactly what you’ve got written here. Is this the pattern?

1001, 1002, 1003, …, 1020?

And then is the question about the sum of all the digits from 1001 to 1020? The sum of all the digits from 1001 to 2020 is a much harder problem, and I doubt it would be an SAT question.

The fastest way to solve the one I think is the real question is to just add up the digits place by place. The ones digits will go 1–9 twice. The tens digits will have ten 1s and one 2. The hundreds digits are always 0 so we can ignore them. The thousands digits will always be 1, and there will be twenty of them.

2(1+2+3+4+5+6+7+8+9) + 10(1) + 2 + 20(1) = 122

unfortunately it was today’s school SAT last grid-in math question and i got 128.. ughh.. close enough. Just wanted to tell u about it. Thank you!!

Another question: PWN The SAT book will cover all the math aspects on the SAT, even the really hard ones, right? Because I’m really enjoying this site and I think the book will be even better.. that might get me to a perfect score in the math section.

there was actually another hard question but it had a drawing so i don’t really know how to put it, I just wanted to know the answer but I guess I have to wait.. Thanks again.

Good day

I’m sorry for being so disturbing, but I just can’t get what’s wrong with this one…

I’m taking SAT tomorrow:((

The answer is (D) because the pattern repeats every 5 days, so you know every 5th day will be Ella. So the 5th day the 10th day, the 15th day, the 50th day, the 100th day—all Ella. Since the 100th day is an Ella day, you know the day before, the 99th day, is a Dilip day.

I’m betting you picked (C) because 99 is a multiple of 3. That’s a common error, but it doesn’t matter what 99 is a multiple of. 6 is also a multiple of 3, but obviously the 6th day is an Alex day, not a Corinne day. The

onlything that matters is where 99 is in relation to multiples of 5. The closest multiple of 5 to 99 is 100.@PWNtheSAT:disqus am sorry i didn’t get it =(

If you’re struggling with this one, it might help you to actually list out the days: A, B, C, D, E, A, B, C, D, E … up to 99.

Can someone show me how to solve this one please?!

I forgot some rules about parallelograms :/

the ans. is B.

sorry if the picture is not well oriented!

The ans is D

i thought it is C. because it also proves the statement is false!!2010 is an even number but the sum of its digits is odd!

can someone make it clearer to me!?

In the future, please type questions like this up and submit them via the Q&A page—they’ll be answered more quickly there. The problem with (C) is that the digits add up to an odd number: 2 + 0 + 1 + 0 = 3. The question is about numbers where the digits add up to even numbers!

(D) works because the sum of the digits is even: 2 + 0 + 1 + 1 = 4, but the number, 2011, is not even. So the “if” was true, but the “then” was false. That disproves the statement.

Okay i will the next time

I got it!👍

thank you so much ! ♡