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TEN FOR TEN®
NUMBER PATTERNS B
1)
The least integer in a set of consecutive integers is –8. If the sum of these integers is 19,
how many integers are in this set?
a) 8
c) 16
b) 9
d) 18
e) 19
The sum of three consecutive even integers is 186.
2)
3)
If y represents the greatest of the three integers, which of the following equations
represents the statement above?
a) 3y = 186
c) 3y – 3 = 186
b) 3y –2 = 186
d) 3y – 6 = 186
e) 3y – 9 = 186
The first two numbers of a sequence are 7 and 9, respectively. The third number is 16, and,
in general, every number after the second is the sum of the two numbers immediately
preceding it. How many of the first 1,000 numbers in this sequence are odd?
a) 333
c) 665
b) 500
d) 666
e) 667
PLEASE READ THE ANSWERS AND EXPLANATIONS FOR PROBLEMS 1 THROUGH 3 NOW
1, 2, 1, -1, -2 …
4)
5)
6)
The first five terms of a sequence are shown above. After the second term, each term can
be obtained by subtracting from the previous term the term before that. For example, the
third term can be obtained by subtracting the first term from the second term. What is the
sum of the first 44 terms of this sequence?
a) 0
c) 7
b) 3
d) 9
e) 14
A paper chain is made by stringing Z individual paper links together in the repeating
pattern orange, white, red, blue, green, and purple. If the paper chain begins with a
orange link and is 73 links long, then the last link is
a) orange
c) blue
b) white
d) green
e) purple
[Grid In] If the sum of the consecutive integers from –23 to p, inclusive, is 75, what is the
value of p?
NUMBER PATTERNS B
2
3, 6, 11, 18 …
7)
The first four terms of a sequence are shown above. Which of the following could be the
formula that gives the nth term of this sequence for all positive integers n?
a) 2n
c) 3n
b) 2n + 1
d) n2 + 1
e)
n2 + 2
44.404004000400004 ...
8)
The decimal number above consists of only 4’s and 0’s to the right of the decimal point.
The first 4 is followed by one 0, the second 4 is followed by two 0’s, the third 4 is followed by
three 0’s, and so on. What is the total number of 0’s between the 74th and 79th “4” in this
decimal number?
a) 380
c) 440
b) 390
d) 443
e)
459
4, 6, 12, 30, 84 …
9)
The first five terms of a sequence are shown above. Each consecutive term in the
sequence is calculated by performing the same operation on the previous term n. What is
the sixth term?
a) 133
c) 176
b) 139
d) 246
e) 252
p, r, s, t, u
10)
An arithmetic sequence is a sequence in which each term after the first is equal to the sum
of the preceding term and a constant. If the list of numbers shown above is an arithmetic
sequence, which of the following must also be an arithmetic sequence?
I.
2p, 2r, 2s, 2t, 2u
II.
p – 3, r – 3, s – 3, t – 3, u – 3
III.
p2, r2, s2, t2, u2
6/15/09
a) I only
c) III only
b) II only
d) I and II
e) II and III
TEN FOR TEN®
ANSWERS AND EXPLANATIONS
NUMBER PATTERNS B
1)
E. Please draw a number line* whenever you’re asked about consecutive integers. As we
proceed upward on this number line, we add all of the negative integers starting with –8.
Since our addition starts in negative territory and we need to come up with a positive sum,
first let’s figure out how to get a sum of zero; we’ll need to add the integers from negative
8 all the way up to … positive 8. OK, let’s keep adding consecutive integers: When we
add the next integer on the number line, 9, to our sum of zero, don’t we get a sum of 9
(0 + 9)? Now, let’s add 10 to that sum of 9; when we do so, we get a sum of 19. Looking
at our number line, how many integers comprise our set? Well, we have 8 negative
integers, 10 positive integers, and zero!
2)
D. First, let’s find calculate the average (it’s 186/3, or 62). So, our three consecutive even
integers must be 60, 62, and 64. At this point, I assume that you know that when we’re
dealing with consecutive numbers, whether they’re consecutive integers or consecutive
multiples of 12, the average is also the median!! Plugging 64 (the greatest of the three
integers) in for y gives us …
3)
E. First, note that we have a three-number pattern that repeats odd, odd, even, odd,
odd, even. So, two numbers out of every three-number sequence are odd. With that in
mind, how many three-number patterns are there in 1,000 numbers? 333.333, says your
calculator. It’s true; we can break 1,000 numbers into 333 complete three-number
patterns. Since each pattern contains two odd numbers, 333 patterns will provide us with
a total of 666 odd numbers. However, what does that annoying decimal (.333) mean? It
means that after we’ve taken out 333 patterns there is a single number left over. Would
such a number be even or odd? If you’ve been paying attention so far, you know that
any such “orphans” will always begin the next pattern. Is the first number in each of our
patterns even or odd? Right, it’s odd. So, 666 + 1 …
PLEASE RETURN AND FINISH PROBLEMS 4 THROUGH 10
4)
B. Did you read the instructions carefully? (“Each term can be obtained by subtracting
from the previous term the term before that …” means that to calculate the 6th term, we
subtract the 4th term from the 5th term.) When we extend the pattern, we find that the sixth
term is –1 [-2 – (-1)]; the seventh term is 1 [-1 – (-2)]; the eighth term is 2 [1 – (-1)]. Since the
pattern repeats beginning with its 7th term, we know that there are six terms in each
repeating set.^ Next: Since we need to come up with an overall sum, first we should
determine the sum of each six-term set. Well, 1 + 2 + 1 + (-1) + (-2) + (-1) = 0, right? So,
given that every six-term set adds up to 0, aren’t we really interested only in the sum of the
remainder when 44 is divided by 6 (7 full sets with a remainder of 2 terms)? When we’ve
*
We recommend using vertical number lines, which eliminate any confusion about which negative
number is “bigger.”
^
Note by how few numbers we had to extend this sequence before it repeated. If you’ve extended a
sequence by five terms and it hasn’t repeated yet, start over because you’ve done the math wrong.
NUMBER PATTERNS B
ANSWERS AND EXPLANATIONS
2
removed those 7 full sets, we still have two terms left over! Wouldn’t they be the first two
terms in the next full set? So, 1 + 2 …
5)
A. The six colors repeat, so we can use the same math that we would use to find the 73rd
term in a 6-term repeating pattern, right? Dividing by 6 (the number of links in the pattern),
we find that the pattern of colors will repeat 12 times with a remainder of 1 link. Wouldn’t
that be the first link in a final, incomplete set?
6)
26. See the explanation for problem 1. Here, starting with -23, how far into positive territory
will we need to add to get a sum of zero? Positive 23, right? So, after we’ve added all the
numbers between negative 23 and positive 23, inclusive, we have a sum of 0. Since we’re
looking for a sum of 75, though, we need to keep adding. Adding the next consecutive
integer, 24, gives us a sum of 24 (0 + 24); adding 25 to that sum gives us 49; adding 26 gives
us 75.
7)
E. It’s important to distinguish between the value of a number and its place in a pattern.
The nth term refers to the (1st, 2nd, 3rd) term in the pattern; here, 3 is the 1st term; 6 is the 2nd
term; 11 is the 3rd. And so on. We need to identify the answer choice that provides us with
a formula we can use to relate 1st (the place in the pattern) to 3 (the number occupying
that place). Eliminate choices (a) and (d), since when we plug in 1 for n into those
formulas, we get 2. Now, we’re down to (b), (c), and (e), so let’s try to relate the 2nd place
in the pattern to the number 6: eliminate choice (b), since plugging 2 for n gives us 5, not
6; let’s try to use our remaining formulas [remember, we’re down to (c) and (e) at this
point] to relate 11: In (c) 3 times 3 (the place in the pattern) gives us 9; so, (e) must be the
answer. Let’s try it: 32 + 2 = 11; 42 + 2 = 18; what would be the 5th term in this pattern?*
8)
A. We’re looking to add the number of zeros that follow the 74th (74), 75th (75), 76th (76),
77th (77), and 78th (78) terms. We end with the 79th term, so any zeros that follow that term
are outside the area we’ve been asked to add. (SAT directions are precise!)
9)
D. When a sequence seems to have inconsistent gaps between the terms, it’s good to
take a minute to examine those gaps to see whether they make up the real sequence!
Here, the gap between 4 and 6 is 2; the gap between 6 and 12 is 6 (3 times 2); the gap
between 12 and 30 is 18 (3 times 6); the gap between 30 and 84 is 54 (3 times 18); so,
mustn’t the next gap be 3 times 54 (or 162)?
10)
*
D. In an arithmetic pattern, the gap between terms remains constant throughout each set.
Let’s plug in numbers 1, 2, 3, 4, and 5 (each succeeding term is equal to the previous term
plus 1). Doing so, we find that both Roman numerals I and II work fine (in Roman numeral I,
the constant is now twice as big—but it’s still a constant!); however, when we try III, we get
1, 4, 9, 16, 25, which means that each term does not relate to the previous one
arithmetically.
27.
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