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28 CONQUERING SAT MATH
12. How many positive integers less than 20 have an odd
number of distinct factors?
A. 12
B. 10
C. 8
D. 6
E. 4
13. In the repeating decimal 0.714285714285 . . . , what
is the 50th digit to the right of the decimal point?
14. How many distinct composite numbers can be formed
by adding 2 of the first 5 prime numbers?
15. A rope is 13 feet long. How many ways can the rope
be cut into more than one piece so that the length of
each piece is a prime number?
A. 4
B. 5
C. 6
D. 7
E. 8
CHAPTER 3 / FACTORS AND MULTIPLES 29
EXPLAINED ANSWERS
1. Answer: D
The factors of 24 are {1,2,3,4,6,8,12,24}.
1 + 2 + 3 + 4 + 6 + 8 + 12 + 24 = 60
2. Answer: C
3. Answer: C
23 and 29 are the only prime numbers between 20 and 30.
This shows that 52 is a factor of 156.
4. Answer: B
Divide the number of people at the party by the number of seats around each table.
Thirteen tables will be needed to seat all 100 people.
5. Answer: 16
The largest integer that evenly divides both 48 and 64 is the GCF of the two numbers.
The factors of 48 are {1,2,3,4,6,8,12,16,24,48}.
The factors of 64 are {1,2,4,8,16,32,64}.
The GCF is 16.
6. Answer: A
Because P is a prime number, the prime factorization of 9 � P is 32 � P. Therefore, 3 and P are the two distinct primes
of 9 � P.
7. Answer: D
Try an example. Choose 3 for x.
3 divided by 8 leaves a remainder of 3.
Multiply 3 × 4 = 12.
12 divided by 8 leaves a remainder of 4.
8. Answer: C
Note that this item asks for the choice that is NOT true.
We know y is a multiple of x and z. That means y times any number is also a multiple of x, so choice A must be true.
Also, because we know y is a multiple of x and z, this means that x and z divide evenly into y, so choices B, D and E
are true.
The process of elimination leaves choice C.
Use an example to check this. Let x = 3, y = 12, and z = 6. y is a multiple of x and z, but x � z = 18 is not a multiple
of y = 12.
100 8 12 5÷ = . .
52 3 156i = .
23 29 52+ = .
3 3 3 2435 = × × 3 × 3 × 3 = .
30 CONQUERING SAT MATH
9. Answer: E
The two-digit numbers that have a remainder of 2 when divided by 5 are
Only choice E is always true, the one’s digit is prime.
Don’t be fooled because some of the answers are true sometimes.
For example, in choice (A) the sum of all digits is odd.
This is true if the two-digit number is 32.
The sum of the digits 3 + 2 = 5 is odd.
But choice (A) is not always true.
If the two digit number is 42, the sum of the digits 4 + 2 = 6 is not odd.
10. Answer: B
The numbers must be 4, 5, and 6.
2,3, and 5 are the three distinct prime factors.
11. Answer: A
The question states that a, b, and c are integers greater than 1. That means b is an integer greater than 1 that divides
evenly into both 21 and 39.
That means that b must be 3.
Divide.
12. Answer: E
The only positive integers less than 20 that have an odd number of factors are the squares 1, 4, 9, and 16. That is be-
cause these numbers, alone, have a factor that is multiplied by itself. The factors for each are as follows:
Number Factors
1: {1}
4: {1,2,4}
9: {1,3,9}
16: {1,2,4,8,16}
13. Answer: 1
In the repeating decimal 0.714285714285 . . . , 5 is the 6th, 12th, 18th, 24th, 30th, 36th, 42nd, and 48th digit.
7 is the 49th digit and 1 is the 50th digit.
You might also realize that there are 6 digits that repeat, divide 50 by 6, and get a remainder of 2. Therefore the 2nd
of the 6 digits that repeat, which is 1, will be the 50th digit.
3 < 7 < 13
< <b a c
a cmust be 7, must be 13,7 3 21 3 13 39i i= =. .
4 5 6 15 4 5 6 120 2 3 53+ + = = =and i i i i
12 17 22 27 32 37 42 47 52 57 62 67 72 77 8, , , , , , , , , , , , , , 22 87 92 97, , , .{ }