Concept Review 7
- Any integer greater than 1 that is divisible only by
1 and itself. - The whole number “left over” when one whole
number is divided by another whole number a
whole number of times. - Any integer that is not divisible by 2. 2n+1.
- Underline and think about those words when you
see them in problems. - Divide the two integers, then multiply the deci-
mal part of the result by the divisor (the number
you divided by). - The least common multiple of k,12, and 35 is
420 k,which is divisible by 3k,10, 28k,2, and 35k. - 0, 1, 2, 3, 4, 5, and 6.
- The pattern is 3 terms long, and 3 divided by
100 has a remainder of 1, so the 100th term is the
same as the 1st.
- The pattern is 3 terms long, and 3 divided by
- The number must be 1 greater than both a
multiple of 7 and a multiple of 9. The only multiple
- The number must be 1 greater than both a
of 7 and 9 that is between 1 and 100 is 63, and
63 + 1 =64.- 1/2. The sequence is
1, 2, 2, 1, 1/2, 1/2, 1, 2, 2, 1, 1/2, 1/2,... , so the
pattern is 6 terms long. 65 divided by 6 leaves a
remainder of 5, so the 65th term is the same as
the 5th, which is 1/2.
- The least common multiple of 6 and 9 is 18,
and 100 ÷ 18 =5.555... , which means that there
are 5 multiples of 18 between 1 and 100.
- The least common multiple of 6 and 9 is 18,
- None. Prime numbers are divisible only by
themselves and 1, but any multiple of 6 must also
be divisible by 2 and 3. - 4z−1 and z^2 +z+1 are the only expressions that
must be odd. Since 4 is even, 4zis even, so 4z− 1
must be odd. z^2 +z+ 1 =z(z+1) +1, and since
either zor z+1 must be even, z(z+1) is even and
z(z+1) +1 is odd. - If adivided by bleaves a remainder of r,then a
must be rgreater than a multiple of b.
Answer Key 7: Divisibility
298 MCGRAW-HILL’S SAT
SAT Practice 7
- 2 Since nleaves a remainder of 7 when divided
by 10, it must be 7 more than a multiple of 10, like
7, 17, 27, etc. When any of these is divided by 5,
the remainder is 2.
2.D Don’t confuse remainder with quotient. Re-
member to think of balloons and kids. If you had
8 balloons to divide among 12 kids, you’d have to
keep all 8 balloons because there aren’t enough to
go around fairly. Also, you can use the calculator
method, and divide 8 by 12, then multiply the
decimal part of the result by 12.
3.B Using p=1 and q=2 rules out II (1/4 is not an
integer, let alone an odd number) and III (4 is
even). p^2 +q^2 will always be odd, because the
square of an odd is always odd and the square of
an even is always even, and an odd plus an even
is always odd.
4.C The sequence that repeats is 5 terms long. 103
divided by 5 leaves a remainder of 3, so the 103rd
term is the same as the 3rd term, which is 5.
5.D For the statement to be correct, b=a(0) +3,
so b=3.
6.A A counterexample to the statement All prime
numbers are oddwould be a prime number that is
not odd. The only even prime number is 2.
7.A If k/7 and k/12 are both positive integers, then
kmust be a common multiple of 7 and 12. The
least common multiple of 7 and 12 is 84. If we
substitute 84 for k,(A) is the only choice that
yields an integer.
8.D Using the example of 2, 3, 4, 5, 6 disproves
statement I. Since multiples of 5 occur every
5 consecutive integers, II must be true (remem-
ber that 0 is a multiple of every integer, including
5). Since bcwill always be an even times an odd
or an odd times an even, the result must always
be even, so bc+1 must be odd.
9.E You might simplify this problem by picking
values for mand nthat work, like 46 and 6.
(When 46 is divided by 6, the quotient is 7 with a
remainder of 4.) If we substitute 6 for n,choice
(E) is the only one that yields 46.
10.D Using a=10 and b=4 disproves statement I.
If a/bequals 5/2 and aand bare both integers,
then a= 5 kand b= 2 k,where kis an integer.
Therefore a+b= 7 k,so II is true. Also, since a/b
equals 5/2, b/a=2/5, so 5b/a=10/5 =2, which is
an integer, so III is true.