260 MCGRAW-HILL’S SAT
b– 1, is 6. The only positive integer pairs with a prod-
uct of 6 are 2 ×3 and 1 ×6, so one possibility is that
a= 2 and b= 4. This gives a(b– 1) = 2(4 – 1) = 2(3) = 6,
and it satisfies the condition that a< b. Now check the
statements. Statement I is true here because 4/2 = 2,
which is an integer. Statement II is also true here
because 4 is an even number. Statement III is also
true because 2 ×4 = 8, which is 6 greater than 2. So
the answer is (E) I, II and III, right?
Wrong. Remember that the question asks what
mustbe true, not just what canbe true. We’ve only
shown that the statements canbe true. We can prove
that statement I mustbe true by testing all the possi-
ble cases. Since there is only one other possible solu-
tion that satisfies the conditions: a= 1 and b= 7, and
since 7/1 = 7 is an integer, we can say with confidence
that statement I must be true. But statement II does-
n’thave to be true because bcan equal 7, which is not
even. We have found a counterexample. Next, we can
prove that statement III must be true by checking both
cases: 2 ×4 is 6 greater than 2, and 1 ×7 is 6 greater
than 1. (We can prove it algebraicallytoo! If we add a
to both sides of the original equation, we get ab= a+ 6,
which proves that abis 6 greater than a.)Process of Elimination (POE)On multiple-choice questions (and especially
“must be true” questions), it helps to cross off
wrong answers right away. Sometimes POE
simplifies the problem dramatically.What if, in the preceding question, the first solu-
tion we found was a= 1 and b= 7. For this solution,
statements I and III are true, but statement II is not.
Therefore, we could eliminate those choices contain-
ing II—(B), (D), and (E). Since the two remaining
choices contain statement I, it must be true—we don’t
even need to prove it!Lesson 7:Thinking Logically
Numerical and Algebraic ProofLogical proofs aren’t just for geometry class.
They apply to arithmetic and algebra, too. In
arithmetic, you often need to apply the laws of
arithmetic (such as odd×even = even, negative
÷positive= negative—see Chapter 9, Lesson 3)
to provewhat you’re looking for. When you
solve an algebraic equation, you use logical
laws of equality (such as the addition law of
equality) to provethe equation you want.“Must Be True” QuestionsLogic is especially useful in solving SAT “must
be true” questions. You know them and hate
them—they usually have those roman numer-
als I, II, and III. To prove that a statement
“must be true,” apply the laws of equality or
the laws of arithmetic. To prove that a state-
ment doesn’thave to be true, just find one
counterexample, a valid example for which the
statement is false.If aand bare positive integers such that a< band
ab– a= 6, which of the following must be true?I. is an integer.II. bis an even number.
III. abis 6 greater than a.
(A) I only
(B) I and II only
(C) I and III only
(D) II and III only
(E) I, II, and III
This requires both numerical and algebraic logic.
First, let’s see how far we can get trying to solve the
equation for aand b.
ab – a= 6
Factor out the a: a(b– 1) = 6
Okay, we’ve got a problem. We have two un-
knowns but only one equation, which means we can’t
solve it uniquely. Fortunately, we know that aand b
must be positive integers, so the equation basically
says that the product of two positive integers,aandb
a