Errata – Number Properties, 4th Edition

Cover for 4th Edition

Edition 4.2	Edition 4.1	Edition 4.0
Release Date: August, 2010 Differentiating Mark: Back Cover, Bottom Right Corner	Release Date: December 1, 2009	Release Date: May 1, 2009

The Number Properties Guide (160 pages) provides a comprehensive analysis of the properties and rules of integers tested on the GMAT. Learn, practice, and master everything from prime products to perfect squares.

4.1

Page	Loc	Description	Erroneous Text	Correction
25	Top	The first line of the explanation is wrong. Substitute 12 for t.	If s is a multiple of t, then so is 7s.	If s is a multiple of 12, then so is 7s.
85	Bot	In the explanation for question #14, the expression udnerneath the first radical in the explanation text is wrong.	…if they are under the same radical: root(y3 + 3x2y3)…	…if they are under the same radical: root(x2y3 + 3x2y3)…
139	Bot	In the paragraph to the left of the lower chart, the second sentence mistakenly says that 20- contains two 3's instead of two 2's.	20 contains two 3's, so x can contain zero, one, or two 2's.	20 contains two 2's, so x can contain zero, one, or two 2's.
156	Mid	The last sentence in the explanation for 5D is incorrect.	…(Odd)(Odd + 2) = (R1)(R3) = R9 = R1 (taking out two 4's).	…(Odd)(Odd + 2) = (R1)(R3) = R3.
176	Mid	There is no Advanced Positives and Negatives question. #152 is an Odds and Evens problem.
170	Mid	In the explanation to question #10, the incorrect values for x and y are listed at the end of the second paragraph.	If y = 1, then x = 5 (odd), but if y = 2, then x = 8 (even).	If y = 1, then x = 1 (odd), but if y = 2, then x = 4 (even).
121	Bot	In the third to last paragraph on the page, the final sentence is missing the word need.	It does not necessarily to contain any twos, so…	It does not necessarily NEED to contain any twos, so…
151	Top	The example problem should include the stipulation that x > 1.	If x^3 – x = p, and x is odd, is p divisible by 24?	If x^3 – x = p, x is odd, and x > 1, is p divisible by 24?
155	Top	Explanation for #1 mistakenly says that q and r must both be odd or both be even.	Similarly, we can evaluate Statement (2) with a scenario table. The variables q and r must either both be odd or both be even, and p can be odd or even.	Similarly, we can evaluate Statement (2) with a scenario table. Either q is even and r is odd or q is odd and r is even, and p can be odd or even.
155	Top	For the explanation to problem #1, scenario 3 in the first table is incorrectly displayed in the final column.	E × O + O = O	E × O + E = E
155	Top	For the explanation to problem #1, scenario 4 in the first table is incorrectly displayed in the final column.	E × O × E + E = E	E × E × E + E = E

4.0

Page	Loc	Description	Erroneous Text	Correction
40	Mid	The product of a set that only contains negative numbers cannot be zero. Drop the “(or zero)” parentheticals in the discussion of Statement 1.	Statement (1) tells us that all of the numbers in the set are negative. If there are an even number of negatives in Set S, the product of its elements will be positive (or zero); if there are an odd number of negatives, the product will be negative (or zero). This also is INSUFFICIENT.	Statement (1) tells us that all of the numbers in the set are negative. If there are an even number of negatives in Set S, the product of its elements will be positive; if there are an odd number of negatives, the product will be negative. This also is INSUFFICIENT.
151	Top	The example problem should include the stipulation that x > 1.	If x^3 – x = p, and x is odd, is p divisible by 24?	If x^3 – x = p, x is odd, and x > 1, is p divisible by 24?
85	Bot	In the explanation for question #14, the expression udnerneath the first radical in the explanation text is wrong.	…if they are under the same radical: root(y3 + 3x2y3)…	…if they are under the same radical: root(x2y3 + 3x2y3)…
139	Bot	In the paragraph to the left of the lower chart, the second sentence mistakenly says that 20- contains two 3's instead of two 2's..	20 contains two 3's, so x can contain zero, one, or two 2's.	20 contains two 2's, so x can contain zero, one, or two 2's.
156	Mid	The last sentence in the explanation for 5D is incorrect.	…(Odd)(Odd + 2) = (R1)(R3) = R9 = R1 (taking out two 4's).	…(Odd)(Odd + 2) = (R1)(R3) = R3.
121	Bot	In the third to last paragraph on the page, the final sentence is missing the word need.	It does not necessarily to contain any twos, so…	It does not necessarily NEED to contain any twos, so…
170	Mid	In the explanation to question #10, the incorrect values for x and y are listed at the end of the second paragraph.	If y = 1, then x = 5 (odd), but if y = 2, then x = 8 (even).	If y = 1, then x = 1 (odd), but if y = 2, then x = 4 (even).
156	Top	Problem 4 explanation: Incorrect result of odd/even equation. However, the statement is sufficient either way, so the problem's answer is unchanged.	4. (A): If both c and d are odd, then c – 3d equals O – (3 × O) = O – O = O.	4. (A): If both c and d are odd, then c – 3d equals O – (3 × O) = O – O = E.
155	Top	For the explanation to problem #1, scenario 3 in the first table is incorrectly displayed in the final column.	E × O + O = O	E × O + E = E
155	Top	For the explanation to problem #1, scenario 4 in the first table is incorrectly displayed in the final column.	E × O × E + E = E	E × E × E + E = E
25	Top	The first line of the explanation is wrong. Substitute 12 for t.	If s is a multiple of t, then so is 7s.	If s is a multiple of 12, then so is 7s.
155	Top	Explanation for #1 mistakenly says that q and r must both be odd or both be even.	Similarly, we can evaluate Statement (2) with a scenario table. The variables q and r must either both be odd or both be even, and p can be odd or even.	Similarly, we can evaluate Statement (2) with a scenario table. Either q is even and r is odd or q is odd and r is even, and p can be odd or even.