GRE_Math_Bible_eBook
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Number Theory |
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30. Since b = a + 3, the digit b is greater than the digit a and a – b = –3.
In a two-digit number, the leftmost digit (the tens-digit) is the “more significant” digit. Since b > a, the number ba (which has b in tens-digit position) is greater than the number ab (which has a in tens-digit position). Hence, Column B is greater than Column A, and the answer is (B).
Method II
Column A: ab = 10a + b.
Column B: ba = 10b + a.
Column A – Column B = 10a + b – (10b + a) = 9a – 9b = 9(a – b) = 9(–3) = –27.
Hence, Column A is 27 units less than Column B. The answer is (B).
31. Since a, b, c, d, and e are consecutive numbers in the increasing order, we have b = a + 1, c = a + 2, d = a + 3 and e = a + 4. The sum of the five numbers is a + (a + 1) + (a + 2) + (a + 3) + (a + 4) = 5a + 10.
Now, we are given that the sum decreased by 20% when one number was deleted. Hence, the new sum should be (5a + 10)(1 – 20/100) = (5a + 10)(1 – 1/5) = (5a + 10)(4/5) = 4a + 8. Now, since New Sum = Old Sum – Dropped Number, we have (5a + 10) = (4a + 8) + Dropped Number. Hence, the number dropped is (5a + 10) – (4a + 8) = a + 2. Since c = a + 2, the answer is (C).
32. The average of the five consecutive positive integers 1, 2, 3, 4, and 5 is (1 + 2 + 3 + 4 + 5)/5 = 15/5 = 3. After dropping 5 (the greatest number), the new average becomes (1 + 2 + 3 + 4)/4 = 10/4 = 2.5. The percentage drop in the average is
Old average – New average 100 = Old average
3− 2.5 100 =
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16.66%
The answer is (E).
33.The term 14! equals the product of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, and 14. Only two
of these numbers are divisible by 7. The numbers are 7 and 14. Hence, 14! can be expressed as the product of k 7 14, where k is not divisible by 7. Now, since there are two 7s in 14!, the numbers 7 and 72 divide
14! evenly. 73 and further powers of 7 leave a remainder when divided into 14!. Hence, the maximum value of m is 2. The answer is (B).
34.The sum of the multiples of 6 between 0 and 100 equals 6 + 12 + 18 + … + 96 = 6(1) + 6(2) + 6(3) +
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The sum of multiples of 8 between 0 and 100 equals 8 + 16 + 24 + … + 96 = 8(1) + 8(2) + 8(3) + … +
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Since 3 16 17 (= 48 17) is clearly greater than 4 12 13 ( = 48 13), Column A is greater than Column B and the answer is (A).
35. A palindrome number reads the same forward or backward. There is no palindrome number between 233 through 239 since none of the numbers read the same both forward and backward. Hence, the palindrome number immediately after 233 is the same as the palindrome number immediate after 239. Hence, Column A and Column B refer to the same number, and the answer is (C).
52 GRE Math Bible
Very Hard
36. We have that the number 2ab5 is divisible by 25. Any number divisible by 25 ends with the last two digits 00, 25, 50, or 75. So, b5 should equal 25 or 75. Hence, b = 2 or 7. Since a is now free to take any digit from 0 through 9, ab can have multiple values.
We also have that ab is divisible by 13. The multiples of 13 are 13, 26, 39, 52, 65, 78, and 91. Among these, the only number ending with 2 or 7 is 52. Hence, ab = 52. The answer is (C).
37. We are given that the numbers m and n, when divided by 6, leave remainders of 2 and 3, respectively. Hence, we can represent the numbers m and n as 6p + 2 and 6q + 3, respectively, where p and q are suitable integers.
Now, m – n = (6p + 2) – (6q + 3) = 6p – 6q – 1 = 6(p – q) – 1. A remainder must be positive, so let’s add 6 to this expression and compensate by subtracting 6:
6(p – q) – 1 =
6(p – q) – 6 + 6 – 1 = 6(p – q) – 6 + 5 = 6(p – q – 1) + 5
Thus, the remainder is 5, and the answer is (E).
38. Since the remainder when m + n is divided by 12 is 8, m + n = 12p + 8; and since the remainder when m – n is divided by 12 is 6,m – n = 12q + 6. Here, p and q are integers. Adding the two equations yields 2m = 12p + 12q + 14. Solving for m yields m = 6p + 6q + 7 = 6(p + q + 1) + 1 = 6r + 1, where r is a positive integer equaling p + q + 1. Now, let’s subtract the equations m + n = 12p + 8 and m – n = 12q + 6. This yields 2n = (12p + 8) – (12q + 6) = 12(p – q) + 2. Solving for n yields n = 6(p – q) + 1 = 6t + 1, where t is an integer equaling p – q.
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by factoring out 6 |
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39. 72 82 = (7 8)2 = 562. |
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The number immediately before 56 that is divisible by 6 is 54. Now, writing 562 as (54 + 2)2, we have
562 = (54 + 2)2 |
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here, the remainder is 4 |
Since the remainder is 4, the answer is (D).
Number Theory |
53 |
40. Choice (A): a + b + c: Suppose a is an even number. Then b, the integer following a, must be odd, and c, the integer following b, must be even. Hence, a + b + c = sum of two even numbers (a and c) and an odd number (b). Since the sum of any number of even numbers with an odd number is odd (For example, if a = 4, then b = 5, c = 6, and a + b + c equals 4 + 5 + 6 = 15 (odd)), a + b + c is odd. Reject.
Choice (B): ab + c: At least one of every two consecutive positive integers a and b must be even. Hence, the product ab is an even number. Now, if c is odd (which happens when a is odd), ab + c must be odd. For example, if a = 3, b = 4, and c = 5, then ab + c must equal 12 + 5 = 17, an odd number. Reject.
Choice (C): ab + d: We know that ab being the product of two consecutive numbers must be even. Hence, if d happens to be an odd number (it happens when b is odd), then the sum ab + d is also odd. For example, if a = 4, then b = 5, c = 6, and d = 7, then ab + d = 3 5 + 7 = 15 + 7 = 23, an odd number. Reject.
Choice (D): ac + d: Suppose a is odd. Then c must also be odd, being a number 2 more than a. Hence, ac is the product of two odd numbers and must therefore be odd. Now, d is the integer following c and must be even. Hence, ac + d = odd + even = odd. For example, if a = 3, then b = 3 + 1 = 4, c = 4 + 1 = 5 (odd) and d = 5 + 1 = 6 (even) and ac + d = 3 5 + 6 = 21, an odd number. Reject.
Choice (E): ac + e: Suppose a is an odd number. Then both c and e must also be odd. Now, ac is product of two odd numbers and therefore must be odd. Summing this with another odd number e yields an even number. For example, if a = 1, then c must equal 3, and e must equal 5 and ac + e must equal 1 3 + 5 = 8, an even number. Now, suppose a is an even number. Then both c and e must also be even. Hence,
ac + e =
(product of two even numbers) + (an even number) =
(even number) + (even number) =
an even number
For example, if a = 2, then c must equal 4, and e must equal 6 and the expression ac + e equals 14, an even number. Hence, in any case, ac + e is even. Correct.
The answer is (E).
54 GRE Math Bible
41. A number less than 500 will be 1) a single-digit number, or 2) a double-digit number, or a 3) tripledigit number with left-most digit less than 5.
Let the compartments shown below represent the single, double and three digit numbers.
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The compartment for the single-digit number can be filled in 4 ways (with any one of the numbers 1, 2, 3, and 5).
Each of the two compartments for the double-digit number can be filled in 4 ways (with any one of the 4 numbers 1, 2, 3, and 5) each. Hence, the two-digit number can be made in 4 × 4 = 16 ways.
Regarding the three-digit number, the left most compartment can be filled in 3 ways (with any one of the numbers 1, 2, and 3). Each of the remaining two compartments can be filled in 4 ways (with any one of the numbers 1, 2, 3, and 5) each. Hence, total number of ways of forming the three-digit number equals 3 × 4 × 4 = 48.
Hence, the total number of ways of forming the number is 4 + 16 + 48 = 68. The answer is (D).
42. We know that 37 = 3 33 33 = 3 27 27 = 3(272). The number immediately before 27 that is divisible by 8 is 24. Hence, replace 27 with 24 + 3. Then we have
37 = 3(272) = 3(24 + 3)2 = 3(242 + 2 24 3 + 32) = 3 242 + 3 2 24 3 + 3 9
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Number Theory |
55 |
43. We are given that the numbers m and n, when divided by 6, leave remainders of 2 and 3, respectively. Hence, we can represent the numbers m and n as 6p + 2 and 6q + 3, respectively, where p and q are suitable integers.
Now, m + n = (6p + 2) + (6q + 3) = 6p + 6q + 5 = 6(p + q) + 5; the remainder is 5. Hence, Column A equals 5.
Now, m – n = (6p + 2) – (6q + 3) = 6p – 6q – 1 = 6(p – q) – 1. A remainder must be positive, so let’s add 6 to this expression and compensate by subtracting 6:
6(p – q) – 1 =
6(p – q) – 6 + 6 – 1 =
6(p – q) – 6 + 5 =
6(p – q – 1) + 5
Thus, the remainder is 5, and Column B equals 5.
Since both columns equal 5, the answer is (C).
Quantitative Comparisons
Quantitative comparisons make up one-half of the math portion of the GRE. This is good news because they are the easiest problems to improve on.
Generally, quantitative comparison questions require much less calculating than do multiple-choice questions. But they are trickier.
Substitution is very effective with quantitative comparison problems. But you must plug in all five major types of numbers: positives, negatives, fractions, 0, and 1. Test 0, 1, 2, –2, and 1/2, in that order.
GENERAL PRINCIPLES FOR SOLVING QUANTITATIVE COMPARISONS
The following principles can greatly simplify quantitative comparison problems.
You Can Add or Subtract the Same Term (Number) from Both Sides of a Quantitative Comparison Problem.
Strategy
You Can Multiply or Divide Both Sides of a Quantitative Comparison Problem by the Same Positive Term (Number). (Caution: This cannot be done if the term can ever be negative or zero.)
Strategy
You can think of a quantitative comparison problem as an inequality/equation. Your job is to determine whether the correct symbol with which to compare the columns is <, =, >, or that it cannot be determined. Therefore, all the rules that apply to solving inequalities apply to quantitative comparisons. That is, you can always add or subtract the same term to both columns of the problem. If the term is always positive, then you can multiply or divide both columns by it. (The term cannot be negative because it would then invert the inequality. And, of course, it cannot be zero if you are dividing.)
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Quantitative Comparisons 57 |
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Example:
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In this form, it is clear that Column B is larger. The answer is (B).
Example:
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58GRE Math Bible
We are also given that 0 < x < 1. So, we may multiply both columns by x to get
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You Must Be Certain That the Quantity You Are Multiplying or Dividing by Can Never Be Zero or Negative. (There are no restrictions on adding or subtracting.)
Watch out!
The following example illustrates the false results that can occur if you don’t guarantee that the number you are multiplying or dividing by is positive.
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Don’t Cancel Willy-Nilly.
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You cannot cancel the x2s from both columns because they do not have the same sign. In Column A, x2 is positive. Whereas in Column B, it is negative.
You Can Square Both Sides of a Quantitative Comparison Problem to Eliminate Square Roots.
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SUBSTITUTION (Special Cases)
We already studied this method in the section Substitution. Here, we will practice more and learn a couple of special cases.
A. In a problem with two variables, say, x and y, you must check the case in which x = y. (This often gives a double case.)
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23 + 23 = 8. In this case, the columns are unequal. This is a double case and therefore the answer is (D).
2
60GRE Math Bible
Example:
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If x ≠ y, then Column A is larger than Column B. x = y = 1, then the columns are equal: 2x + y = 21+1 enough information to decide.
(Plug in a few numbers until you are convinced.) But if = 22 = 4 and 2x + 2y = 21 + 21 = 4. Hence, there is not
B. When you are given x < 0, you must plug in negative whole numbers, negative fractions, and –1. (Choose the numbers –1, –2, and –1/2, in that order.)
Strategy
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decide and the answer is (D).
C. Sometimes you have to plug in the first three numbers (but never more than three) from a class of numbers.
Strategy
Example:
x is both an integer and greater than 1. Let x stand for the smallest positive integer factor of x not equal to 1.
Column A |
Column B |
x |
x3 |
Choose x = 2, 3, and 4. If x = 2, then x = 2 and x3 = 8 = 2. So, for this choice of x, the two columns are
equal. If x = 3, then x = 3 and x3 = 27 = 3, again the columns are equal. Finally, If x = 4, then x = 2
and x3 = 64 = 2, still again the columns are equal. Hence, the answer is (C). Note, there is no need to check x = 5. The writers of the GRE do not change the results after the third number.