ORIGINAL WORKED EXAMPLE
Determine what must be odd
Integers a and b satisfy ab odd. Is a+b+1 odd or even?
- An odd product requires both a and b to be odd.
- Their sum is even; adding 1 makes it odd.
- No sign assumption is needed.
Answer Odd
GRE · Arithmetic · TOPIC LESSON
Integers, divisibility, primes and remainders
Use structure, not large calculations, to reason about divisibility and possible integers.
BY THE END, YOU CAN
- Translate divisibility into prime exponents and remainder equations.
- Distinguish necessary conclusions from possible examples.
- Solve factor, multiple and parity problems without assuming positivity.
BEFORE YOU START
- Integer addition and multiplication
- Meaning of quotient and remainder
Original educational lesson. Work through the examples before revealing the practice answers. This original lesson supports current GRE General Test study; it does not reproduce official questions, certify a writing score or predict test content.
LEARN THROUGH ORIGINAL PRACTICE
Original GRE-style worked examples and practice
All teaching passages, writing prompts, examples and exercises in this lesson are original. No official ETS questions, prompts or explanations are reproduced. Current structure and observed public sample metadata are separate from this original teaching; no official score or question probability is implied.
01 / UNDERSTAND & APPLY
Start with the allowed numbers
Integers are …, −2, −1, 0, 1, 2, … . Positive integers exclude zero; nonnegative integers include it. A variable described only as a number is not automatically an integer. Parity and divisibility arguments require integer inputs. Zero is even because it equals 2 times an integer, but zero is neither positive nor negative.
An even integer has form 2k; an odd integer has form 2k+1. Negative integers fit these forms too. Consecutive integers can be written n, n+1, n+2; consecutive even integers are 2k, 2k+2, 2k+4. The variable need not be positive unless the question says so.
02 / UNDERSTAND & APPLY
Parity is a structural shortcut
Even plus even and odd plus odd are even; even plus odd is odd. A product is odd exactly when every factor is odd. One even factor makes the product even. Squaring preserves parity. These facts follow by substituting 2k and 2k+1, and apply equally to subtraction.
The converse matters: an even product does not tell you which factor is even. Likewise, an even sum may come from two odd or two even numbers. Testing a few positive examples does not prove a conclusion about all integers.
03 / UNDERSTAND & APPLY
Factors and divisibility tests
For nonzero integer d, d divides n when n=dk for an integer k. A factor divides a number; a multiple is obtained by multiplying it by an integer. Every nonzero integer divides zero, but division by zero is undefined. When a problem asks for the number of factors, it usually specifies positive factors; read that restriction.
Base-ten shortcuts: divisibility by 2 uses the final digit; by 3 or 9 uses the digit sum; by 5 uses a final 0 or 5; by 4 uses the final two digits; by 8 uses the final three. Divisibility by 6 needs both 2 and 3. Divisibility by 12 can be tested with 3 and 4, because these are relatively prime. Testing 2 and 6 would only guarantee a multiple of 6, not 12.
04 / UNDERSTAND & APPLY
Prime powers record exactly what is available
A prime is an integer greater than 1 with precisely two positive factors. One is not prime, and 2 is the only even prime. Factor a positive integer into prime powers. To test whether d divides n, compare every prime exponent: n must contain at least as many copies as d requires.
For n=p^a q^b with distinct primes, its positive factors number (a+1)(b+1): each factor independently chooses an exponent from 0 through a and 0 through b. A perfect square has even exponents for every prime; an odd number of positive divisors occurs exactly for a perfect square.
ORIGINAL WORKED EXAMPLE
Count factors and complete a square
Find the number of positive factors of 360 and the least positive integer m for which 360m is a square.
- 360=2³·3²·5. The factor count is (3+1)(2+1)(1+1)=24.
- The exponents of 2 and 5 are odd; multiply by one additional 2 and one additional 5.
- m=10 gives 3600=60². No smaller positive multiplier can supply both required primes.
Answer 24 factors; m=10
05 / UNDERSTAND & APPLY
Common factors use minima; common multiples use maxima
The greatest common divisor (GCD) takes the minimum exponent of each prime in two positive integers. The least common multiple (LCM) takes the maximum. Do not multiply the original numbers unless they are relatively prime. For positive a and b, GCD(a,b)·LCM(a,b)=ab.
Use GCD to split quantities into equal largest-size groups with no remainder; use LCM to find the next simultaneous occurrence of repeating schedules. A schedule question also needs a common starting point.
ORIGINAL WORKED EXAMPLE
Two repeating timers
Timers sound together at noon, then every 18 and 30 minutes. When do they next sound together?
- 18=2·3² and 30=2·3·5.
- The LCM is 2·3²·5=90 minutes.
- Add 90 minutes to noon; the next shared sounding is at 1:30 p.m.
Answer 1:30 p.m.
06 / UNDERSTAND & APPLY
Use the division equation, including for negative numbers
For positive integer divisor d, write n=dq+r with integers q,r and 0≤r<d. The remainder cannot be d or negative. Knowing r restricts n to a family, not to one value. Arithmetic with remainders works because adding or multiplying multiples of d produces another multiple of d.
If a leaves remainder r and b leaves remainder s on division by d, reduce r+s or rs again modulo d. Do not leave a remainder outside 0 through d−1. For powers, look for a repeating remainder cycle and align the exponent with the correct cycle position.
ORIGINAL WORKED EXAMPLE
Negative dividend
What remainder does −23 leave on division by 6?
- Write −23=6(−4)+1.
- The remainder 1 meets 0≤r<6.
- Using quotient −3 would give −5, which is not an allowed remainder.
Answer 1
ORIGINAL WORKED EXAMPLE
A power cycle
Find the remainder when 3²⁷ is divided by 7.
- Successive remainders for powers 1 through 6 are 3,2,6,4,5,1.
- The six-term cycle repeats since multiplication by 3 starts it again.
- 27=6·4+3, so use the third entry, 6.
Answer 6
07 / UNDERSTAND & APPLY
Choose a method and answer the requested question
All examples and practice in this lesson are original. For quantitative comparison (QC), use A if Quantity A is greater, B if Quantity B is greater, C if they are equal, and D if the relationship cannot be determined from the information given. A, B or C must hold for every permitted case. To establish D, give two allowed cases producing different relationships; an inability to calculate is not itself evidence for D.
A single-answer question requires one choice. For multiple-answer questions, obey the stated selection count or select every correct option if instructed. Test each option separately; do not stop after finding one. Multiple-answer questions receive credit only when the complete correct set is selected, with no incorrect choices. Numeric entry requires the requested value, with units handled in the calculation; round only when the question directs you to. A single numeric-entry box accepts an integer or decimal; a pair of fraction boxes accepts an integer numerator and a nonzero integer denominator. Fractions need not be reduced, but reducing makes checks easier. Do not enter a percent sign when the requested number is a percentage. Preserve exact fractions until the last step.
A calculator can check arithmetic but cannot decide the domain, interpretation or denominator for you. Estimate sign and size first, enter parentheses carefully, and compare the result with the original conditions. Cancel factors or exploit structure when doing so is faster and more reliable than decimal entry.
08 / UNDERSTAND & APPLY
Original practice: prove, calculate and reject traps
Use the stated number domain in each question. For QC use the four choices defined above.
ORIGINAL PRACTICE · 1
Single answer: Which must be even for every integer n? A n²+1; B n(n+1); C 2n+1; D n+3; E n²−1.
Reveal answer & reasoning
B
Consecutive integers include an even factor, so n(n+1) is even. A and E are odd when n is even; C is always odd; D is odd when n is even.
ORIGINAL PRACTICE · 2
Numeric entry: What is the least positive integer divisible by 14 and 35?
Reveal answer & reasoning
70
14=2·7 and 35=5·7. Take each highest prime power: 2·5·7=70. The product 490 unnecessarily counts 7 twice.
ORIGINAL PRACTICE · 3
Select all that apply: Which divide 2⁴·3²·7? A 16; B 27; C 42; D 63; E 49.
Reveal answer & reasoning
A, C and D
16 needs four 2s; 42 needs one each of 2,3,7; 63 needs 3²·7. All are present. 27 requires three 3s and 49 two 7s, more than available.
ORIGINAL PRACTICE · 4
QC: n is an integer and n² is divisible by 2. Quantity A: remainder of n on division by 2. Quantity B: 0.
Reveal answer & reasoning
C
An odd integer has an odd square, so n must be even. Its remainder is 0. This works for negative n and zero as well.
ORIGINAL PRACTICE · 5
QC: positive integer n leaves remainder 2 on division by 5. Quantity A: n. Quantity B: 10.
Reveal answer & reasoning
D
n=2 gives A<B; n=12 gives A>B. The remainder condition does not fix the quotient, so neither universal inequality is justified.
ORIGINAL PRACTICE · 6
Numeric entry: How many positive divisors does 72 have?
Reveal answer & reasoning
12
72=2³·3², giving (3+1)(2+1)=12. Counting prime factors or adding exponents does not count the independent exponent choices.
ORIGINAL PRACTICE · 7
Single answer: An integer leaves remainder 4 when divided by 9. What remainder does twice that integer plus 5 leave? A 0; B 1; C 4; D 5; E 8.
Reveal answer & reasoning
C: 4
Write n=9q+4. Then 2n+5=18q+13=9(2q+1)+4. Reporting 13 violates the remainder range.
ORIGINAL PRACTICE · 8
Numeric entry: The GCD of two positive integers is 6 and their LCM is 180. One integer is 30. Find the other.
Reveal answer & reasoning
36
Their product is 6·180=1080, so the other integer is 1080/30=36. Check GCD(30,36)=6 and LCM(30,36)=180; product identity plus the check confirms the result.
KEEP THE BIG IDEAS
Quick revision
- Write the domain before applying parity or factor rules.
- Use prime-exponent minima for GCD and maxima for LCM.
- Remainders obey 0≤r<d, even for negative dividends.
- Two permitted cases with different comparisons establish QC choice D.
READ WITH CONTEXT
References & scope
The teaching notes, passages, worked explanations and practice are original. Official references identify the assessment framework and skill scope; they do not endorse this lesson or license official question reproduction.
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