GMAT · Arithmetic and number properties · TOPIC LESSON
Original GMAT Problem Solving instruction: reason from constraints, calculate efficiently without a calculator, and check the requested quantity.
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01 / UNDERSTAND & APPLY
An integer is a whole number, positive, negative or zero. Even means divisible by 2, so zero and negative multiples of 2 are even. Odd integers have the form 2k+1 for an integer k. A negative value is not automatically odd, and a positive value is not automatically even.
Sums are even when the two summands have the same parity. A product is odd only when every factor is odd. For signs, count negative factors: an even count gives a positive product, an odd count a negative product, unless a zero factor makes the product zero. These rules establish certainty across all permitted integers, whereas one convenient example only suggests a pattern.
ORIGINAL WORKED EXAMPLE
If n is an odd integer, what is the parity of n(n+3), and can its sign be determined?
Answer Always even; its sign is not determined.
ORIGINAL PRACTICE · 1
Yes; neither must be positive.
An odd product forces both factors to be odd. The sum of two odd integers is even. Taking a=−3 and b=−5 shows that neither integer needs to be positive. Confusing sign with parity is the trap.
02 / UNDERSTAND & APPLY
A nonzero integer d divides n when n=dk for an integer k. The positive factors of a positive integer form a finite list; its positive multiples continue indefinitely. Pair factors around the square root to avoid omissions. A perfect square has one unpaired square-root factor and hence an odd number of positive factors.
For nonzero d, d divides 0. Zero itself must not be used as a divisor. When a question asks for a number of factors, establish whether it means positive factors; counting negative partners would double that number for a nonzero integer.
ORIGINAL WORKED EXAMPLE
List the positive factors of 36 and count them.
Answer 1, 2, 3, 4, 6, 9, 12, 18, 36: nine factors.
ORIGINAL PRACTICE · 1
24.
Multiples of 12 are 12, 24, 36, …; 24 is also divisible by 8. The sum 20 need not be a common multiple. The product 96 is a common multiple but is not the least.
03 / UNDERSTAND & APPLY
A prime has exactly two positive divisors, 1 and itself. The number 1 is neither prime nor composite; 2 is the only even prime. A positive integer greater than 1 has a unique prime factorization, apart from factor order. Repeated factors are tracked with exponents.
If N=p^a q^b for distinct primes p and q, its positive divisors are formed by independently selecting exponent 0 through a and 0 through b. Thus it has (a+1)(b+1) divisors. A square has only even prime exponents; a cube has exponents divisible by 3.
ORIGINAL WORKED EXAMPLE
What is the least positive integer k such that 72k is a perfect square?
Answer 2.
ORIGINAL PRACTICE · 1
12.
150=2×3×5². Choose an exponent for 2 in 2 ways, for 3 in 2 ways and for 5 in 3 ways: 2×2×3=12. Adding exponents counts neither divisors nor prime factors correctly.
04 / UNDERSTAND & APPLY
An integer is divisible by 2 when its last digit is even, by 5 when its last digit is 0 or 5, and by 10 when its last digit is 0. Divisibility by 4 depends on the final two digits; divisibility by 8 depends on the final three. Earlier place values are already multiples of those divisors.
Because 10 leaves remainder 1 on division by 3 or 9, a number and its digit sum have the same remainder modulo 3 or 9. Divisibility by 6 requires both 2 and 3. Do not combine overlapping tests indiscriminately: divisibility by 4 and 6 guarantees divisibility by 12, not necessarily 24.
ORIGINAL WORKED EXAMPLE
Find the digit d for which the four-digit number 5d24 is divisible by 9.
Answer 7.
ORIGINAL PRACTICE · 1
Yes.
The final two digits 28 are divisible by 4. The digit sum 15 is divisible by 3. Since 3 and 4 share no prime factor, both tests imply divisibility by 12; indeed 1,428=12×119.
05 / UNDERSTAND & APPLY
The greatest common factor takes each shared prime to the smaller exponent; the least common multiple takes every prime that occurs to the larger exponent. This follows from the requirement that a factor fit inside both factorizations and a multiple contain both.
For positive integers a and b, GCF(a,b)×LCM(a,b)=ab. Use GCF to find the greatest possible number of identical packs when every category total must be divided evenly among the packs; use LCM for the earliest repeated coincidence. Read what the answer measures before selecting a formula.
ORIGINAL WORKED EXAMPLE
Two alarms sound every 18 and 24 minutes. If they sound together now, after how many minutes will they next sound together?
Answer 72 minutes.
ORIGINAL PRACTICE · 1
42 packs; each has 2 blue and 3 green tokens.
The pack count must divide both totals. GCF(84,126)=42, found by 126−84=42 and 84=2×42. Dividing totals by 42 gives contents. Using the LCM would exceed the available tokens.
06 / UNDERSTAND & APPLY
For a positive divisor d, write n=dq+r with integer q and 0≤r<d. This representation also handles negative integers: −7=5(−2)+3, so its remainder on division by 5 is 3. A negative leftover must be adjusted into the allowed range.
For sums and products, replace each integer by its remainder, calculate, then reduce again. Powers often repeat. Find a complete cycle and use the exponent position within it; a zero remainder when dividing the exponent selects the last term of the cycle, not the first.
ORIGINAL WORKED EXAMPLE
An integer n leaves remainder 4 when divided by 7. What remainder does 3n+5 leave?
Answer 3.
ORIGINAL PRACTICE · 1
2.
Powers of 2 have remainders 2, 4, 1 and then repeat. Since 10=3×3+1, the exponent occupies the first cycle position. The remainder is 2, not the exponent remainder 1.
07 / UNDERSTAND & APPLY
Consecutive integers can be written n,n+1,n+2. Consecutive even or consecutive odd integers differ by 2, so write n,n+2,n+4 with the relevant parity restriction. Confusing these models changes the sum and product.
In an odd-length block of consecutive integers, the mean equals the central integer because terms pair symmetrically. Among k consecutive integers, exactly one is a multiple of k. A product of consecutive integers therefore carries useful divisibility information, but individual terms need not have it.
ORIGINAL WORKED EXAMPLE
Five consecutive integers have sum 85. What is the product of the least and greatest?
Answer 285.
ORIGINAL PRACTICE · 1
24.
The middle integer is 66/3=22, so the list is 20,22,24. Answering 23 mistakenly uses consecutive integers with spacing 1.
08 / UNDERSTAND & APPLY
A units-digit question is a remainder question with divisor 10. Last digits of products depend only on last digits of factors, so large exact powers are unnecessary. Bases ending in 0, 1, 5 or 6 keep their units digit for every positive exponent; 4 and 9 have two-term cycles; 2, 3, 7 and 8 have four-term cycles.
The exponent must be positive for these quoted cycles. An exponent of zero gives 1 for every nonzero base, including bases whose positive-power cycle is constant. Parentheses matter for negative bases: (−3)² is positive, but −3² means −(3²).
ORIGINAL WORKED EXAMPLE
What is the units digit of 7²³+3¹⁴?
Answer 2.
ORIGINAL PRACTICE · 1
6.
The 8 cycle is 8,4,2,6; exponent 20 selects its fourth term, 6. Every positive power of 6 ends in 6. The product ends as 6×6=36 does: 6. Selecting cycle position zero as the first term produces a wrong result.
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