Numerical Relations and Number Series

Linear inequality solution set

A linear inequality in one variable has the form ax+b \\le c (or with <, >, \\ge) where a \\ne 0. It is solved by isolating x with the same operations used for equations, with one essential exception concerning the sign. The solution set is the collection of all real numbers that make the statement true. For a single linear inequality, the solution set is always a half-line, an interval unbounded in one direction, whose endpoint is the value at which the corresponding equation holds with equality.

Several rules keep an inequality valid. Adding or subtracting the same real number on both sides preserves the direction of the inequality. Multiplying or dividing both sides by a positive number also preserves it, whereas multiplying or dividing by a negative number reverses it. When a fraction appears, multiply both sides by the positive common denominator first, so that no sign reversal is needed and the coefficients become integers. For example, from \\frac{x}{2}-1 \\ge 3 we get x-2 \\ge 6 and hence x \\ge 8. Forgetting to flip the sign after multiplying by a negative coefficient turns a correct half-line into the opposite half-line.

On the number line, the endpoint of the solution is drawn as a filled dot when equality is included, as in \\le or \\ge, and as an open circle when the inequality is strict, as in < or >. The bold ray runs toward the side where the solution lies: x \\ge c extends to the right toward larger values, and x \\le c extends to the left. Such diagrams are usually not drawn to scale, so the position of the dot relative to zero must be found by computing the endpoint first. A fraction such as \\frac{2}{7} is a small positive number, so its dot sits just to the right of zero. Students often choose a figure with the correct endpoint type but the wrong ray direction, or a figure whose dot lies on the negative side even though the computed endpoint is positive.

When two conditions must hold at the same time, the solution set is the intersection of the two individual solution sets. Solve each inequality separately and then take the overlap. For instance, x>2 together with x<7 gives the open interval 2<x<7. If the two solution sets do not overlap, no real number satisfies both. Strictness matters at the ends of the overlap: a value that makes one inequality an equality fails if that inequality is strict. A quick check is to test a value inside the proposed interval in both original inequalities, and to test each endpoint. Any option that includes a value violating either condition is wrong, even if it also contains the correct interval.

When two inequalities in two variables point in the same direction, both of the form less than, they can be added term by term, and the sum is again a valid inequality of the same type. Adding 3p-q<4 and 3q-p<12 gives 2p+2q<16, so every pair satisfying both must satisfy p+q<8. Such a sum is a necessary consequence of the system, not a full description of all admissible pairs, so it tells you which bound must hold without showing that every value below it is reached. Inequalities cannot be subtracted term by term in general, and combining inequalities that point in opposite directions by addition gives no valid conclusion.

To find a bound on a combination of variables, isolate one variable from each inequality to obtain a lower bound and an upper bound, and then require that the lower bound be less than the upper bound. For example, q>3p-4 and q<\\frac{p+12}{3} together force 3p-4<\\frac{p+12}{3}, which simplifies to p<3. Substituting the resulting bounds into an expression such as p+q gives a fixed upper limit that holds for every feasible pair. Multiplying one inequality by a positive constant before adding it to another is a systematic way to eliminate a variable.

A strict inequality excludes its boundary, so an option asserting that a sum equals the bound is false, even though the sum can approach the bound arbitrarily closely. A fast check is to substitute a simple point such as the origin into both original inequalities. If it satisfies both, any option requiring the sum to be at least a positive threshold is refuted immediately, so the test eliminates options without full algebra.

When the variables are restricted to positive integers, the solution is a finite set rather than an interval. First find bounds for each variable from the inequalities, then test the few integer candidates that remain. Since a positive integer is at least 1, a sum constraint with a small upper bound can force both variables to equal 1. Every condition must be checked with the candidate pair, because a pair that satisfies one inequality may still fail another.

Word problems often give a total that must lie between a lower and an upper limit, together with fixed amounts or minimum amounts for each category. To find the maximum possible value of one quantity, write every condition as an equation or inequality, set the other quantities to their most favorable extreme values, and keep the total within its limits. If the other categories have fixed amounts F and minimum amounts M, and the total is bounded above by T, then the largest possible value of the remaining quantity is T-F-M, provided the lower limit on the total is still met. Fixed amounts are subtracted directly, and minimum amounts are subtracted at their least allowed value.

Clock hand coincidence count

Clock hand coincidence count asks how many times two hands of an analogue clock point in the same direction during a given interval. Two hands coincide when their angles, measured from the 12 o'clock mark, are equal modulo 360^\circ. Every such problem follows three steps: find the relative angular speed, list the coincidence times as an arithmetic progression, and count the terms that lie inside the window.

The second hand turns 6^\circ per second, the minute hand turns 0.1^\circ per second, and the hour hand turns 0.5^\circ per minute. The second hand therefore gains 354^\circ per minute, or 5.9^\circ per second, on the minute hand, and the minute hand gains 5.5^\circ per minute on the hour hand. Since a coincidence requires a relative gain of exactly 360^\circ, successive second-minute coincidences are \frac{360}{354}=\frac{60}{59} minutes apart, about 61.02 seconds. Starting from an alignment, the second hand overtakes the minute hand 59 times in the open interval of one hour.

When the window does not begin at an alignment, the progression is shifted. Let \theta_0 be the minute-hand angle minus the second-hand angle at the start, in degrees. Coincidences then occur at t=\frac{\theta_0+360k}{5.9} seconds for integers k, measured from the start. Beginning at a quarter past the hour, \theta_0=90^\circ and the first meeting comes only about 15 seconds in. A 30-minute window then holds 29 or 30 meetings depending on the phase, so the count must be computed from the exact terms rather than estimated from the average of 29.5.

For a window that starts at an alignment, meeting times are t_k=\frac{3600k}{59} seconds with k=0,1,2,\dots. The count is the number of integers k with t_{\text{start}}\le t_k\le t_{\text{end}}. Divide both endpoints by \frac{3600}{59}, round the lower bound up and the upper bound down, and subtract. For example, from 12:10:00 to 12:40:00 the bounds are 600\cdot\frac{59}{3600}\approx 9.83 and 2400\cdot\frac{59}{3600}\approx 39.33, so k=10,\dots,39 and there are 30 meetings. A 50-minute window has length 50\cdot\frac{59}{60}\approx 49.2 in units of meeting intervals, so it contains 49 or 50 meetings, and only the exact range of k decides which.

The hour and minute hands use the same method with their own relative gain. Starting from an alignment at 12:00, they meet at t=\frac{720k}{11} minutes, so successive meetings are \frac{720}{11}=65\frac{5}{11} minutes apart, giving 11 meetings in 12 hours and 22 in a day. Because a one-hour window is shorter than this interval, it contains at most one meeting; between 6 and 7 o'clock the hands meet once, at 6{:}32\frac{8}{11}. When someone sees the hands exactly aligned at two moments that are consecutive meetings, the elapsed time is precisely 65\frac{5}{11} minutes, not 60 minutes and not a rounded 65 minutes.

• cross each other — for two clock hands, to point in the same direction at one instant; each such instant counts once as a coincidence.
• opposite — hands pointing in exactly opposite directions; this is a different event and is not a coincidence.
• between A and B — the endpoints must be checked against the meeting formula rather than assumed to be meeting times or excluded by default.

Custom binary operator evaluation

A custom binary operator is a symbol such as \oplus, \odot, \Delta or \nabla that the question defines to combine two operands by a rule of its own. Whenever the symbol appears, the rule replaces it, so the work is mechanical once the substitution is done correctly. For an operator \ast defined by a \ast b = f(a,b), the value p \ast q means f(p,q), with p in the first slot and q in the second. The order of the operands therefore matters, and in general p \ast q \neq q \ast p, because rules such as subtraction and division are not symmetric.

Evaluation proceeds in three steps. First, replace every symbol by its definition, or by the ordinary arithmetic sign it stands for when the statement gives a word-to-symbol mapping. Second, evaluate the innermost brackets first, and keep each intermediate value with its correct sign. Third, work outward, applying operations at the same level from left to right unless the statement or the usual precedence rules say otherwise. Each bracket is a separate sub-evaluation whose result feeds the next outer step, so nested chains of three or more operands must be reduced from the inside out.

Operators are generally not associative, so the grouping changes the result. Using a \star b = a + 2b and a \circ b = ab - 1, we get (3 \star 4) \circ 2 = 11 \circ 2 = 21, while 3 \star (4 \circ 2) = 3 \star 7 = 17. A single misplaced bracket therefore changes the answer, so copy the grouping exactly as written. A subtraction can produce a negative intermediate value, and losing that minus sign is a common slip that leads to a distractor differing only in sign.

Some items give an equation with operators on both sides and ask for an unknown. Substitute both definitions, clear the denominators, and simplify. If the result is a perfect square equal to zero, the two variables must be equal. For example, if a \diamond b = \frac{a}{b} + \frac{b}{a} and a \diamond b = 2, then a^2 + b^2 = 2ab, so (a-b)^2 = 0 and hence a = b.

When the symbols themselves are unknown, the question supplies several relations among them. Treat each unknown symbol as a variable, write each relation as an equation, and solve the resulting system. Then substitute the solved values into the requested expression. If \alpha - \beta = 3 and \alpha + \beta = 11, then \alpha = 7 and \beta = 4, so (\alpha - \beta)^2 = 9.

Operators defined through ratios often reward squaring. Since \left(\frac{p}{q}+\frac{q}{p}\right)^2 = \frac{p^2}{q^2} + \frac{q^2}{p^2} + 2, a known value s of \frac{p}{q}+\frac{q}{p} gives \frac{p^2}{q^2}+\frac{q^2}{p^2} = s^2 - 2. Recognising this cross term is the key step once an operator has been unpacked into such a symmetric expression.

Some questions define symbols with words rather than formulas. A statement such as 'the symbol means subtraction' assigns each symbol an ordinary operation, and the expression must be rewritten with those signs before any calculation. For example, if \heartsuit means subtraction and \spadesuit means division, then 10 \heartsuit 4 \spadesuit 2 = 10 - (4 \div 2) = 8. A symbol placed before the first operand can act as a unary sign on that operand rather than as a binary operation, so read its position carefully. After rewriting, apply the usual order: brackets first, then multiplication and division, then addition and subtraction, each left to right at its own level.

Many choice lists include values produced by stopping early, such as the result of the first operation alone, or by applying one operation in the wrong order. Before selecting an answer, recompute the entire chain from the rewritten expression.

Arithmetic progression term count

An arithmetic progression (AP) is a sequence in which each term differs from the previous one by a fixed common difference d. If the first term is a, the nth term is a_n = a + (n-1)d. If l denotes the last of n terms, then l = a + (n-1)d, and solving for n gives the term-count formula n = \frac{l-a}{d} + 1. The +1 accounts for both endpoints. For example, the terms 4, 9, 14, \ldots, 99 have n = \frac{99-4}{5} + 1 = 20 terms. The common error is to omit the +1, which undercounts by one. Checking a small case helps: the even numbers from 2 to 20 contain \frac{20-2}{2} + 1 = 10 terms.

The sum of the first n terms is S_n = \frac{n}{2}(a + l) = \frac{n}{2}\{2a + (n-1)d\}. When each term is a fixed multiple of its position, the terms form an AP and the sum can be found without listing them. For example, the sum of 3k over k = 1 to 6 has a = 3, d = 3, l = 18, so S_6 = \frac{6}{2}(3 + 18) = 63. Quantities such as the perimeter of a square whose side is k, summed over consecutive values of k, follow this same pattern.

Two sequences of equal length with the same common difference differ by a constant amount term by term. If the first terms differ by c, every pair of corresponding terms differs by c, so the two sums differ by exactly nc. Consecutive odd integers starting at an odd number a and consecutive even integers starting at a+1 form a concrete case. The odd-side sum is n(a+n-1), the even-side sum is n(a+n), and their difference is n. Thus a given difference between the two sums fixes the number of terms directly. To find the count, write the difference as nc = D, solve for n, and verify by substituting back into both sums.

Number series of powers of two minus one

A number series can follow a closed-form rule rather than a simple additive or multiplicative step. One common rule is 2^{n}-1, in which every term is one less than a power of two. For consecutive values of n, the terms are 1, 3, 7, 15, 31, 63, \ldots, so the exponent increases by exactly one at each position.

When a term is unknown, the fastest test is to add one to the candidate. A value v belongs to the family exactly when v+1 is a power of two. For example, 47+1=48 is not a power of two, so 47 cannot belong to the series, whereas 63+1=64=2^{6} does.

To fill a blank, first determine which exponent the blank must carry by counting its position from a known term. In 3, 7, x, 31, for example, the blank follows 2^{3}-1=7 and precedes 2^{5}-1=31, so it must be 2^{4}-1=15. Check the neighbours on both sides before deciding, because a value that fits only one side is not the answer.

Since the exponent steps by one, the pattern is rigid: a candidate that breaks the rule on either side of the blank is ruled out at once. Keeping the index consistent in this way leads to a unique answer.

Calendar weekday reasoning

Calendar problems rest on the fact that the week repeats every seven days, so the weekday of any date is found by counting days modulo 7. Each full group of seven days leaves the weekday unchanged, and only the remainder shifts it forward.

Month lengths govern how often a weekday appears within a month. A common year has 28 days in February, which contains exactly four of every weekday, while a leap year has 29 days. A 31-day month gives five occurrences of its first three weekdays, a 30-day month gives five occurrences of its first two weekdays, and a 29-day February gives five occurrences of only its first weekday.

Therefore, if February contains a fifth occurrence of some weekday, the year must be a leap year and 1 February must fall on that weekday. This fixes the weekday of every other date in February, and the day after 1 February is the 2nd. Any weekday other than the one that starts the month appears only four times.

To count the days between two dates in different months, add the days remaining in the first month to the day number of the second date. For example, from 10 March to 4 April the gap is (31-10)+4=25 days. Since 25 = 3\times 7 + 4, a Friday shifts forward by 4 days and becomes a Tuesday.

A frequent slip is to stop at the same day number in the following month. The gap from a date to the same day number in the next month equals the length of the first month, so the day one later in the next month is one more day beyond that. Thus the count is the month length plus one, and it is this extra day that changes the remainder modulo 7.

Infinite series of reciprocal powers

Infinite series of reciprocal powers can often be split into simpler series whose sums are already known. When the denominators come from two different bases, such as powers of 2 and powers of 3, separate the terms into one family per base. The families share no denominator, because no power of 2 greater than 1 is also a power of 3, so no term is counted twice.

Each family is a geometric series \sum_{k\ge 1} r^{k} with 0<r<1, whose sum is \dfrac{r}{1-r}. Thus \sum_{k\ge 1} 1/2^{k} = 1 and \sum_{k\ge 1} 1/3^{k} = \dfrac{1}{2}. Any constant term written before the series is simply added to these partial sums. For instance, \sum_{k\ge 1} 1/5^{k} = \dfrac{1/5}{4/5} = \dfrac{1}{4}.

A second family uses reciprocals of factorials. The Maclaurin expansion e^{x}=\sum_{n=0}^{\infty} x^{n}/n! at x=1 gives \sum_{n=0}^{\infty} 1/n! = e \approx 2.718. A series that starts at n=0 therefore sums to e.

The starting index changes the result. A series beginning at n=1 omits the term 1/0! = 1, so its sum is e-1 \approx 1.718, and a series beginning at n=2 equals e-2. Adding 1 to e overstates the sum by exactly the omitted term, which is a common error when the index is misread.

Recurrence relation sequence terms

A recurrence relation defines each term of a sequence from the terms before it. It defines the sequence completely only together with its initial terms. The usual two-term form F_{n+1}=F_n+F_{n-1} says that each term is the sum of the two terms immediately before it. Because the relation is a linear equation, it can be read in both directions. When two consecutive terms are known, every earlier term follows from F_{n-1}=F_{n+1}-F_n. This backward computation is the standard method when a question gives later terms and asks for an early one. Verify the result by running the rule forward once.

The Fibonacci series is the standard example, in which each term is the sum of the two preceding terms. In a series such as 1, 1, 2, 3, 5, 8, 13, 21, every position satisfies the sum rule. If one entry in such a series is wrong, find the first position where the sum rule fails and replace that entry with the value the rule requires. The corrected value must also agree with every later term.

Tables that describe an iterative process use the same idea in a different notation. An output column X(i) may satisfy X(i)=X(i-1)+I(i), so each output combines the previous output with the current input by addition. Another column may satisfy Y(i)=Y(i-1)\cdot I(i), which combines them by multiplication. Different columns may use different operations, so each column must be tested separately against every row. A common trap is an index shift. A rule that uses I(i-1) instead of I(i) gives a different value at the first step, so check the rule from i=1 onward before accepting it.

• recurrence relation — an equation that defines a term from earlier terms. Example: with a_1=3 and a_n=2a_{n-1}, the sequence is 3, 6, 12.
• initial condition — the starting value or values that fix the sequence. Without them, the recurrence does not determine a unique sequence.
• iterative table — a table in which each row is computed from the previous row. Example: Z(i)=Z(i-1)-I(i) gives Z(1)=5 when Z(0)=8 and I(1)=3.

Prime number set membership

A prime number is a natural number greater than 1 whose only positive divisors are 1 and itself. The number 1 is not prime, and 2 is the only even prime. An odd-one-out question over a set of numbers is solved by testing each member for primality and finding the single member that fails. Trial division by primes up to the square root of the number is sufficient for this test. Quick checks such as divisibility by 3, which holds when the digit sum is divisible by 3, often reveal a composite number at once. A composite number has at least three positive divisors. For example, 21 equals 3\times 7 and is composite, while the other numbers in a set of two-digit primes have no factors other than 1 and themselves.

For a statement that must hold for all primes, the product of two primes p_1p_2 always has the divisors 1, p_1, p_2 and p_1p_2. If p_1=p_2, the divisors are 1, p and p^2. In either case the product has at least three divisors, so it is always composite.

Sums and expressions with an added 1 are not guaranteed to be prime. Since 2+3=5 is prime, a sum of two primes can be prime. A single counterexample disproves a universal claim, so choose small primes and test each option. For instance, 3+5+1=9 and 3\cdot 5+1=16 are both composite, and 3+2+1=6 is composite.

• prime — a natural number greater than 1 with exactly two positive divisors.
• composite — a natural number greater than 1 that is not prime.
• counterexample — one specific case that makes a universal statement false. Example: 2+2=4 shows that not every sum of two primes is prime.

Custom function evaluation

A custom function, also called a defined operation, is given by its own rule. The question asks for its value at specific arguments, and no standard formula can be assumed. A piecewise definition chooses a formula according to a condition on the arguments. For example, \lambda(p,q)=(p-q)^2 when p\ge q, and \lambda(p,q)=p+q when p<q. Evaluate such a function in three steps. First, simplify each argument completely, including signs and nested operations. Second, decide which branch the simplified arguments fall into. Third, substitute and compute the result. Note that the condition p\ge q includes equality, so equal arguments always use the first branch, and \lambda(1,1)=0.

A symbol may also be defined only through given relations among unknown values. In that case, express every unknown in terms of one of them, substitute into the target expression, and solve. For example, if \alpha=2\beta and \alpha+\beta=9, then 3\beta=9, so \beta=3 and \alpha=6.

An algebraic identity can simplify an equation before it is solved. The difference of squares a^2-b^2=(a-b)(a+b) reduces an expression to a linear one when a-b is constant. For example, x^2-(x-2)^2=2(2x-2)=4x-4. Solve the simplified equation and check the root by substitution.

In an operation defined on letters arranged on a circle, assign each letter an index. In a 12-letter clock, the letters O to Z receive indices 0 to 11 in order. A step rule such as adding 2 or subtracting 5 is applied to the index modulo 12, that is, (k+s)\bmod 12. Before using a step rule on a new input, confirm it on the given example, since a single wrong step changes every later letter.

• piecewise definition — a function whose formula depends on which condition the arguments satisfy.
• argument simplification — evaluating the inner expressions first, as the outer operation can only be applied to numbers.

Letter-number grid pattern

In a letter-number grid, letters and numbers are linked by a fixed code. The most common code is the alphabetical position, with A = 1, B = 2, and so on up to Z = 26. Find the rule from the entries already filled in, check it on every filled pair, and then apply it to the blanks. In a paired-cell arrangement, the number written below a letter equals the alphabetical position of the letter in the horizontally adjacent cell. The relation runs in both directions, so the number under one letter and the letter beside it point to each other. A blank letter can therefore be found from the number beneath it, and a blank number from the letter it sits under.

A different arrangement codes each letter pair by a product. Here the code of a pair is the product of the alphabetical positions of its two letters, for example E=5 and F=6 give EF=30. To decode a given number, factor it into two positions between 1 and 26. A code such as 35 can be split as 5\times 7, which gives the letters E and G. Check that the factorization is unique within the range, and then use the sector opposite each code to decide which letter pair it represents.