Number Theory

Welcome to Number Theory — and let's begin with one of its most practical tools: finding numbers that must be divisible by two different values at the same time. This is where the concept of the Lowest Common Multiple (LCM) earns its keep.

Divisibility and LCM questions feature in virtually every OC MR paper. They are usually placed early in the test and are among the most reliably scorable questions for students who know the LCM shortcut.

The examiner is checking whether you can translate "divisible by both A and B" into the single condition "divisible by LCM(A, B)", and then efficiently count how many such multiples fall within a given range — without listing every single number.

You are given a range (e.g. 1 to 100) and two divisors. Your task is to count how many numbers in that range are divisible by both. The trap is trying to check each number individually — that is too slow and error-prone.

What number is six hundreds less than 74 251? The trap is translating 'six hundreds' correctly — it means 6 × 100 = 600, not 6000 and not 60. Once you have the right amount to subtract, borrowing across the hundreds digit gives the answer in two steps.

Place value subtraction questions using word descriptions ('six hundreds', 'four thousands', 'nine tens') appear in the easy band of OC MR. They are designed to test whether students can translate number words into the correct digit value before calculating. Option E (subtracting six tens = 60) and option A (subtracting four thousands = 4000) are the most common wrong answers.

The examiner is checking whether students can correctly translate a number-word expression ('six hundreds') into a numerical value (600) before subtracting. Both distractor A (70 251 = −4000) and distractor E (74 191 = −60) are designed for students who misread the place value by one or two columns.

A five-digit number is given. The question asks what is 'N [place value units] less than' or 'more than' that number. Students must convert the word expression to a number (e.g. four thousands = 4000, six hundreds = 600, nine tens = 90), then perform the subtraction or addition. Borrowing across columns is often required.

Five numbers are each rounded to the nearest 10. Which one changes the most? The key is understanding that 'changes the most' means the biggest gap between original and rounded — and that the worst-case change when rounding to the nearest 10 is exactly 5. One of these five numbers achieves that maximum.

Rounding comparison questions appear in the easy band of OC MR. The most common mistake is choosing 99 (because 99 'feels' far from 90) or 80 (because it rounds to itself with 0 change). Students who calculate each change systematically never get these wrong.

The examiner is testing whether students know that rounding change = |rounded − original| and can apply this to all five options. The trap options are 99 (only changes by 1, despite being close to 100) and possibly 54 (changes by 4, the second-largest). Only 45, with a units digit of exactly 5, achieves the maximum possible rounding change of 5.

Five numbers are listed. Each is rounded to the nearest 10 (or 100). Students must calculate the absolute difference between each original number and its rounded equivalent, then identify which difference is largest. Numbers are chosen so that the one with a units digit of 5 has the maximum change of 5, while larger-looking numbers (like 99) have much smaller changes.

Jo has three digit cards: 4, 2, and 9. She arranges all three to make different three-digit numbers. The question asks for the gap between the biggest number she can make and the smallest. The trap: option C (675) appears if you accidentally use 924 as the largest instead of 942.

Digit arrangement and difference questions appear regularly at the easy end of OC MR. They test two skills in sequence: ordering digits to form extreme values, then performing a subtraction. Students who rush often mis-order the hundreds digit and calculate 924 − 249 = 675 instead of 942 − 249 = 693.

The examiner is testing whether students can systematically arrange digits to produce the largest and smallest possible numbers, then accurately subtract them. Option C (675) specifically traps students who write 924 instead of 942 for the largest number — a digit transposition error.

A set of three (or four) distinct digit cards is given. Students must form the largest number (digits descending) and the smallest number (digits ascending), then find their difference. Occasionally a zero digit is included to add a layer of difficulty (leading zero rule).

Replace the digit 8 with 9 in each of five numbers — which increases the most? The answer is whichever number has 8 in the highest place value position. In 81 234, the 8 is in the ten-thousands place, so replacing it adds 10,000. Students who read the numbers carefully rather than choosing the largest-looking option get this right every time.

Place value questions where a digit replacement determines the magnitude of change appear regularly at the easy level of OC MR. They test whether students can identify the position of a digit within a large number and understand that the increase = (new digit − old digit) × place value.

The examiner is testing pure place value understanding. Option B (48 001) is a trap for students who see '48' as a large part of the number and assume a big change. Option E (99 899) traps students who focus on the size of the number rather than the position of the 8.

Five numbers are given, each containing the digit 8 in a different place value position. Students must identify where 8 sits in each number, calculate the increase from replacing 8 with 9 (= 1 × place value), then identify the maximum.

Which mixed number is closest to 2? Find the gap for each: 1¾ → 0.25; 1½ → 0.5; 2⅔ → 0.667; 2⅓ → 0.333; 2½ → 0.5. Smallest gap = 0.25 = option A. Option D (2⅓, gap 0.333) is the most tempting wrong answer for students who confuse 'closer above' with 'closest overall'.

Closest-to-a-target questions with mixed numbers appear at the easy level of OC MR. Students convert each option to a decimal (or compare fractions), compute the absolute gap from the target, and select the minimum.

The examiner tests whether students can compare fractions by converting to decimals and calculating distance from a whole number. Option D (2⅓) is the trap for students who only look above 2 and pick the smallest value there, without comparing against 1¾ below.

Five mixed numbers are given, some below and some above the target. Students compute |option − target| for each and identify the minimum.

11 thousands + 11 hundreds + 11 tens + 11 ones = 11000 + 1100 + 110 + 11 = 12,221. The key mistake (option A = 11,111) comes from thinking each digit place just gets an '11', forgetting the column additions carry over into adjacent places.

Expanded-form place-value addition is an easy-to-medium OC MR staple. When the multiplier for each place value group is the same (here, all 11), students must still properly add the resulting values rather than just reading off the multiplier in each column.

The examiner tests whether students can convert 'N thousands + N hundreds + N tens + N ones' into actual values and sum them with carrying. Option A (11,111) is the most common wrong answer — it comes from writing N directly into each digit slot without doing the addition.

A number is described as 'X thousands + Y hundreds + Z tens + W ones' where X, Y, Z, W may be greater than 9 (forcing carries). Students expand and sum.

Maisy lays 6 bricks per layer and has 26 bricks total. 26 ÷ 6 = 4 remainder 2. She finishes 4 full layers (24 bricks) and then starts a 5th — her last 2 bricks go there. Option A (4th) traps students who stop at the last complete layer.

Division-with-remainder problems framed as grouping (layers, boxes, rows) are an easy-level OC MR staple. The key step is recognising that any non-zero remainder means one more group is needed.

The examiner tests whether students understand that a remainder means the last brick is on the next (incomplete) layer, not the last complete layer. Option A (4th layer) is the most common wrong answer — students who get 4 full layers and stop there miss the remaining 2 bricks.

Items are distributed into equal-sized groups one at a time. The total is not a perfect multiple of the group size. Students use division with remainder to find which group the final item lands in.

Now let's step into one of the most elegant corners of Number Theory: puzzles where the number itself is unknown and the clues are hidden in its individual digits. These questions reward students who can translate plain English into a pair of equations.

Digit and number problems appear consistently in the mid-to-upper difficulty range of OC MR tests. They are a reliable separator — students who know the algebraic representation of a two-digit number solve them in under a minute; everyone else guesses.

The examiner wants to see whether you can represent a two-digit number as 10a + b (where a is the tens digit and b is the units digit), write two separate equations from the two clues given, and then solve the system without resorting to trial and error.

Two conditions about a two-digit number are stated — one typically involves the sum of its digits, the other describes what happens when the digits are reversed (usually that the number increases or decreases by a fixed amount). You must find the original number.

Hassan visits several shops. The first takes 1/3 of his pocket money; every other shop takes 1/6. He has 1/6 left. How many shops in total? The trick is setting up one equation: first-shop fraction + (n × each-other-shop fraction) + leftover = 1, then solving for n and adding 1 for the first shop.

Fraction-sharing word problems where a fixed amount is distributed across several recipients (or events) appear regularly in OC MR at the medium difficulty level. Converting all fractions to a common denominator before solving is the fastest approach.

The examiner tests whether students can translate a multi-step fraction scenario into one equation and solve it, rather than guessing by trial. Option A (3) traps students who solve for n = 3 correctly but forget to add the first shop; option C (5) traps students who include the leftover as an additional shop.

A total quantity (money, juice, ribbon) is divided: one recipient gets one fraction, several others each get a different fraction, and a specified remainder is left. Students must find how many recipients there are in total. The setup always means: first fraction + n × repeat fraction + leftover = 1.

Panu counts halves in 2½ → 5. Quinn counts thirds in 2⅓ → 7. Rita counts quarters in 1½ → 6. Order: 5, 6, 7 = Panu, Rita, Quinn. The trick: 'number of unit fractions that fit' is just the mixed number divided by that unit fraction, which simplifies to multiplying by the denominator.

Counting unit fractions inside a mixed number is a medium-level OC MR pattern. It tests whether students know that dividing by a fraction is the same as multiplying by its reciprocal. The ordering step at the end adds a comparison layer.

The examiner tests two skills in sequence: (1) can the student interpret 'how many halves/thirds/quarters fit into X' as a division, and (2) can the student convert mixed numbers to improper fractions and divide. Option C (Quinn, Rita, Panu — descending order) catches students who order largest to smallest.

Three people each define their number as 'the count of a specific unit fraction needed to build a given mixed number'. Students compute each count and then compare them.

Which combinations of three whole numbers can add to 100? The key: 100 is even. Even+Even+Even = Even ✓. Even+Even+Odd = Odd ✗. Odd+Odd+Even = Even ✓. Odd+Odd+Odd = Odd ✗. So only statements 1 and 3 work. Statement 2 is the classic trap — students assume two evens 'fix' the odd, but they don't.

Parity (odd/even) rule questions appear regularly in OC MR. The statement-combination format (which of 1/2/3/4 are true?) is common. Students must know: odd+odd=even and even+odd=odd, then chain these to check each three-number combination.

The examiner tests whether students know the parity addition rules and can apply them to a three-addend sum. Statement 2 (even+even+odd) is the trap: students who think 'two evens make it balanced' will incorrectly include it. Option D (1, 2 and 3 only) traps those who include statement 2 by mistake.

Four statements each describe a way to pick three numbers (by odd/even type). Students check each against the parity of the target sum. A statement-combination option list (1 only, 1 and 3 only, etc.) is used.

Charles counts 4→7 in jumps of ⅓: range=3, jumps=3÷⅓=9. Julie counts 6→8 in jumps of ¼: range=2, jumps=2÷¼=8. Charles makes 1 more jump. The key insight: number of jumps = range ÷ jump size = range × denominator.

Counting fractional jumps over a range is a medium-level OC MR pattern. Two people with different ranges and different jump sizes are compared. Students must correctly compute range ÷ (unit fraction) = range × denominator for each person.

The examiner tests whether students can convert a 'count jumps over a range' problem into range ÷ jump size. Option C (same number) traps students who compare only the ranges (3 vs 2) or only the denominators (3 vs 4) without combining them correctly.

Two people each count from one number to another using a given fractional jump size. Students compute the number of jumps for each (range ÷ fraction = range × denominator) and then compare.

Smallest 3-digit number with 3 different digits: hundreds=1 (can't use 0), tens=0, units=2 → 102. Largest: hundreds=9, tens=8, units=7 → 987. Sum = 102 + 987 = 1089. The classic mistake is thinking 100 or 101 qualify — they only have two distinct digits.

Finding extremes of digit-constrained numbers is a medium OC MR pattern. Questions ask for the smallest or largest number satisfying a condition on the digits (all different, exactly two the same, etc.). A two-person question then asks for the sum.

The examiner tests whether students understand 'three different digits' strictly (no repeats). The traps: thinking 100 is the smallest (repeated 0), or thinking 998 is the largest (repeated 9). Option B (1098) catches students who use 111 as Sean's number.

Two people each construct a number at the extreme (smallest/largest) subject to a digit constraint. Students find both numbers and compute a combined value (sum, difference, or product).

Six children each play every other child once in a tennis tournament. How many matches in total? The trick: if you just multiply 6 × 5 = 30, you count every match twice (A vs B AND B vs A). You must divide by 2 to avoid double-counting.

Round-robin / handshake combinatorics problems (n people, each pair meets once) appear regularly in OC MR at medium difficulty. They are solved either by systematic listing (5+4+3+2+1+0) or by the formula n(n−1)÷2. The double-counting trap (answer D = 30) catches a large proportion of students.

The examiner is testing whether students understand that a match between A and B is the same as a match between B and A — and so they must avoid counting it twice. Option D (30) is the most common trap: 6 × 5 = 30 gives every ordered pair, but matches are unordered, so you divide by 2.

A group of n people (or teams, or objects) each interacts with every other exactly once. Students count the total number of interactions (pairs). Typical n values at OC level: 5, 6, or 7. Sometimes phrased as 'every team plays every other team once' or 'every person shakes hands with every other person'.