Number Theory

Smallest 4-digit number = 1000. Largest 2-digit number = 99. 1000 × 99 = 99 000.

Place-value identification questions (smallest/largest N-digit number) appear regularly in the easy band of Selective MR. Students who confuse “smallest 4-digit” with 9999 or “largest 2-digit” with 100 lose marks on very straightforward calculations.

The examiner tests whether students correctly identify 1000 as the smallest 4-digit whole number and 99 as the largest 2-digit whole number, then carry out the multiplication accurately.

Students must recall the boundary values of number ranges (smallest/largest N-digit numbers) and multiply them together.

7/10 = 0.700. The closest decimal is whichever has the smallest gap to 0.700. Compare gaps: 0.705 is only 0.005 away, while 0.68 is 0.020 away and 0.73 is 0.030 away.

Fraction-to-decimal proximity questions appear in the easy band of Selective MR. Students who can’t convert the fraction first often guess or pick a decimal that looks similar but ignores place value.

The examiner checks whether students (1) convert 7/10 to 0.700 correctly, (2) compare all five decimals to 0.700 (not to each other), and (3) avoid the trap of picking 0.072 or 0.069 which look related to ‘7’ and ‘10’.

A simple fraction is given and students must identify which of five decimals is numerically closest to it.

Kim’s multiples of 5 (0–50): largest = 45. Jamie’s even multiples of 7 (0–50): 14, 28, 42 — smallest = 14. Difference = 45 − 14 = 31.

Constrained-multiple questions with a largest/smallest twist appear regularly in the easy–medium band of Selective MR. The two-condition filter for Jamie (even AND multiple of 7) is where many students go wrong by forgetting one constraint.

The examiner checks whether students (1) list multiples of 5 correctly up to 49, (2) filter multiples of 7 by evenness, (3) correctly identify largest vs smallest, and (4) subtract in the right order.

Two people each pick a number satisfying different constraints within a fixed range. Students find the largest valid number for one person and the smallest valid number for the other, then find their difference.

4 850 000 rounded to the nearest million = 5 000 000 (the 8 in the hundred-thousands place rounds up). Rounded to the nearest hundred thousand = 4 900 000 (the 5 in the ten-thousands place rounds up). Difference = 100 000.

Two-person rounding comparison questions appear regularly in the easy band of Selective MR. Students who mix up rounding levels (nearest million vs nearest hundred thousand) almost always pick 150 000 or 50 000.

The examiner checks whether students (1) correctly identify which digit to look at for each rounding level, (2) apply the ‘round half up’ rule correctly for both 8 and 5, and (3) subtract the two rounded values in the right order.

Two people round the same large number to different place values. Students find both rounded answers and calculate the difference.

In 30 060, the digit 3 sits in the ten-thousands column (value 30 000) and the digit 6 sits in the tens column (value 60). Divide: 30 000 ÷ 60 = 500.

Place-value comparison questions appear in the easy-to-medium band of Selective MR. Students who confuse the digit’s face value with its positional value consistently pick 5 or 50 instead of 500.

The examiner checks whether students can identify the positional value (not just the digit) of each named digit, then correctly compute how many times larger one value is than the other.

A multi-digit number is given. Students must find the value each of two named digits represents (using place value), then divide the larger by the smaller.

Number-theory questions often start with listing multiples in a range — write the valid set for each person before you compare or combine them.

Multiple-constraint whole-number items appear in the medium band of Selective MR — solid marks when you filter by “multiple of” and “odd” systematically.

The examiner checks whether you can list all valid multiples under a cap, apply extra filters (odd/even), and then use largest/smallest from each list correctly.

Two people each pick a different whole number below a limit, with separate divisibility rules. You find an extreme value (largest minus smallest, etc.).

Digit-reversal puzzles always become algebra: call the tens digit a and units b, then turn “reversed number is 72 more” into 9b − 9a = 72.

Two-digit reversal items with a worked mini-example appear on Selective MR — the template is identical once you see the 9(b − a) shortcut.

The examiner tests whether you can generalise from a numeric example (35 → 53, +18) to unknown digits and solve for a specific digit asked in the question.

A short example shows how reversing digits changes the value. You apply the same logic to a new two-digit number with a stated increase after reversal.

Valerie subtracts a positive multiple of 7 from her secret number and gets 59. So her number = 59 + 7k for some whole number k ≥ 1. List every valid value of k that keeps her number strictly between 0 and 100, and count them.

Counting valid unknowns from a constraint (reverse arithmetic + range limit) appears regularly in the medium band of Selective MR. The method is always the same: express the unknown in terms of a parameter, then count the integers in range.

The examiner tests whether students can (1) rearrange the clue to express the favourite number as 59 + 7k, (2) identify k must be a positive integer (≥ 1), (3) list multiples of 7 until they exceed the upper bound, and (4) count the valid values rather than stopping at the first.

A person describes a number using arithmetic with a multiple constraint. Students must find how many values of the original number are possible given a range.

Three fraction statements — test each one separately. Statement Y is the trickiest: 1 − 3/8 = 5/8, which is bigger than 3/8, not smaller.

Multi-statement fraction comparison questions appear in the medium band of Selective MR. Statement Y (subtraction result compared with subtracted fraction) is a classic examiner trap — students who don’t compute the result will guess it’s less than 3/8.

The examiner tests whether students (1) correctly add 3/4 + 3/4 = 6/4 = 1½ and compare with 1¼, (2) compute 1 − 3/8 = 5/8 and compare it with 3/8 (not less), and (3) recognise that a smaller denominator means a larger fraction (1/6 > 1/10).

Three fraction statements make claims about addition, subtraction, and size comparisons. Students identify which are correct.

Convert everything to fifths: Amina = 4/5, Fred = 6/5, Sally = 3/5. Sum = 13/5. Quoc = 4 − 13/5 = 20/5 − 13/5 = 7/5. Quoc − Amina = 7/5 − 4/5 = 3/5.

Multi-person fraction sharing questions appear in the medium band of Selective MR. Students who forget to convert mixed numbers to improper fractions (Fred = 1 1/5 = 6/5) consistently get the wrong total and pick a wrong answer.

The examiner checks whether students (1) correctly convert 1 1/5 to 6/5, (2) add all three known portions as fifths, (3) subtract from 4 to find Quoc’s share, and (4) subtract Amina’s share from Quoc’s (not from the total).

Four people share a whole-number amount. Three amounts are given (one as a mixed number); the fourth person takes all the rest. Students find the difference between two specific people’s shares.

Count multiples of 3 up to 100 (there are 33), then subtract those that are also multiples of 8 — which means multiples of LCM(3, 8) = 24. There are 4 of those, so the answer is 33 − 4 = 29.

Set-subtraction divisibility questions (‘multiples of A but not B’) appear in the medium band of Selective MR. Students who forget to subtract the overlap (multiples of both 3 and 8) arrive at 33 — the most common wrong answer.

The examiner checks whether students can (1) correctly count multiples of 3 in a range, (2) identify that ‘multiples of both 3 and 8’ means multiples of LCM(3, 8) = 24, and (3) subtract the overlap.

Students are given a range and must count numbers that satisfy one divisibility condition but not another. The key tool is LCM to find the overlap set.

Rounding to the nearest hundred splits every range into a band of 100 numbers. Find both players’ complete ranges first, then pick the extreme values that give the biggest gap.

Rounding-range maximisation questions appear in the medium-to-difficult band of Selective MR. Students who only think about the “target” (800 or 1000) without finding the full range will be drawn to the trap answer of 200.

The examiner tests whether students (1) know that rounding to the nearest hundred gives a 100-wide band centred on the target (e.g. 750–849 rounds to 800), (2) identify the extreme values in each range, and (3) subtract min from max to find the largest possible difference.

Two people each pick a whole number that rounds to a given value. Students must find the complete rounding range for each person and compute the extreme difference.

Label each straw E (even) or O (odd). For 3 numbers to add to an ODD total, you need either 3 odds or 1 odd + 2 evens. Count valid combos from each case — no need to add the actual numbers.

Odd/even parity selection problems (how many groups of k sum to odd?) appear in the medium-to-difficult band of Selective MR. Students who try to enumerate all C(5,3)=10 triples without parity reasoning waste time and make arithmetic errors.

The examiner tests whether students know the rule 'sum of three is odd when parity is OOO or OEE', can apply it to classify available items, identify that 3-odd is impossible (only 2 odds exist), and correctly count 1-odd-2-even combinations using multiplication.

A set of numbers (or lengths) with a mix of odd and even values is given. Students pick a subset of a fixed size and count how many subsets have an odd (or even) sum.

Work backwards: ■ is even, so ■÷2 is a whole number. That whole number rounds to 50, meaning it’s somewhere between 45 and 54. List every possible ■, compute each digit sum, and find the largest.

Multi-constraint maximisation questions (rounding range + digit property) appear in the difficult band of Selective MR. The key steps are: (1) find the valid range, (2) enumerate all candidates, (3) compute the target quantity for each and pick the max.

The examiner tests whether students (1) correctly identify the rounding range (45–54 for rounding to 50), (2) multiply by 2 to get ■’s range (90–108), (3) realise ■ must be even so only even numbers in that range qualify, and (4) compare digit sums to find the maximum. The trap is 108 (digit sum 9) — larger number but smaller digit sum than 98 (digit sum 17).

A number is constrained by being even, having its half round to a given value, and the question asks for the largest possible digit sum. Students must enumerate the valid range and check digit sums.

Only 4 numbers, choose 3 — that’s just 4 possible combinations. List all four sums first, then test each statement against every sum. If even one sum breaks a statement, that statement is false.

Multi-statement number property questions (can/cannot/always) appear in the difficult band of Selective MR. Students who don’t enumerate all combinations will miss counterexamples. With only C(4,3)=4 options, listing them all is fast and reliable.

The examiner tests whether students (1) correctly list all 4 sums (14, 16, 17, 19), (2) know that 16 is a multiple of 8 (disproving Statement 1), (3) check all four sums for divisibility by 3 (none qualify, disproving Statement 2), and (4) notice 14 and 16 are even (disproving Statement 3).

Three numbers are chosen from a small set and summed. Three statements make claims about divisibility or parity of all possible totals. Students identify which are correct.

Green+blue share a cycle every LCM(3,5)=15 min. Red joins every LCM(2,15)=30 min. Count the 15-min hits in the window, then subtract the 30-min hits.

Multi-light / multi-event LCM questions appear in the difficult band of Selective MR. Students who only find one LCM but forget to subtract the overlap (red also on) will over-count and pick 11 instead of 6.

The examiner tests whether students (1) find LCM(3,5)=15 for green+blue, (2) list or count all 15-min multiples in the window 7:01–9:59, (3) identify that red flashes at every even multiple of 15 (i.e. every 30 min), and (4) subtract those overlap times.

Three lights flash at different intervals and start together. Students count how many times two specific lights coincide without the third also being on, within a given time window.

Even number ÷ 8: the remainder must be even because even − even = even. Only 0, 2, 4, 6 are possible. Since 0 doesn’t appear in the options, the answer is 2, 4, 6.

Remainder-parity questions appear in the difficult band of Selective MR. Students who just list all remainders 1–7 for division by 8 miss the key constraint: the number is even, so the remainder must also be even.

The examiner tests whether students (1) know that even − multiple-of-8 = even remainder, (2) correctly list only the even remainders 0, 2, 4, 6, and (3) choose C (2,4,6) rather than E (all 1–7) or D (odd only).

A number with a specific property (even, odd, multiple of something) is divided by another number. Students find all possible remainders.

Write both numbers as 4m and 4n (since HCF = 4). LCM = 4mn = 40 so mn = 10. The coprime factor pairs of 10 are (1,10) and (2,5). The first gives (4,40) — already given. The second gives (8,20), summing to 28.

HCF/LCM pair-finding questions appear in the difficult band of Selective MR. Most students know how to find HCF and LCM of a given pair but struggle to reconstruct a pair from its HCF and LCM.

The examiner checks whether students understand the relationship HCF × (product of coprime quotients) = LCM, and can systematically enumerate factor pairs of LCM/HCF while enforcing the HCF constraint.

One pair of numbers with a given HCF and LCM is shown. Students must find a second, distinct pair with the same HCF and LCM.