AMC Y5-6 Advanced Course
For Years 5–6 with secure foundations. Develop upper-primary problem-solving through the Australian AMC pathway.
First 3 starter lessons or units free with an accountSee curriculum & course details →Register & try 3 free →Translate every restriction into a separate check, then use the strongest one to create a short candidate list. A condition on each digit is different from a condition on the whole number. Remainders, divisibility and distinct digits must all be satisfied.
The official samples combine constraints involving digits, divisibility, remainders and patterns. Knowing a rule in isolation does not identify which rule is useful in a longer statement. Begin by separating what is allowed for the complete number from what is allowed in each position.
For example, 123 is divisible by 3 even though some of its digits are not. Conversely, the fact that a particular digit is not divisible by 3 does not settle whether the full number is divisible by 3. Write the condition in plain language and test a small example before applying it to many candidates.
Check 243 against the original statement: 243/9 = 27, 243 = 5×48 + 3, and its digits differ. This final substitution guards against solving the simplified list rather than the actual question. The answer is the smallest valid number, not the number of valid candidates.
Now change the required remainder to 2. The last digit is 2 or 7. The corresponding multiples are 207, 252, 297, 342 and 387. Removing the repeated-digit case 252 leaves four candidates, and the smallest is 207. This changed example exposes an easy omission: do not start your list at the candidate nearest the previous answer.
In a model where every r-th square of row r is marked, a marked square in column c means that r divides c. This changes a picture into a question about divisors. It is a useful type of representation seen in the official samples; the following numbers are our own practice version.
For column 36, the pairs are 1×36, 2×18, 3×12, 4×9 and 6×6. The repeated central divisor 6 is counted once, giving 9 positive divisors. This is why a square number can have an odd divisor count.
Use a changed condition after teaching, then ask the child which part of the method remains valid. Memorising a candidate list is not transferable. The useful learning is knowing how to construct a new one without skipping a case.
Explore the complete curriculum, then start the free lessons in your own learning account. These existing AMC and AIMO courses develop related skills; they are not dedicated AJMO courses or official competition entry.
For Years 5–6 with secure foundations. Develop upper-primary problem-solving through the Australian AMC pathway.
First 3 starter lessons or units free with an accountSee curriculum & course details →Register & try 3 free →For Years 7–8 ready to extend beyond foundations. Build geometry, number-theory and counting techniques.
First 3 starter lessons or units free with an accountSee curriculum & course details →Register & try 3 free →For students ready for deeper problems and written reasoning. Inspect the level before choosing the stretch.
First 3 starter lessons or units free with an accountSee curriculum & course details →Register & try 3 free →It may end in 3 or 8. Both last digits leave remainder 3 on division by 5.
No. Apply digit-by-digit conditions separately from a divisibility test on the complete number.
A candidate can satisfy a simplified rule while failing another restriction, or it can be valid without answering the smallest/largest/count question asked.
In a factor pair such as 6×6, both entries are the same divisor. Counting them twice would duplicate one value.