NUMBER RESTRICTIONS WORKSHOP

AJMO Number Problems: Use Divisibility, Digits and Remainders Together

By Ace Achievers · Published 16 October 2026 · Official information checked 14 October 2026

The short answer

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.

Why do familiar number rules become difficult?

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.

How do we build and filter a short candidate list?

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.

When does a grid pattern become a factor question?

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.

What should my child write on the page?

  1. List the restrictions: interval, divisibility, remainder, digit condition, and whether to count or maximise.
  2. Choose a filter that reduces the search substantially.
  3. Organise the remaining candidates in increasing order or a table.
  4. Test every condition and check the requested quantity.

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.

Give practice a clear learning pathway

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.

Questions parents ask

Does remainder 3 on division by 5 mean the number ends in 3?

It may end in 3 or 8. Both last digits leave remainder 3 on division by 5.

Are digit restrictions the same as divisibility of the whole number?

No. Apply digit-by-digit conditions separately from a divisibility test on the complete number.

Why do we check the answer after finding a candidate?

A candidate can satisfy a simplified rule while failing another restriction, or it can be valid without answering the smallest/largest/count question asked.

Why is the middle factor counted once for a square number?

In a factor pair such as 6×6, both entries are the same divisor. Counting them twice would duplicate one value.