Overview
An integer problem often becomes small when its restrictions are combined in the right order. A bounded interval may contain only a few multiples; a digit relation may determine both digits; a divisor count may identify an entire class of numbers. The goal is not to guess quickly but to create a short, justified candidate set and then apply every remaining condition.
Use the strongest filter to make a short list
An integer n lies strictly between 40 and 80, is divisible by 9 and leaves remainder 4 on division by 5. The multiples of 9 in the interval are 45, 54, 63 and 72. Checking their remainders modulo 5 leaves only 54.
This list is complete because consecutive multiples of 9 before and after it lie outside the strict interval. The distractor 63 satisfies the divisibility condition but leaves remainder 3, so it fails the second filter. Starting with the bounded multiples is more efficient than checking every integer in the interval. At the end, substitute the surviving value into all original conditions. This confirms not only that the arithmetic works but that an endpoint or a remainder interpretation was not silently changed along the way.
Translate place value rather than guessing digits
A two-digit positive integer has digits summing to 9, and its tens digit is twice its ones digit. Let the ones digit be u; then the tens digit is 2u. Their sum gives 3u = 9, so u = 3 and the tens digit is 6. The integer is 63.
The number 36 reverses the digit roles even though its digits have the correct sum. This makes it a plausible distractor for a solver who stops after finding the two digits. State the place-value expression—ten times the tens digit plus the ones digit—before composing the number. Digit variables also have restrictions: they are integers from 0 through 9, and a leading digit cannot be zero. These restrictions may become important in harder digit equations where ordinary algebra produces a value that cannot occupy a digit position.
Recognize numbers with exactly three divisors
A positive integer has exactly three positive divisors precisely when it is the square of a prime. Its divisors are then 1, p and p². If the square root were composite, additional factors would create more divisors; if two different prime factors were present, their combinations would also give too many.
Below 200, the eligible prime squares arise from p = 2, 3, 5, 7, 11 and 13. The extra condition is that n + 1 be divisible by 5. Checking the six squares leaves n = 4, 9, 49 and 169, so there are four. The answer six counts the prime squares before applying the second condition. This problem shows why a structural fact can be more useful than a long divisor search: it reduces the possible integers to a small, organized list while still explaining why no other form can work.
The prime-square characterization also gives a quick reason the candidate list stops at 13. A prime larger than the square root of 200 would have a square outside the required interval. You need not test every integer below 200 for divisors once this structure is proved. In harder bounded problems, choosing which condition to apply first can reduce the work far more than any arithmetic shortcut.
Find the least value and justify where you start
To find the least positive n for which n + 1 is divisible by 4 and n − 1 by 3, the first condition gives n = 3, 7, 11, … . The first candidate 3 fails the second condition, while 7 gives n − 1 = 6 and succeeds. Because the candidates were tested in increasing order from the least positive possibility, 7 is proved least.
For transfer, choose a representation suited to the strongest condition: multiples for divisibility, place-value variables for digits, prime powers for divisor counts, and repeating residue classes for shifted divisibility. Then apply all remaining restrictions without expanding the search unnecessarily. In practice, add a sentence explaining completeness or minimality. Finding a working number proves existence; counting all working numbers or proving the least requires an argument that the search has not skipped any eligible case.
Apply the method to a new case
Find the smallest positive integer n that leaves remainder 2 when divided by 4 and remainder 3 when divided by 5. The first condition gives candidates 2, 6, 10, 14, 18, ... . Their remainders modulo 5 cycle as 2, 1, 0, 4, 3, so 18 is the first match. Alternatively, write n = 4a + 2. Then 4a + 2 ≡ 3 mod 5, giving a ≡ 4 mod 5; the smallest nonnegative such a is 4 and n = 18. Reporting 8 meets the second condition but not the first. Verify 18 = 4·4 + 2 = 5·3 + 3. The combined pattern repeats every least common multiple, 20, so all solutions are 18 + 20k; positivity selects k ≥ 0. Use the repeating structure to justify why the finite search establishes the smallest solution.