Represent consecutive (or consecutive even/odd) integers algebraically and solve.
Free, timed SHSAT practice sets on consecutive-integer setups— every question includes a full worked explanation, and there's no account required.
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Let the smallest integer be \(n\) and write the others from it.
Add them, set the total equal to the given sum, and solve.
Example: three consecutive integers have a sum of \(21\). Then \(n + (n + 1) + (n + 2) = 3n + 3 = 21\), so \(n = 6\). The integers are \(6, 7, 8\).
Shortcut: for an odd number of consecutive integers, the middle one is the sum divided by how many there are. Here, \(21 \div 3 = 7\).
Common traps:
Let the integers be \(n, n+1, n+2\). Their sum is \(3n + 3 = 36\), so \(n = 11\). The integers are 11, 12, 13. The sum of the smallest and largest is 24. (Stopping at \(n\) gives \(11\). Adding all three integers gives \(36\). Adding the smallest and the middle integers gives \(23\).)
Let the odd integers be \(n, n+2, n+4\). Their sum is \(3n + 6 = 57\), so \(n = 17\), giving 17, 19, 21. The smallest is 17. (Dividing the sum by 3 gives the middle integer, \(19\). \(21\) is the largest of the three, not the smallest. Spacing the integers 1 apart, \(n + (n+1) + (n+2) = 57\), gives \(18\), which is not odd.)
Let the integers be \(n, n+1, n+2, n+3\). Their sum is \(4n + 6 = -22\), so \(n = -7\), giving -7, -6, -5, -4. The second smallest is -6. (Stopping at \(n\) gives the smallest integer, \(-7\). \(-5\) is the second largest integer, not the second smallest. Stopping at \(4n = -22 - 6\), before dividing by 4, gives \(-28\).)