One letter for the smallest, and every other number in the problem is built from it.
Find the largest
The sum of three consecutive integers is 54. Find the largest.
Call them x, x + 1, x + 2. The sum is 3x + 3 = 54, so x = 17 and the three are 17, 18, 19 — the largest is 19. B is the middle one, which is just 54 ÷ 3; a tempting shortcut that answers a different question.
1st
n
+1
2nd
n+1
+1
3rd
n+2
n + (n + 1) + (n + 2) = 54
the sum, as given
3n + 3 = 54
collect
3n = 51, n = 17
the smallest
17, 18, 19 → largest 19
what was actually asked
The question asked for the largest, so 17 is a wrong answer arrived at by correct algebra. Write the three numbers out before choosing — that one habit is worth several marks a paper.
This is the single biggest error in the topic.
| Run | Write them as | Example |
|---|---|---|
| integers | n, n+1, n+2 | 17, 18, 19 |
| even integers | n, n+2, n+4 | 24, 26, 28 |
| odd integers | n, n+2, n+4 | 17, 19, 21 |
| multiples of 5 | n, n+5, n+10 | 15, 20, 25 |
The next odd number after 17 is 19, which is two more — not one, and not three. Odd and even runs are written identically; the only difference is the value n turns out to have.
n + (n+2) + (n+4) + (n+6) = 80
four terms, stepping by 2
4n + 12 = 80
2 + 4 + 6 = 12
4n = 68, n = 17
odd, as required
17, 19, 21, 23
all four
A run of odd integers must start odd. An even or fractional n means the sum you were given is impossible for that many terms — or the +2 steps got written as +1.
When the condition compares terms, translate each phrase into the n's. The sum of three consecutive even integers is 26 more than the largest. Find them.
n + (n + 2) + (n + 4) = (n + 4) + 26
“the largest” is n + 4, “26 more than” adds 26
3n + 6 = n + 30, so 2n = 24, n = 12
n on both sides: collect it on one
12, 14, 16
check: 42 = 16 + 26 ✓
The same translating works for “twice the smaller” (2n), “three times the middle” (3(n + 2)) or “the sum of the other two”.
For an odd count of terms, the average is the middle number.
The terms are evenly spaced, so they balance around the centre: the amount the first is below the middle is exactly the amount the last is above it. With an even count the centre falls in the gap, which still locates the run immediately.
On a multiple-choice paper this is often a five-second question. Divide, then step outward.
n(n + 1) = 156
two consecutive integers
n² + n − 156 = 0
standard form
(n − 12)(n + 13) = 0
factors of 156 differing by 1
n = 12 or n = −13
both are genuine
12 and 13, or −13 and −12
the wording decides
If the question says “positive integers”, keep 12 and 13 and drop the other pair. If it does not, both answers are correct — and 12 × 13 = 156 is worth knowing on sight, along with 13 × 14 = 182 and 14 × 15 = 210.
When the condition involves squares or a product of three, naming the middle term n makes the cross terms cancel.
The squares of three consecutive integers add to 149. Find the integers.
(n − 1)² + n² + (n + 1)² = 149
n is the middle one
3n² + 2 = 149
the −2n and +2n cancel
n² = 49, n = 7 → 6, 7, 8
36 + 49 + 64 = 149 ✓
The squares of two consecutive odd integers add to 202. Find them.
n² + (n + 2)² = 202 → 2n² + 4n + 4 = 202
two terms: start at the smaller
n² + 2n − 99 = 0 → (n + 11)(n − 9) = 0
divide by 2, factor
n = 9 → 9 and 11
81 + 121 = 202 ✓ (or −11 and −9)
The middle term also tames a product of three: if three consecutive integers multiply to 8 times their sum, (n − 1)n(n + 1) = 8 × 3n, so n² − 1 = 24, n = 5, and the run is 4, 5, 6 (120 = 8 × 15 ✓).
The run is negative: five consecutive odd integers add to −35. Find the largest.
middle = −35 ÷ 5 = −7
odd count, so the average is the middle
−11, −9, −7, −5, −3
step out by 2 each way
largest = −3
the one nearest zero
What is different: with negatives, “largest” is the one with the smallest size. The algebra (5n + 20 = −35, n = −11) gives the same run.
−11 looks biggest because 11 is the biggest digit pair, but it is furthest below zero. On the number line the largest number is the one furthest right.
No answer at all: can three consecutive even integers add to 50?
n + (n + 2) + (n + 4) = 50
the usual set-up
3n + 6 = 50, 3n = 44
collect
n = 443 — no such integers
n must be a whole even number
What is different: the equation solves, but the answer is not an integer. Three consecutive evens add to 3 × (an even middle), a multiple of 6, and 50 is not one.
The squares of two consecutive odd integers differ by 56. Find the integers.
(n + 2)² − n² = 56
larger squared minus smaller squared
4n + 4 = 56, n = 13
the n² terms cancel — no quadratic
13 and 15
225 − 169 = 56 ✓
a² − b² = (a − b)(a + b). The two integers differ by 2, so 56 = 2 × (sum), and the sum is 28 — which is 13 + 15. For integers 1 apart, the difference of squares is the sum.
These items usually ask for the sum, the product or the larger integer, and n² (49 above), n itself, and the other integer are all printed as choices. Write the run out in full first.
Consecutive-integer items set a sum, a product, or a condition on squares. Divide a sum by the count to find the middle; name the middle term n when squares or a product of three appear; and remember that odd and even runs both step by 2.
Find the smallest
The sum of five consecutive integers is 105. Find the smallest.
The five add to 5x + 10 = 105, so x = 19 and they run 19 to 23. C is the middle term (105 ÷ 5 = 21) — with an odd number of terms the mean lands exactly on the middle one, never on the smallest.
