Question

The sum of 18 consecutive positive integers is a perfect square what is the smallest possible value of this sum

Answer 1

Step-by-step explanation:

1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18=171

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Answer 2

Answer:

225

Step-by-step explanation:

Denoting the first term of the sequence by a, then tis arithmetic sequence

with common difference d = 1 has the sum,

a + (a + 1) + · · · + (a + 17) = 18a + (1 + 2 + · · · + 17)

= 18a +1/2(17 · 18) = 9(2a + 17) .

For the sum to be a perfect square, the term 2a + 17 must be a perfect square. By

inspection, we can see that this first occurs when a = 4. This gives the minimal

sum, which is 9 · 25 = 225.