If the odd natural numbers are arranged as shown, then the extremities of the row containing the number 2009 are _____<br>1<br>3 5<br>7 9 11<br>13 15 17 19<br>21 23 ...

PLACEHOLDER ANSWER - the answer is two numbers. Row $$k$$ has $$k$$ numbers, so the first $$k$$ rows hold $$\frac{k(k + 1)}{2}$$ odd numbers. 2009 is the 1005th odd number, and since $$\frac{44 \times 45}{2} = 990 < 1005 \le \frac{45 \times 46}{2} = 1035$$, it lies in row 45. That row runs from the 991st to the 1035th odd number, so its extremities are $$2 \times 991 - 1 = 1981$$ and $$2 \times 1035 - 1 = 2069$$.