1990 AHSME Problems/Problem 13

Problem

If the following instructions are carried out by a computer, which value of $X$ will be printed because of instruction $5$ ?

1. START  AT  AND  AT .  
2. INCREASE THE VALUE OF  BY .   
3. INCREASE THE VALUE OF  BY THE VALUE OF . 
4. IF  IS AT LEAST ,   
       THEN GO TO INSTRUCTION ;  
       OTHERWISE, GO TO INSTRUCTION .  
       AND PROCEED FROM THERE.  
5. PRINT THE VALUE OF .  
6. STOP.

$\text{(A) } 19\quad \text{(B) } 21\quad \text{(C) } 23\quad \text{(D) } 199\quad \text{(E) } 201$

Solution

Looking at the first few values, it becomes clear that the program stops when $5+7+9+11+13+\ldots+(x-2)+x\ge 10000$ which is to say $1+3+5+7+9+11+13+\ldots+(x-2)+x\ge 10004$ However, the left hand side is now simply the square $\frac{(x+1)^2}4$ . Multiplying out, we get $x+1\ge \sqrt{40016}\approx 200.039996...$

So the correct answer is $201$ , which is $\fbox{E}$

1990 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
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1990 AHSME Problems/Problem 13

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