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Difference between revisions of "1990 AHSME Problems/Problem 13"

(Created page with "== Problem == If the following instructions are carried out by a computer, which value of <math>X</math> will be printed because of instruction <math>5</math>? 1. START <math>...")
 
Line 11: Line 11:
 
         AND PROCEED FROM THERE.   
 
         AND PROCEED FROM THERE.   
 
  5. PRINT THE VALUE OF <math>X</math>.   
 
  5. PRINT THE VALUE OF <math>X</math>.   
−
  6.STOP.  
+
  6. STOP.  
  
 
<math>\text{(A) } 19\quad
 
<math>\text{(A) } 19\quad
Line 20: Line 20:
  
 
== Solution ==
 
== Solution ==
−
<math>\fbox{D}</math>
+
Looking at the first few values, it becomes clear that the program stops when <cmath>5+7+9+11+13+\ldots+(x-2)+x\ge 10000</cmath> which is to say <cmath>1+3+5+7+9+11+13+\ldots+(x-2)+x\ge 10004</cmath>
 +
However, the left hand side is now simply the square <math>\frac{(x+1)^2}4</math>. Multiplying out, we get <cmath>x+1\ge \sqrt{40016}\approx 200.039996...</cmath>
 +
 
 +
So the correct answer is <math>201</math>, which is <math>\fbox{E}</math>
  
 
== See also ==
 
== See also ==

Revision as of 11:43, 4 February 2016

Problem

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

1. START $X$ AT $3$ AND $S$ AT $0$.  
2. INCREASE THE VALUE OF $X$ BY $2$.   
3. INCREASE THE VALUE OF $S$ BY THE VALUE OF $X$. 
4. IF $S$ IS AT LEAST $10000$,   
       THEN GO TO INSTRUCTION $5$;  
       OTHERWISE, GO TO INSTRUCTION $2$.  
       AND PROCEED FROM THERE.  
5. PRINT THE VALUE OF $X$.  
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 $\fbox{E}$

See also

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
All AHSME Problems and Solutions

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