1990 AMC 12 Problem 13

Question

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.

19

21

23

199

201

After

k

passes through instructions

2

and

3,

find

X

and

S

The values added to

S

are consecutive odd numbers beginning with

5

After

k

passes,

X=2k+3,

and

\begin{aligned} S&=5+7+\cdots+(2k+3)\\ &=k(k+4). \end{aligned}

Now

98(102)=9996\lt10000,

while

99(103)=10197\ge10000.

Thus the loop stops at

k=99,

when

X=2(99)+3=201.

Thus the correct answer is E.

Problem 13 in other years