Problem:
P ( x ) = ( x − 1 2 ) ( x − 2 2 ) ⋯ ( x − 10 0 2 ) P ( x ) = ( x − 1 2 ) ( x − 2 2 ) ⋯ ( x − 1 0 0 2 )
How many integers n n P ( n ) ≤ 0 P ( n ) ≤ 0
Answer Choices:
A. 4900 4 9 0 0 4950 4 9 5 0 5000 5 0 0 0 5050 5 0 5 0 5100 5 1 0 0
Solution:
We perform casework on P ( n ) ≤ 0 P ( n ) ≤ 0
P ( n ) = 0 P ( n ) = 0 100 1 0 0 n n
1 2 , 2 2 , 3 2 , … , 10 0 2 1 2 , 2 2 , 3 2 , … , 1 0 0 2
Interval of x # of Negative Factors Valid? ( − ∞ , 1 2 ) 100 ( 1 2 , 2 2 ) ′ 99 ✓ ( 2 2 , 3 2 ) 98 ( 3 2 , 4 2 ) 97 ✓ ( 4 2 , 5 2 ) 96 ( 5 2 , 6 2 ) 95 ✓ ( 6 2 , 7 2 ) 94 ( 9 9 2 10 0 2 ) 1 j ( 10 0 2 ⋅ ∞ ) 0 Interval of x ( − ∞ , 1 2 ) ( 1 2 , 2 2 ) ′ ( 2 2 , 3 2 ) ( 3 2 , 4 2 ) ( 4 2 , 5 2 ) ( 5 2 , 6 2 ) ( 6 2 , 7 2 ) ( 9 9 2 1 0 0 2 ) ( 1 0 0 2 ⋅ ∞ ) # of Negative Factors 1 0 0 9 9 9 8 9 7 9 6 9 5 9 4 1 0 Valid? ✓ ✓ ✓ j
P ( n ) < 0 P ( n ) < 0 100 1 0 0 P ( x ) P ( x )
Note that there are 50 5 0 x x
( 2 2 − 1 2 − 1 ) + ( 4 2 − 3 2 − 1 ) + ( 6 2 − 5 2 − 1 ) + ⋯ + ( 10 0 2 − 9 9 2 − 1 ) = ( 2 2 − 1 2 ) ⏟ ( 2 + 1 ) ( 2 − 1 ) + ( 4 2 − 3 2 ) ⏟ ( 4 + 3 ) ( 4 − 3 ) + ( 6 2 − 5 2 ) ⏟ ( 6 + 5 ) ( 6 − 5 ) + ⋯ + ( 10 0 2 − 9 9 2 ) ⏟ ( 100 + 99 ) ( 100 − 99 ) − 50 = ( 2 + 1 ) + ( 4 + 3 ) + ( 6 + 5 ) + ⋯ + ( 100 + 99 ) ⏟ 1 + 2 + 3 + 4 + 5 + 6 + ⋯ + 99 + 100 − 50 = 101 ( 100 ) 2 − 50 = 5000 ( 2 2 − 1 2 − 1 ) + ( 4 2 − 3 2 − 1 ) + ( 6 2 − 5 2 − 1 ) + ⋯ + ( 1 0 0 2 − 9 9 2 − 1 ) = ( 2 + 1 ) ( 2 − 1 ) ( 2 2 − 1 2 ) + ( 4 + 3 ) ( 4 − 3 ) ( 4 2 − 3 2 ) + ( 6 + 5 ) ( 6 − 5 ) ( 6 2 − 5 2 ) + ⋯ + ( 1 0 0 + 9 9 ) ( 1 0 0 − 9 9 ) ( 1 0 0 2 − 9 9 2 ) − 5 0 = 1 + 2 + 3 + 4 + 5 + 6 + ⋯ + 9 9 + 1 0 0 ( 2 + 1 ) + ( 4 + 3 ) + ( 6 + 5 ) + ⋯ + ( 1 0 0 + 9 9 ) − 5 0 = 2 1 0 1 ( 1 0 0 ) − 5 0 = 5 0 0 0
In this case, there are 5000 5 0 0 0 n n
Together, the answer is 100 + 5000 = ( E ) 5100 1 0 0 + 5 0 0 0 = ( E ) 5 1 0 0
OR OR
Notice that P ( x ) P ( x ) x x 10 0 2 1 0 0 2 9 9 2 , 9 8 2 9 9 2 , 9 8 2 9 7 2 , ⋯ , 2 2 9 7 2 , ⋯ , 2 2 1 2 1 2
( ( 100 + 99 ) ( 100 − 99 ) + 1 ) + ( ( 98 + 97 ) ( 98 − 97 ) + 1 ) + ⋯ + ( ( 2 + 1 ) ( 2 − 1 ) + 1 ) ( ( 1 0 0 + 9 9 ) ( 1 0 0 − 9 9 ) + 1 ) + ( ( 9 8 + 9 7 ) ( 9 8 − 9 7 ) + 1 ) + ⋯ + ( ( 2 + 1 ) ( 2 − 1 ) + 1 )
This reduces to
200 + 196 + 192 + ⋯ + 4 = 4 ( 1 + 2 + ⋯ + 50 ) = 4 ⋅ 50 ⋅ 51 2 = ( E ) 5100 2 0 0 + 1 9 6 + 1 9 2 + ⋯ + 4 = 4 ( 1 + 2 + ⋯ + 5 0 ) = 4 ⋅ 2 5 0 ⋅ 5 1 = ( E ) 5 1 0 0
The problems on this page are the property of the MAA's American Mathematics Competitions