SW-MOARML Practice Questions
12-13-09
1. Determine the distance from the point (1, 1) to the line x + 2y = 5.
x + 2y = 5  x = 5 – 2y. The coordinates of the points on the given line are in the form
 5  2 y 1   y 1 . To
2
2
minimize the distance, we could minimize  5  2 y  1   y  1 = 5y2 – 18y + 17 instead.
(5 – 2y, y) and the distance to the given point (1, 1) would be
The minimum of 5y2 – 18y + 17 occurs when y 
2
2
(18) 9
 and the minimum value is
2(5)
5
81  162  85 4
9
9
5    18    17 
 , so the distance from the point (1, 1) to the line
5
5
5
5
2
x + 2y = 5 is
2 2 5
.

5
5
In general, the distance from a point (, ) to the line ax + by + c = 0 is equal to
a  b  c
(1)(1)  (2)(1)  5 2
2


. In the case above, the distance is therefore
.
2
2
2
2
5
5
a b
1 2
2. Given the sequence 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, … (where the integer n is repeated n times),
compute the sum of the first 150 terms.
We must first determine the value of the 150th term. To do that, we must find the largest
n(n  1)
 150 . The largest such n is 16, so the segment of sequence
integer n for which
2
16(17)
 1  137 . Thus, the sum of the first 150 terms is
comprised of 17’s starting at term
2
16(16  1)(2 16  1)
 (17)(14)  1496  238  1734 .
12 + 22 + 32 + … + 162 + 1417 =
6
3. The two-digit integers from 19 to 92 are written consecutively to form the large integer N =
19202122…909192. If 3k is the highest power of 3 that is a factor of N, then k = ?
Since 0 + 1 + 2 + … + 9 = 45 and the digit sum of N is (1+ 9) + (10)(2) + 45 + (10)(3) + 45 +
… + (10)(8) + 45 + (3)(9) + 3 = 3610 + 745 + 27 + 3 = 9(….) + 3 is divisible by 3 but not
9, k = 1.
4. What is the largest subset S of F = {1, 2, 3, … 50} such that no pair of distinct elements of S
has a sum divisible by 7?
Partition F into seven subsets Fi, 0  i  6, so that all elements in each Fi are congruent to i
mod 7. Note that S can contain at most one number of F0, but that if S contains some
member of any of the other subsets, then it can contain all of the members of that subset.
Also, S cannot contain members of both F1 and F6, or both F2 and F5, or both F3 and F4.
Since F1 contains 8 elements and each of the other subsets contains 7 elements, the largest S
can be constructed by selecting one element of F0, all the members of F1, all the members of
either F2 or F5, and all the members of either F3 or F4. Thus, the largest S contains 1 + 8 + 7
+ 7 = 23 elements.
5. Let a1, a2, ….. be an arithmetic sequence with a4 + a7 + a10 = 17 and a4 + a5 + a6 + a7 + a8 +
a9 + a10 + a11 + a12 + a13 + a14 = 77. If ak = 123, then k = ?
17
77
7, d =
a7 = , a9 =
3
11
17
3  2  123 = 7 + (k – 9) 2  k = 183.
3
2
3
7
6. Find the largest positive value attained by the function f  x   8x  x 2  14 x  x 2  48 ,
where x is a real number where both square roots are defined.
f  x   16  ( x  4)2  1  ( x  7)2 The first term 16  ( x  4)2
represents the upper semicircle centered at (4, 0) with radius 4, and the second term
Completing the squares, we have
1  ( x  7) 2 the upper semicircle centered at (7, 0) with radius 1. The largest positive value
is attained when the distance between these semicircles is the longest, which occurs at x = 6.
Thus, the largest value is 48  36  12 .
7. First a is chosen at random from the set {1, 2, 3, …, 99, 100}, and then b is chosen at
random from the same set. What is probability that the integer 3a + 7b has units digit 8?
3a  3, 9, 7, 1; 7b  7, 9, 3, 1. Note that 9 + 9  7 + 1  1 + 7  8 (mod 10). The probability
(1/4)(1/4) + (1/4)(1/4) + (1/4)(1/4) = 3/16.
log n if log8 n is rational
8. For any positive integer n, let f  n    8
, Find
otherwise
 0
log8 n is rational whenever n is a power of 2.
1 2
log8 1024 = 0    1 
3 3
3

10 1  2  3 

3
3
2009
n 1

2009
n 1
f ( n) .
f (n) = log8 1 + log8 2 + log8 4 + … +
 10

55
.
3
 n 
9. Compute the smallest positive integer n > 100 such that 
 is divisible by
101
equal to it.
 n 

 , but not
 100 
 n 
n!


101 101!(n  101)! n  100


 2 . Therefore, the smallest possible n is 302.
We have 
n!
101
 n 


100  100!(n  100)!
10. If (x2 + x + 1)(x6 + x3 + 1) =
10
, compute the real value of x.
x 1
We have 10 = (x – 1)(x2 + x + 1)(x6 + x3 + 1) = (x3 – 1) )(x6 + x3 + 1) = x9 – 1, which implies
that x9 = 11, so x = 9 11 .