Download Solutions - Missouri State University

SW-ARML Practice 10-14-12 Solutions
1.
The rational number r is the largest number less than 1 whose base-7 expansion consists of two
p
distinct repeating digits, r = (0.ABABAB…)7. Written as a reduced fraction, r = . Compute
q
p + q. [Hint: If it were base-10, r would be 0.989898….]
In base-7, the value of r must be 0.656565… = 0.657 .
Then, 1007r = 65.657 , and
(1007  1)r = 657.
47
Converting to base-10, 48r = 47, so r =
and p + q = 95.
48
2.
Compute the least positive integer n such that the set of angles {123, 246, …., n123} contains
at least one angle in each of the four quadrants.
The first angle is 123 which is in Quadrant II, the second angle (246) is in Quadrant III, and the
third angle (369  9 mod 360) is in Quadrant I.
The missing quadrant is IV which is 270  246 = 24 away from the second angle in the
sequence. Because 369  9 mod 360, the terminal ray of the (n + 3)rd angle is rotated 9
counterclockwise from the nth angle.
Thus, three full cycles are needed to reach Quadrant IV from the second angle: the fifth is  255,
the eighth is  264, and the eleventh angle is  272. So, n = 11.
3.
Let set A be a 90-element subset of {1, 2, 3, 4, …., 98, 99, 100} and let S be the sum of the
elements of A. Find the number of possible values of S. [Hint: Find the largest and smallest
sums first.]
90  91
 4095 .
2
90 111
 4995 .
The largest S is 11 + 12 + 13 +  + 100 =
2
All numbers between 4095 and 4995 are possible values of S, so the number of possible values of
S is 4995 – 4095 + 1 = 901.
The smallest S is 1 + 2 + 3 +  + 90 =
4.
Find the least positive integer such that when its leftmost digit is deleted, the resulting integer is
1
of the original integer. [Hint: An integer can be expressed as 10na + b if a is the leftmost
29
digit.]
The number can be represented as 10na + b, where a is the leftmost digit and b is the rest of the
number.
1
10n a  b
Then, b 
29
n
 29b = 10 a + b
 28b = 10na
a=7
 4b = 10n
 b = 25 & n = 2
 The original number is 725.
5.
Let (a1, a2, a3, …, a12) be a permutation of {1, 2, 3, …, 12} for which a1> a2 > a3 > a3 > a5 > a6
and a6 < a7 < a8 < a9 < a10 < a11 < a12. An example of such a permutation is
(10, 7, 5, 3, 2, 1, 4, 6, 8, 9, 11, 12). Find the number of such permutations. [Hint: What must a6
be?]
Clearly, a6 = 1. Now, we may select any 5 of the remaining 11 numbers to form (in descending
order) {a1, a2, a3, a4, a5} and let the remaining 6 to form (in ascending order)
{a7, a8, a9, a10, a11, a12}.
11 1110  9  8  7
 11 3  2  7  462 such permutations.
Hence, there are   
 5  1 2  3  4  5
6.
 200 
 200 
200!

What is the largest 2–digit prime factor of the integer 
? [Hint: 
]


 100 
 100  100!100!
 200 
200!

Since 
, every prime factor between 1 and 100 appears twice in the denominator.

 100  100!100!
So, we need to find a 2–digit prime p that appears three times in the numerator 200!, or 3p < 200.
The largest such prime is 61.
7.
Let S be the set of real numbers that can be represented as repeating decimals of the form 0.abc ,
where a, b, c are distinct digits. Find the sum of the elements of S. [Hint: All 0.abc have the same
denominator as fractions.]
Numbers of the form 0.abc can be written as
abc
.
999
There are 10  9  8 = 720 such numbers.
Each digit will appear in each place value
720
 72 times, and the sum of the digits, 0 through 9,
10
is 45.
So the sum is
45  72  (100  10  1)
 360 .
999
2
8.
The lengths of the sides of a triangle with positive area are log 12, log 75, and log n, where n is a
positive integer. Find the number of possible values of n. [Hint: The Triangle Inequality.]
By the Triangle Inequality, we must have log 12 + log n > log 75 and log 12 + log 75 > log n.
12 + log n > log 75
 log 12n > log 75  12n > 75  n > 7.25
log 12 + log 75 > log n
 log 900 > log n  n < 900
Therefore, there are 900 – 7 = 893 possible values for n.
9.
Let P be the product of the first 100 positive odd integers. Find the largest integer k such that P is
1 2  3 10
10!
 5
divisible by 3k. [Hint: For example, 13579 =
.]
2  4  6 10 2  5!
Note that P = 1357195197199 =
1 2  3 200
200!
 100
.
2  4  6 200 2 100!
Hence, we are looking for the number of 3’s in 200! minus the number of 3’s in 100!, which is
 200   200   200   200  100  100  100  100 
equal to 







 3   9   27   81   3   9   27   81 
= 66 + 22 + 7 + 2 – 33 – 11 – 3 – 1
= 97 – 48
= 49
10. A sequence is defined as follows:
a1 = a2 = a3 = 1, and for all positive integers n, an+3 = an+2 + an+1 + an. Given that a28 = 6090307,
a29 = 11201821, and a30 = 20603361, find the remainder when

28
a  a1  a2 
k 1 k
 a28 is
divided by 1000.
Denote the sum by S.
a1 = 1
a2 = 1
a3 = 1
a4 = a3 + a2 + a1
a5 = a4 + a3 + a2
……
a27 = a26 + a25 + a24
a28 = a27 + a26 + a25
a29 = a28 + a27 + a26
a30 = a29 + a28 + a27
S + a29 + a30 = 1+ S + a29 + S  1 + S  a28
 a28 + a30 = 2S
a a
307  3361 3668

 1834  834 (mod 1000)
So, S  28 30 
2
2
2
3