SW-ARML Practice 3-3-13 Solutions
Individual Problems
1.
How many positive perfect square less than 106 are multiples of 24?
Since 24 = 233, each square divisible by 24 must have at least 4 factors of 2 and 2 factors of
3. This means that each square is in the form (12c)2, where c is any positive integer such
1000 
that 12c < 103 = 1000. Since there are 
 83 possible values for c, there are 83
 12 
positive perfect squares less that 106 that are multiples of 24.
2.
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.
For every number 0.abc , there will be exactly one another number 0.(9  a)(9  b)(9  c)
abc
such that their sum is 0.999  1 . Since 0.abc 
, there are 10  9  8 = 720 such
999
numbers, so the sum is 360.
3.
Let N = 1002 + 992 – 982 – 972 + 962 +  + 42 + 32 – 22 – 12, where the additions and
subtractions alternate in pairs. Find the remainder when N is divided by 1000.
We may ignore 1002, since it’s divisible by 1000. Then,
992 – 982 – 972 + 962 + 952 – 942 – 932 + 922 + + 42 + 32 – 22 – 12
= (99 + 98)(99 – 98) – (97 + 96)(97 – 96) + (95 + 94)(95 – 94) –  + (3 + 2)(3 – 2) – 1
= 197 – 193 + 189 – 185 +  + 5 – 1
= 4  25 = 100.
4.
Hexagon ABCDEF is divided into five rhombuses, P, Q, R, S, and
T as shown. Rhombuses P, Q, R, and S congruent, and each has
area 2013 . Let k be the area of rhombus T. Given that k is a
positive integer, find the number of possible values for k.
(Note: 2013  44.8)
Let x denote the common side length of the rhombuses and y the smaller interior angle of
rhombus P. Then, x2sin y = 2013 and k = x2sin 2y = 2x2sin y cos y = 2 2013 cos y.
Therefore, 0 < k < 2 2013  89.6, and k can be any integer between 1 and 89, inclusive.
The number of possible vlaues for k is 89.
1
5.
Find the ordered pairs of positive integers (a, b) such that a + b = 1000 and neither a nor b
has a zero digit.
Let a = mnp and b = rst be 3-digit numbers.
mnp
p and t must add up to 10, n and s must add up to 9, and m and r must add up to 9.
+rst
Since none of the digits can be 0, there are 9  8  8 = 576 possibilities if both
1000
numbers are 3 digits.
There are two other scenarios: a 3-digit number and a 2-digit number, or a 3-digit number
and a 1-digit number. The former has 9  8  2 = 144 possibilities (the 2 accounting for
either a or b is 3-digit), and the latter has 9  2 = 18 possibilities.
In all, there are 576 + 144 + 18 = 738 possibilities.
6.

Let N   k 1 k log
1000
2

k   log 2 k  , where  x  (the ceiling function) is the least integer
greater than or equal to x, and  x  (the floor function) is the greatest integer less than or
equal to x. Find the remainder of N when divided by 1000.
log 2 k   log 2 k   0 if an only if log 2 k is an integer, and log 2 k   log 2 k   1 if
and only if log 2 k is not an integer.
log 2 k is an integer means k is a power of
2 . Since k is an integer, it must be a power of
2. If k ≤ 1000, then k = 20, 21, …., 29.
Therefore, N =
(1  2 
 1000)   20  21 
 29  
1000(1001) 210  1

 500500  1023  499477 , and
2
2 1
the remainder is 477.
7.
The polynomial P(x) is cubic. What is the largest value of k for which the polynomials
Q1(x) = x2 + (k – 29)x – k and Q2(x) = 2x2 + (2k – 43)x + k are both factors of P(x)?
Q1(x) and Q2(x) must have a root in common for them to be factors of a same cubic
polynomial. Let this root be a.
Then a is also a root of Q2(x) – 2Q1(x) = 15x + 3k, so 15a + 3k = 0 and a =
k
.
5
 k 
 k 
Substituting into Q1, we have     k  29    k  0 , k2 – 5(k – 29)k – 25k = 0,
 5 
 5 
2
2
–4k + 120k = 0, k = 30k. So, k = 30 is the largest value.
2
2
 k 
 k 
(Substituting into Q2, we have 2     2k  43    k  0 , 2k2 – 5(2k – 43)k + 25k =
 5 
 5 
2
2
0, –8k + 240k = 0, k = 30k, giving k = 30 again.)
2
8.
There exist unique positive integers x and y such that x2 + 84x + 2013 = y2. Find x + y.
x2 + 84x + 422 + 249 = y2  383 = 249 = y2 – (x + 42)2 = (y – x – 42)(y + x + 42).
So, y – x – 42 = 1, y + x + 42 = 249 OR y – x – 42 = 3, y + x + 42 = 83.
The first case implies y = 125 and x = 82, while the second case yields y = 43 and x = –2.
So, the answer is 207.
9.
Let Sk be the set of all integers n such that 100k ≤ n < 100(k + 1). For example, S4 is the set
{400, 401, 402, …, 499}. How many of the sets S0, S1, S2, …., S999 do not contain a perfect
square? (Note: 100000  316.2)
The difference between consecutive squares is (x + 1)2 – x2 = 2x + 1, which means all
squares above 502 = 2500 are more than 100 apart. Then, the first 26 sets S0, S1, S2, …., S25
each has at least one perfect square. Since 3162 < 1000000 < 3172, there are 316 – 50 = 266
perfect squares after 2500 and thus there are 266 other sets Sk after S25 that have a perfect
square. Therefore, there are 1000 – 266 – 26 = 708 sets without a perfect square.
1
1
1
1 π
10. Find the positive integer n such that arctan  arctan  arctan  arctan  .
3
4
5
n 4
ab
(Hint. It’s known that arctan a  arctan b  arctan
for all a, b such that ab ≠ 1.)
1  ab
1 1
1
1
4  arctan 7
arctan  arctan  arctan 3
3
4
11
1 1  1
3 4
7 1
7
1
46
23
arctan  arctan  arctan 11 5  arctan
 arctan
11
5
48
24
1 7  1
11 5
23  1
23
1
24
n  arctan 23n  24  π
arctan  arctan  arctan
24
n
24n  23 4
1  23  1
24 n
23n  24
 1, 23n  24  24n  23  n  47
So,
24n  23
3