SW-ARML Practice 9-22-13
In ARMLopolis, every house number is a positive integer, and City Hall’s address is 0.
However, due to the curved nature of the cowpaths that eventually became the streets of
ARMLopolis, the distance d(n) between house n and City Hall is not simply the value of n.
Instead, if n = 3kn, where k ≥ 0 is an integer and n is an integer not divisible by 3, then
d(n) = 3−k. For example, d(18) = 1/9 and d(17) = 1. Notice that even though no houses have
negative numbers, d(n) is well-defined for negative values of n. For example, d(−33) = 1/3
because −33 = 31 ·(−11). By definition, d(0) = 0. Following the dictum “location, location,
location,” we will refer to “houses” and “house numbers” interchangeably. Curiously, the
arrangement of the houses is such that the distance from house n to house m, written d(m, n), is
simply d(m − n). For example, d(3, 4) = d(−1) = 1 because −1 = 30 ·(−1). In particular, if m = n,
then d(m, n) = 0.
1.
(a) Compute d(6), d(16), and d(72).
Solution.
Factoring, 6 = 2 · 31, 16 = 16 · 30, and 72 = 8 · 32, so d(6) = 1/3, d(16) = 1, and d(72) =
1/9.
(b) Of the houses with positive numbers less than 100, find, with proof, the house or houses
which is (are) closest to City Hall.
Solution.
If n = 3km where 3 | m, then d(n) = 1/3k. So the smallest values of d(n) occur when k is
largest. The largest power of 3 less than 100 is 34 = 81, so d(81) = 1/34 = 1/81 is
minimal.
(c) Find four houses at a distance of 1/9 from house number 17.
Solution.
If d(n, 17) = 1/9, then 9 | (n − 17) and 27 | (n − 17). So n = 17 + 9k where k ∈ Z is not
divisible by 3. For k = −1, 1, 2, 4, this formula yields n = 8, 26, 35, and 53, respectively.
(d) Find an infinite sequence of houses {hn} such that d(17, h1) > d(17, h2) > · · · > d(17, hn)
> d(17, hn+1) > · · ·.
Solution
There are many possible answers; the simplest is hn = 17+3n, yielding the sequence {20,
26, 44, 98, . . .}.
(Continued) The neighborhood of a house n, written N(n), is the set of all houses that are the
same distance from City Hall as n. In symbols, N(n) = {m | d(m) = d(n)}. Geometrically, it may
be helpful to think of N(n) as a circle centered at City Hall with radius d(n).
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2.
(a) Find four houses in N(6).
Solution.
Because d(6) = 1/3, N(6) is the set of all houses n such that d(n) = 1/3, equivalently,
where n = 3k and k is not divisible by 3. So N(6) = {3, 6, 12, 15, 21, 24, . . .}.
(b) Suppose that n is a house with d(n) = 1. Find four houses in N(n).
Solution.
Using similar logic above, N(n) = {m positive integer | 3 | m}. So N(n) = {1, 2, 4, 5, 7,
8, . . .}.
(c) Suppose that n is a house with d(n) = 1/27. Determine the ten smallest positive integers
m (in the standard ordering of the integers) such that m ∈ N(n), i.e. m is in the
neighborhood of n.
Solution.
Here, N(n) = {m | m = 27k, where 3 | k}. The ten smallest elements of N(n) are 27, 54,
108, 135, 189, 216, 270, 297, 351, and 378.
3.
Compute the smallest n such that in the regular n-gon A1A2A3 · · ·An, m∠A1A20A13 < 60◦.
Solution.
If the polygon is inscribed in a circle, then the arc A1 A13 intercepted by ∠A1A20A13 has
measure 12(360◦/n), and thus m∠A1A20A13 = 6(360◦/n). If 6(360/n) < 60, then n > 6(360)/60
= 36. Thus the smallest value of n is 37
4.
Compute the largest base-10 integer A B C D, with A > 0, such that A B C D = B! + C! + D!.
(Hint: 7! = 5040, 6! = 720, 5! = 120, 4! = 24, 3! = 6, 2! = 2)
Solution.
Let A B C D = N. Because 7! = 5040 and 8! = 40,320, N must be no greater than 7! + 6! + 6!
= 6480. This value of N does not work (because B = 4), so work through the list of possible
sums in decreasing order: 7! + 6! + 5! = 5880, 7! + 6! + 4! = 5784, 7! + 6! + 3! = 5766, etc.
The first value that works is N = 5762 = 7! + 6! + 2!.
5.
Let x be the smallest positive integer such that 1584x is a perfect cube, and let y be the
smallest positive integer such that xy is a multiple of 1584. Compute y. (Note: 1584 =
243211)
Solution.
In order for 1584 · x to be a perfect cube, all of its prime factors must be raised to powers
divisible by 3. Because 1584 = 24·32·11, the smallest positive x must be 22·3 ·112 = 1452.
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But in order for xy to be a positive multiple of 1584, xy must be of the form 2a·3b·11c·d,
where a ≥ 4, b ≥ 2, c ≥ 1, and d ≥ 1. Thus y must equal 22·31 = 12.
6.
A set S contains thirteen distinct positive integers whose sum is 120. Compute the largest
possible value for the median of S.
Solution.
Let SL be the set of the least six integers in S, let m be the median of S, and let SG be the set
of the greatest six integers in S. In order to maximize the median, the elements of SL should
be as small as possible, so start with SL = {1, 2, 3, 4, 5, 6}. Then the sum of SL’s elements is
21, leaving 99 as the sum of m and the six elements of SG. If m = 11 and SG = {12, 13, 14,
15, 16, 17}, then the sum of all thirteen elements of S is 119. It is impossible to increase m
any further, because then the smallest set of numbers for SG would be
{13, 14, 15, 16, 17, 18}, and the sum would be at least 126. To get the sum to be exactly
120, simply increase either 6 to 7 or 17 to 18. The answer is 11.
7.
Emma goes to the store to buy apples and peaches. She buys five of each, hands the
shopkeeper one $5 bill, but then has to give the shopkeeper another; she gets back some
change. Jonah goes to the same store, buys 2 apples and 12 peaches, and tries to pay with a
single #10 bill. But that’s not enough, so Jonah has to give the shopkeeper another $10 bill,
and also gets some change. Finally, Helen goes to the same store to buy 25 peaches.
Assuming that the price in cents of each fruit in an integer, compute the least amount of
money, in cents, that Helen can expect to pay.
Solution.
Let a be the price of one apple and p be the price of one peach, in cents. The first transaction
shows that 500 < 5a + 5p < 1000, hence 100 < a + p < 200. The second transaction shows
that 1000 < 2a + 12p < 2000, so 500 < a + 6p < 1000. Subtracting the inequalities yields 300
< 5p < 900, so 60 < p < 180. Therefore the price of 25 peaches is at least 25 · 61 = 1525
cents.
8.
Compute the least positive integer b such that, when expressed in base b, the number 11!
ends in exactly two zeros. (Hint: the highest power k of any prime p such that pk | n! (any
n  n   n 
integer n) is equal to     2    3   ; note that this is a sum of only finitely many
 p  p   p 
nonzero summands.)
Solution.
For any integers n and b, define d(n, b) to be the unique nonnegative integer k such that
bk | n and bk+1 | n. That is, d(n, b) is the highest power of b dividing n. For example,
d(9, 3) = 2, d(9, 4) = 0, and d(18, 6) = 1. So the problem asks for the smallest value of
b such that d(11!, b) = 2. If p is a prime and p | b, then d(11, b) ≤ d(11, p), so the least value
of b such that d(11, b) = 2 must be prime. Also, if b is prime, then d(11, b) =
3
11  11   11 
11 
 b    b 2    b3   . The only way that d(11, b) can equal 2 is if the first term  b 
equals 2 and all other terms equal zero. (If 11 ≥ b2 would mean the first two terms by
11 
themselves would have a sum of at least 3.) The smallest prime b such that   = 2 is
b
b = 5.
9.
10,000
Let X be the number of digits in the decimal expansion of 1001,000 , and let Y be the
100,000
number of digits in the decimal expansion of 100010,000
. Compute log X Y  (= the
greatest integer less than or equal to log X Y .) (Hint #1: The number of digits of a positive
log Y
integer n is log n  1 .) (Hint #2: Change-of-Base Formula: log X Y 
)
log X
Solution.
10,000
Because 1001,000
 102 
1,00010,000
 1021,000
10,000
, X = 2· 1,00010,000 +1. Similarly,
Y = 3· 10,000100,000 + 1. Therefore,
 log Y   log  3 10000100,000  1   log  3 10000100,000    log  3  log 10000100,000  



log X Y   
≈
 log X   log  2 100010000  1   log  2 100010000    log  2   log 100010000  




 log 10000
   100, 000  log(10000)    400, 000    40   13

10000
 log 1000
   10, 000  log(1000)   30, 000   3 


100,000
10. Compute the number of quadratic functions f(x) = ax2 + bx + c with integer roots and integer
coefficients whose graphs pass through the points (0, 0) and (15, 225).
Solution.
Because the graph passes through (0, 0), conclude that c = 0 and f(x) = ax2 + bx.
Then, f(15) = 225 ⇒ a(15)2 + b(15) = 225a + 15b = 225, from which b = 15 − 15a.
On the other hand, f can be factored as f(x) = ax(x + b/a), so if the roots are integers, b/a
must be an integer. Divide both sides of the equation b = 15 − 15a by a to obtain
b/a = 15/a −15. Thus 15/a must be an integer, and a  {±1,±3,±5,±15}.
Because b = 15 − 15a is linear, each of these values for a yields a unique integer value for b,
so there are 8 such ordered pairs.
FYI. The values of a, b, and the
nonnegative root are given in the table:
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