I
MATHCOUNTS
1993-94
I National Competition I
«A
,.
Answer Key
.s,
sum-a«».,
-xi
Optional units are indicated in parentheses. Standard
abbreviations have been used for units of measure. Complete
words or symbols are also acceptable. Square/cube units
may also be expressed as units? or units3.
MATHCOUNTS is a cooperative project of the National Society of Professional Engineers,‘ the CNA Insurance
I
Companies, the Cray Research Foundation, the General Motors Foundation, the Intel Foundation, Texas Instruments
Incorporated,the National Council of Teachers of Mathematics, and the National Aeronautics and Space Administration.
Sprint Round
1, Z 5_ 210 13_ 19(integers) 17_ Oor-—% 25, 150
N.
(square inches)
18. 65
14. O
7, ($)32.50(do1lars) 26.___......__._.._..__....__(1' °>5-()()
(d°‘1‘*“‘)
19_ 2(units)
15. 4
2. fa
8. 32
20. -4 .
27. 3.6 (% or percent
128“ 9. 194
3.
(cubiccentimcters) 21. 8
%_
22.__6._<§.i_<.*:=«2>_______. 23_ 1,920 (tiles)
8__1LOr
16.______l2_____12.___
4 .
17 (1,.Get )
(cubiccentimeters)
10 1 +‘/3 (units)
23.___1_.l;.L_._...__
290 625
11. 4
80 (°or degrees)
5
12.__§__%__........__.__ 30. <$> 68,000
MC (dollars)
Target Round
7. 27 + 2.2511;
(inches)
2. 12 (teachers) 4_ 12 6_ 100 (° or degrees) 8_ 1 12.5 (square meters)
1ban1Round
2+¢T§
L -1
35
2.07 (inches)
23
1% (minutes)
10. 6 -
3 3\/Z (inches)
6666
Countdown Round
1_ d or $80 Ber day 10, \/T5‘ (centimeters) 19. *?rs2 or 9:?-
':w~¢W;/31-r (square units)
9 (square units) 11.
D or
(-31,-)“4 0
2_ 20.
12. 45 .5 (square units) 21. *3"
13.
22. 24
14.
4. 2(~’7+x/7)
or 2\/'7 +2\/?:
5_ 8 (feet)
23_ 35 (zeros)
15. 0.189
6_ 48 (numbers) 24_ 12 (gaths)
16.
7‘ 4V/3 (units)
17. y=—x+1
ory=-—1x+1
36 (feet)
25. 343 (cubic centimeters)
?‘ .
Masters Round
l ,
i3, .4
Part I
The evolution of the various configurations are given beiow. We stop at the point that we find a
previous configuration. Continued shuffling will produce cyclic behavior.
1. (3)->(2. 1)->(2. 1)
2. (4)->(3.1)->(2.2)->(2,1.1)->(3.1)
3. (5)—>(4,1)->(3,2)->(2,2,1)->(3,1,1)—+(3,2)
(6)—-=-(5,1)—>(4,2)-—->(3,2,1)->(3,2, 1)
(7)~>(6, 1)->(5,2)-a-(4,2,1)->(3,3,1)—>(3,2,2)e>(3,2,1,1)->(4,2,1)
(8)->(7, l)—->(6, 2)->(5, 2, 1)—->(4, 3, 1)->(3, 3, 2)—>(3, 2, 2, 1)-.>-(4, 2, 1, 1)->(4,3, 1)
(9)-—>(8, 1)—>(7, 2)—->(6,2, 1)—>(5,3, 1)->(4, 3, 2)->(3, 3,2, 1)->(4, 2, 2, 1)->(4, 3, 1, 1)->(4, 3, 2)
(10)—>(9, 1)—>(s, 2)->(7, 2, 1)->(6, 3, i»)—>(5, 3, 2)-+(4, 3, 2, 1)——>(4, 3, 2, 1)
Part II
4. The starting heights of l, 3, 6, and 10 lead to the stationary configurations (l), (2, l), (3, 2, l), and (4, 3, 2, l).
5. The stationary configurations are all of the form (n, n l, n 2, — -s
3, 2, 1).
. . .
,
6. Using the observations above, it is clear that certain configurations of the form (n, n 1, n 2, 3, 2, 1) -— —
. . .
,
are stationary. It is reasonable to assume that all configurations of this form are stationary using
the following argument: given the configuration (n, n l, n 2, —
3, 2, 1), when we remove
——
..
. .
,
one chip from each stack, we are left with stacks of height 1 through 12 l, but, since we had n -
stacks before the removal, we now have n chips to form a next stack. This results in an identical
configuration. Thus the cycle has length 1 and the configuration is stationary.
Based on the preceding observations, the next stationary configuration should occur when the
stacks are in the form (5, 4, 3, 2, 1). Therefore a single stack of height 15 should be the next stack
that evolves into a stationary configuration.
The starting heights for all single stacks that evolve into
stationary configurations are the
numbers (1, 3, 6, 10, 15, 21, 28, 36, .). . . These numbers
called the triangular numbers and
are
are of the form
T” = 1’5-"—2+-9-
wheren is a positive integer and
T” is the 1'2"‘ triangular number.
A rigorous proof for this can be developed but is beyond the scope of this discussion.
Part III
9. If you visualize the cutting and stacking that is being carried out, you see that the white chips
arebeing shuffled just as if the red chip was not there. But with the red chip included, there
are six chips being shuffled and they evolve into the stationary configuration (3, 2, 1), with the
red chip on top. Let us describe a stack containing a red chip by writing “c-+1” to indicate a stack
with c white chips and one red chip on top of them. Eventually the six chips reach the cycle
(2+l, 2, 1) -> (3, 1+1, 1) -> (3,. 2, 0+1)-> (2+l, 2, 1). If we disregard the red chip, we observe the
cycle (2, 2, 1) —> (3, l, 1) ——> (3, 2) -2» (2, 2, 1). We can visualize this cycle of five chips as being the
stationary cycle of six chips except that one of the six chips is invisible (or red but not counted)
and this invisible chip moves among the stacks in a cycle of length 3.
10. If we start with nine chips we add a red chip to get ten (a triangular number). Eventually the chips
evolve into the cycle (3+1, 3, 2, l)—>(4, 2+1, 2, 1)->(4, 3, 1+1, I)-a~(4, 3, 2, 0+1)->(3+1, 3, 2, l).
As we did in the above argument, we can visualize this cycle as being the stationary cycle for 10
chips except that one chip is invisible and moves among the stacks in a cycle of length 4.
11. A repeatedly shuffled, single stack of height 14 will evolve into a cycle of length 5.
12. There are two configurations using exactly 18 chips that are cyclic of length 2: (6, 4, 4, 2, 2) and
(5.5,3,3.1.1)-
