Mathcounts Chapter Competition Solutions Worksheet - Middle School - 2014 Page 12
ADVERTISEMENT
1
2
3
4
5
6
7
8
9
10
11
12Q is a consonant worth 2.
U is a vowel worth 1.
E is an adjacent vowel worth 2.
U is an adjacent vowel worth 4.
E is an adjacent vowel worth 8.
This problem can be solve using mass-point
I is an adjacent vowel worth 16.
geometry to solve this problem. Let D serve
N is a consonant worth 2.
as the fulcrum balancing segments BC.
G is a consonant worth 4.
Likewise, E and F each serve as the fulcrum
QUEUEING is worth a total of 2 + 1 + 2 +
balancing segments AC, and AB,
4 + 8 + 16 + 2 + 4 = 39 points.
respectively. So the product of the mass at B
Following are the point values for each letter
and the length of segment BF must be equal
in SYZYGY.
to the product of the mass at A and the length
S is a consonant worth 2.
of segment AF. Since AF = 3 and BF = 2, we
Y is an adjacent consonant worth 4.
assign a mass of 3 to B, which leads to
Z is an adjacent consonant worth 8.
assigning a mass of 2 to A. Now since the
Y is an adjacent consonant worth 16.
mass at C must be such that the product of
G is an adjacent consonant worth 32.
the mass and the length of segment CD must
Y is an adjacent consonant worth 64.
equal the product of the mass at B and the
SYZYGY is worth a total of 2 + 4 + 8 + 16 +
length of segment BD. Since BD = CD = 2,
32 + 64 = 126 points. That gives us an
we assign a mass of 3 to C. Therefore, the
absolute difference of 126 – 39 = 87 Ans.
mass at E (the sum of the masses at A and C)
is 5.
9. Using each of the digits 1 to 6 exactly once,
how many six-digit integers can be formed
that are divisible by 6?
An integer is divisible by 6 if it is divisible by
both 2 and 3. So, for an integer to be divisible
by 6, it must be even (divisible by 2), and the
sum of its digits must be divisible by 3. Well,
Finally, if G serves as a fulcrum balancing
1 + 2 + 3 + 4 + 5 + 6 = 21, which is
segment BE, the product of the mass at B
divisible by 3. But only half of the integers
and the length of segment BG must equal the
composed of these six digits are even. There
product of the mass at E and the length of
are a total of 6! = 720 different ways to
segment EG. While we don’t know the exact
arrange the digits 1, 2, 3, 4, 5 and 6 to form a
lengths of segments BG and EG, we do know
six-digit integer. Therefore, the number of
that in order for segment BE to remain
these integers that are divisible by 6 is 720/2
balanced on G, the segments must be in the
= 360 Ans.
ratio BG/EG = 5/3 Ans.
10. Points D, E and F lie along the perimeter of
∆ABC such that AD, BE and CF intersect at
point G. AF = 3, BF = BD = CD = 2 and
AE = 5. Find BG/EG.
ADVERTISEMENT
0 votes
Related Articles
Related forms
Related Categories
Parent category: Education