i N filil
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Q1997 Wright Group/McGraw-Hill 24
PCopyright
uzzle
In the Balance is a trademark of Creative Publications.
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Puzzle
. 8.
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. . . 18
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Puzzle 23 . .
Puzzle 22 . .
Puzzle 6 . . . . . . . . . . . . . 14
Puzzle 7 . . . . . . . . . . . . . 16
Puzzle 21 ...
Puzzle 5 . . . . . . . . . . . . . 12
Puzzle 20 . .
Puzzle 4 . . . . . . . . . . . . . 10
ALL rights reserved. No part of this publication may be
reproduced, stored in a retrieval system, or transmitted,
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photocopying, recording, or otherwise, without the
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Puzzle 19 ...
8
Puzzle 3......... .
Puzzle 18 ...
Puzzle 17 . .
Puzzle 1.. .
4
Puzzle 16 . . .
Strategy SL ggesti ons . . . . . 3
.......
Puzzle 15 . .
Puzzle 14 . .
Puzzle 13 ...
Puzzle 12 ...
Samples and Solutions . . . . 2
Directions . . . . . . . . . . . . 1
Acknowled(jments........ 1
To the Tea ( her .......... 1
6
Cover Design: Al Golden
Translation Services: Sybil de Munsuri
Production Services: Dovetail Publishing Services
Production Coordinator: Kate Rapson
Designers: Gregg McGreevy, Meg Saint-Loubert
Editor: John Lanyi
Senior Editor: Glenda Stewart
ONTENTS
Puzzle 2.. .
dents is to have them design original puzzles.
One suggestion to extend the thinking of
these puzzles for highly able or motivated stu-
more difficult mathematically, a puzzle is more
difficult because its complexity requires more
sophisticated organization.
will always find certain puzzles easier or harder
than others. Sometimes, rather than being
The puzzles are arranged in an approximate
easy-to - harder progression. Individual students
by exploring relationships between numbers.
My own students have completed them independently and in cooperative groups.
These activities allow students to apply
mathematical reasoning skills and problemsolving strategies and to develop number sense
Lou Kroner
Cincinnati, Ohio
Thanks to Tandy Lingo (J. F. Dulles School),
Cheryl Tallman (Our Lady of Visitation School)
and Todd Forman (Summit Country Day School)
for their prof(,ssional encouragement and
feedback.
Most of all, thanks to my family-Mary Kay,
Katie, and Alin. Without their patience and
support, this project would have faded away
long ago.
colleagues.
As I worked on this project, I received lots
of encouragement from family, friends, and
Mobiles have always fascinated me. As a
child, I watched Alexander Calder's mobile "20
Leaves and an Apple" spin a leisurely course in
the air currents at the Cincinnati Art Museum.
But this book is not about Calder's creative
genius or the delicate balance of the kinetic
sculptor's art. The mobiles in this book were
designed with math teachers and their pupils
in mind.
AC NOWLEDGMENTS
TO THE TEACHER
,
DI
^
does not affect th
and right arms. Al
own definite wei
• A piece hanging d
• All weights are ei
tive, whole numbe
• There are no "use
clue says that the
ple of the triangle
assume that the t
• Each shape has a
within a puzzle, an
• The right and left
beam must balanc
Use your sense
experience with nu
you discover the val
work, remember the
In these puzzles,
number value or "w
zles, you must mak
of the mobile balan
as a sculptor must
mobile.
h
Solution
Each half weighs half of the 24,
or 12. Therefore, each of the Triangles must weigh 6. That also means
that there could be the folloving
Since the whole of the left side is
taken up by the circle, it must weigh
8. Since half of the right side, which
also weighs 8, is taken up by the
Clue: , - Q = 2
Total = 24
Solution
Total = 16
SAMPLES AND SOLUTIONS
The problem here is that, because of the diam
the values of the two arms. We do know that eac
weigh 15. But 15 cannot be halved to suit the rig
each of the arms must weigh 12, 8, or 4. (Since th
Solution
Clues: ®-1 = ♦ 4
Total = 31
2
3
4
5
6
7
8
9
12
10
8
6
4
2
Using the clue provided, students can now eliminate all but three possible combinations. Of course, a multi-columned table may be necessa y or
helpful for more comp
Clue: 0+w<14
Total = 20
1
18
16
14
0- -
Although each person's strategy for solving these problems will d-fer,
one that I have found helpful is the If-Then table. Using this simple
strategy, students prepare a table that lists all the possible combinations
of weights. This table helps students organize their work, maintains a
record of their trials, and presents the weights in a visual format. HE re is
a sample:
If-Then Table
STRATEGY SUGGESTIONS
1. As mentioned above, there are severalsolve these puzzles. The solutions that follow a
sibilities. An attempt has been made to include
so you'll notice that some of the solutions see
white others seem more involved. Sometimes,
putting solutions into word form makes a soluti
2. To identify shapes from the mobiles in th
convention has been used: R indicates right ar
and C indicates center arms. To get to a shape l
take the right arm from the top, then the right
right arm of that arm, and then (finally!) the l
Two Final Notes
Another important strategy involves learning
sents an even or an odd number. Discovering (
ships like "an even number subtracted from an
even number" will help as a pupil analyzes a pr
strategy.
Even or Odd Number?
Name
6. Knowing the above shapes forces circle (7) in several locations
and hexagon (16) in L.R.
5. Knowing pentagon forces rectangle (10) in RLLL.
4. Knowing trapezoid forces pentagon (9) in LLLR.
3. Knowing triangle and ellipse forces square (5) in LLRR and
trapezoid (8) in LLRLR.
2. Halving the weights of the arms, you learn ellipse (11) in F LRL.
subtract the triangle weight from the puzzle weight (356). You
know that triangle is even (subtracted from an even numb(!r,
356, it yields a number that can be further halved) and less
than 5 (given in clue). The value 2 doesn't work because, when
subtracted from 356, its difference (354) can only be furth ?r
halved one time (177). So you force triangle (4). (356 - 4 =
352. This difference can be halved to 176 to 88 to 44 to 2
to 11.)
1. To determine the total weights of the two main arms, you nust
Solution
PUZZLE 1
Key
9
L = Left arm
C = center arm
R = right arm
0=
4
Values
0
Sample
Q =
o
q
right arm of that a
Left arm of that ar
RRRL = (from the
= 10
= 11
Name
5. Knowing this, you can list possible diamond:e[lipse:penta ]on
using information from step 2): 3:6:9 and
4. In RRL, you learn that diamond + diamond = triangle + hexagon.
Using this knowledge, return to the final equation in step 3 and
substitute diamond + diamond for the triangle + hexagon, leaving one pentagon equals three diamonds.
3. In LR, you see that (rectangle + pentagon) = (diamond +
triangle + hexagon + rectangle). By eliminating the rectargle on
either side, you learn that pentagon = (diamond + triangle +
hexagon).
shape greater than 11.
2. From the clues, you know that two diamonds equal one ellipse.
Knowing this, list diamond:ellipse possibilities: 3:6 and 6:12. If
you continue any higher, you once again have more than )ne
1. From the clues, you know that two diamonds equal three trapezoids. Using an If-Then table to list those possible combinations, you get these diamond:trapezoid values: 3:2, 6:4, 1;'.:8.
Continuing any higher would violate the clue that only oni!
shape is greater than 11.
Solution
PUZZLE 2
R = right arm
Key
0 =8
0=3
Values
A
=7
® _
=6
Sampl
q
RRRL = (from t
I
0
10. Knowing other shapes from above forc
9. Knowing hexagon, you learn star (30)
8. Knowing ellipse and hexagon forces tr
triangle (5) in RRRR.
7. Knowing the total weight at RLL and
mond, you learn hexagon (1) in RLR.
3J1Ijl U1.flyJUUU00000oAl 11^1J 0J 0 0 ) J J J 1911
Name
Discover the
PUZZLE 3
0 1 1 I I I'II
5. Knowing the above shapes, you learn that the combined wt?ight
of square plus trapezoid in RR equals 20. Using an If-Then table
to list combinations that are still possible for the square:tr,ipe
4. Knowing rectangle forces ellipse (5) at LRLL.
3. Knowing pentagon forces rectangle (4) at LRRRL.
2. From the breakdown of the weights on LR, you can learn that
pentagon must be an even number (when subtracted from an
even weight, it yields an even weight). That, when used w th
the second clue, forces pentagon (2). That in turn forces d 1amond (3), circle (6), and hexagon (12).
A second method for Step 1 might begin in this manner: Square
must be an odd number; when subtracted from 147, it yields an
even number. No odd number less than 19, when subtracted
from 147, leaves a number that can be redivided as many -imes
as are needed in LR.
diamond:circle:hexagon combinations in RL: 2:4:8, 3:6:12.
(Because of the given clues, diamond cannot be 1-it weighs
more than something else-and hexagon cannot be 16.)
Solution
1. Begin by using an If-Then table to list possible
PUZZLE 3
IJ
0=2
Values
I =5
=3
[]=1
®=
7. Knowing square and trapezoid forces tri
The first difference can be halved only
67). But you know that, in LR, the rem
able to be halved four more times. The
be halved any further. But the other p
uation: 128 can be further divided to fi
You now know square (19) and trapezoi
147 - 19 = 128
147 - 16 = 131
147 - 13 = 134
position in the center arm. If you subt
ties, you [earn
3 . a 3J tplDII
l v0 0ki
Name
7. Knowing the above forces pentagon (5) at LL and star (9) at L.R.
6. Knowing diamond, you return to the information in step i . Only
circle (3) and rectangle (10) will fit the clue given.
5. In only one of the possible weights (in step 3) was triangle
even. So you know triangle (8) and diamond (7).
4. At LR, the total must be an even number since it can be evenly
divided at LRR and LRL. If LR is even, then so is LL. But i t LL, a
triangle is subtracted leaving an even number (it is split
between two pentagons). Therefore, triangle must be ever .
3. Knowing the above two possibilities, you list possible
diamond:triangle combinations at RL, and find 7:8 and 1: 11.
2. Using an If-Then table to list possible circle: rectangle conibinations at RR, you find 3:10, 7:8, 11:6, 15:4, and 19:2. But only
3:10 and 7:8 could fit the clue given. In the other combinations, circle cannot be subtracted from rectangle.
8
R = right arm
L = left arm
C = center arm
Key
5
A=
1. Circle (at RR) and diamond (at RL) must be odd and recta igle
(at L) must be even.
Sample
* =
RRRL = (from th
right arm of that
left arm of that a
♦ = 9
7
1IItoI1 v L1 1v 0 0 0 it 0 kv
Values
PUZZLE 4
I 1v01.100 Ll1WI1
Solution
j ► )jjjLlj
Name
6.
Knowing trapezoid and diamond forces hexagon (35) on Ll.
5. Knowing star and circle forces diamond (9) on RLLL and ellipse
(10) on RLL.
4. Knowing trapezoid forces star (5) on RLRR and that forces circle
(4) on RLR.
3. Knowing rectangle forces trapezoid (12) on RRRL.
2. On LRRR, you know that the triangle (7), the rectangle, ar id
the two squares weigh 14. That gives you three sets of po;sibilities for the rectangle and the squares: 5 + 1 + 1, 3 + 2 + ;', or
1 + 3 + 3. The clue given eliminates two possibilities and Forces
square (3) and rectangle (1).
12
C = center arm
R = right arm
L = left arm
Key
= 10
► =7
1. Breaking down the weights of the arms into halves, you learn
the values of the triangle (7) in RRLRL and the pentagon ;14)
in RRLL.
Values
14
Sample
0=
W =
RRRL = (from th
right arm of that
left arm of that a
S =35
0=
I §'II I j JJLf I II 11vIAAq
Solution
PUZZLE 5
I jjjj1. 0 1 1. W1O0 .1111 t
Name
3^ t1 tv3
PUZZLE 6
3 U001.
6. Knowing the above forces pentagon (4) at L. From above, you
know that RL equals 46 and, therefore, R equals 92. You a so
know that LR equals 44, and therefore so does LL. This he ps
5. Knowing circle forces square (21) at RLL.
4. Knowing diamond and rectangle forces ellipse (11) at RLR t and
circle (1) at RLR.
3. Knowing diamond forces rectangle (5) at RRLL and hexagcn (10)
at RRLR.
2. Using an If-Then table to list possible diamond:RRL combinations, you find 2:22, 4:21, 6:20, 8:19, .... But RRL is h;ilved
(RRLR) and then halved again (at RRLL), limiting the possibilities to 6:20, 14:16, 22:12, 30:8, and 38:4. But a diamonc value
of 14 or greater would weigh too much for RLRL. Therefor!, you
force diamond (6).
tracted from even numbers, they each yield numbers that can be
halved. Circle (RLR) and square (RLL) are odd because, wl en
subtracted from an odd number (23), they each yield eve i
numbers.
C = center arm
R = right arm
L left arm
Key
=2
Sample
• =
hr
Liu
RRRL = (from th
right arm of that
left arm of that
q = 21
0 = 1
A
M °5
0-6
Values
0 l9I,31, 1 1 1.i 1. dl_11_ ol_0J1vLV1 tv
1. Pentagon (L) and diamond (RR) are even because, when 5 ub-
Solution
atS'
Name
PUZZLE 7
5. To figure possible diamond values at L, notice that L is halved
four times after the diamond. Using an If-Then table to list
possible diamond:L combinations, you find 1:52, 3:50, 5:48,
7:46. .... Of the possible L weights in this series, only 48 and
32 can be halved four times. Therefore, diamond must weigh 5
or 21. The second clue tells you that diamond is prime, forcing
diamond (5).
4. Since LLRL can be halved, it is even. You know square is odd
(step 2) and star is even (step 3). Therefore, the totals of LLRLL
and LLRLR (the circle) must be odd.
3. LLR can be halved, so it is even and so must be LLL. Therefore,
star must be even because when subtracted from an ever number at LLL, it yields an even number.
2. Since LL can be halved, it is even and so, therefore, mus: be LR.
If LR is even, then square must be odd because, when acded to
the ellipse (which is odd), it yields an even number.
R = right arm
L = left arm
C = center arm
Key
D = 17
= 23
♦=5
Values
Sampl
a
RRRL = (from t
right arm of tha
left arm of that
=3
11. Knowing the above forces triangle (1
10. Knowing hexagon forces rectangle (7)
9. Knowing trapezoid forces hexagon (4)
1. Each main arm weighs 53 (clue 1). Diamond, triangle, and
ellipse are odd because, when subtracted from 53, each 0elds
an even number.
8. Knowing ellipse forces trapezoid (9)
Solution
PUZZLE 7
Name
6. Knowing diamond and triangle forces ellipse (9) in RRRR.
5. Knowing diamond and square forces rectangle (7) in RRLR
4. Knowing the above forces trapezoid (14) in RLR.
3. Knowing square and triangle forces diamond (11) in LLLL ind
circle (4) in LLRL.
2. Halving the arms, you learn the value of hexagon (23) in ZLLR
and pentagon (10) and triangle (5) in RLLL.
total, it yields an even number) and less than 19 (clue).
Therefore, square could be 1, 3, 5, 7, 9, 11, 13, 15, or 17 When
subtracted from the total, only one of these (3) will leave a difference (368) that can be halved four more times, as seer in
the arms of the right side. Therefore, square equals 3.
1. Square is an odd number (when subtracted from the odd puzzle
Solution
PUZZLE 8
=3
a
R = right arm
L = left arm
C = center arm
Key
=9
4 = 11
0
Values
RRRL = (from th
right arm of that
left arm of that a
Sample
Q=
9
=4
Q=
5=23
Name
9. Knowing the above forces circle (44) at RRR and trapezoid (40)
8. Knowing pentagon forces square (3) at LLLLRL.
7. Knowing the above forces diamond (4) at RLRRR and penlagon
(6) at LLR.
6. Knowing hexagon forces rectangle (14) at LLLLL.
5. Knowing triangle forces hexagon (2) at LLRRL.
Therefore, you learn triangle equals 11.
4. Using the possible rectangle values from step 3, list possible
rectangle:triangle combinations at LLLR: 14:11, 10:13, an l 2:17.
But, at LLRR, the possible triangle values of 13 and 17 pave
too great. (They would exceed the known weight of 72 at LLR.)
3. In LLLLL, list possible rectangle:hexagon combinations: 1'F:2,
10:4, and 2:8.
2. Hexagon is even (clue). Trapezoid (RL), circle (RR), pentaton
(LLR), and rectangle (LLLLL) also are even because when subtracted from an even weight, they each yield even weight;.
Sample
M
q
1 =
RRRL = (from th
right arm of that
left arm of that
. = 11
^r ^ r r,r^r^r r
R = right arm
L = left arm
C = center arm
Key
0 =40
0
=9
Values
u^r.r.^.^ ► ^ty
1. Halving the weights of the arms, you learn ellipse (9) in l LLLRR.
Solution
PUZZLE 9
JUUJ
Name
6. Knowing the above forces square (3) at LLL and diamond 9) at
5. Since you have only two possibilities, you might try a "guess
and check" strategy: Trying ellipse as 24 would make the ;quare
at LLL weigh 6. But that leaves a remainder of 9 to be evenly
split between the diamonds LLLL and LLLR. Eliminating the
24:15 possibility forces ellipse (12) and hexagon (21).
4. But ellipse weights of 48 and 36, with related rectangle weights
of 16 and 12, would prove too heavy at R, so you are left with
ellipse:hexagon possibilities of 12:21 and 24:15.
3. Using an If-Then table to list possible ellipse:hexagon weights
at LL and LLR, and knowing (from the clue) that ellipse must be
divisible by both 3 and 4, you find 12:21, 24:15, 36:9, and
48:5.
2. Ellipse (LL), triangle (LR and R), and rectangle (RL) must )e
even. When subtracted from even weights, they each yielc
weights that can be halved.
weight of LL and LR (54 each).
1. Breaking the weights of the arms into halves, you learn tF e
Solution
PUZZLE 10
R = right arm
L = left arm
C = center arm
Sample
•
RRRL = (from th
right arm of that
left arm of that
Q=7
I =4
Key
Q = 21
0 = 12
Values
10. Knowing ellipse and rectangle at RLR
9. Knowing circle and diamond at RRL for
and RRRL.
t1JJUJU00 I 'll _ I II I I jJ 111 t L,1 ^VA 0 0 0 00
Name
PUZZLE 11
J
U U
UU
J
PUZZLE 11
5. Knowing the above forces ellipse (6) at CL and circle (29) a : C.
tagon (8).
4. Using square and triangle possibilities from above, list possi ale
triangte:square:diamond:pentagon combinations at R and fird
3:19:11:8 and 7:17:12:5. Knowing, from the clue, that pentagon
is even forces triangle (3), square (19), diamond (11), and pen-
3. Using these triangle possibilities, list possible triangle:squar!
combinations that fit at L. Since square is odd, 3:19 and 7:17
are the only two possibilities.
and 9:5:18.
2. To work at CR, triangle must be 1, 3, 5, 7, or 9. If it weighed
more, it would prove too heavy for the total of the center arns.
Using an If-Then table to list possible triangle:circle: ellipse
combinations at C, you find 1:37:2, 3:29:6, 5:21:10, 7:13:14,
0
UU U UU U UUU 'U UU U
1. Triangle, circle, and square are odd because when subtracted
from the odd total, they each yield even numbers.
Solution
"U
U
C = center arm
R = right arm
L= left arm
Key
6
. =3
Values
U U U U U
= 19
J
U
Sample
0 =
1_1 U1
RRRL = (from the t
right arm of that a
left arm of that ar
QQ = 29
0
UU
Name
PUZZLE 12
5. Listing possible square:pentagon:circle combinations that work
at L and LL, you find 2:6:3, 10:4:2, and 18:2:1. (Other e'en val-
ellipse (4).
4. Extending the above to include ellipse, you find the folloNing
possibilities-21:26:13:4, 29:22:11:9, 37:18:9:14, 45:14:7:19,
and 53:10:5:24. The first clue forces diamond (13) and forces
3. The least that C can weigh is 5. You know that diamond is half
of R and that R is half of the total minus the center arm, C.
Listing possible C:R:diamond combinations, you get five working
possibilities-21:26:13, 29:22:11, 37:18:9, 45:14:7, and
53:10:5. (Combinations 5:34:17 and 13:30:15 were excluded
because diamond weighs more than C!)
2. The right arm must be even because it is halved. If R is even,
then L is even. Square at L must be even because, when s ubtracted from an even number, it yields an even number.
be odd.
1. The total of the center arm (C) must be odd because, when subtracted from an odd total (73), it yields a number that can be
halved. Since C is divided below the diamond, diamond mast
Solution
I - ^ - - -
Key
3
13
R = right arm
L = left arm
C = center arm
0-
Values
=4
RRRL = (from t
right arm of th
left arm of that
Sampl
Q
L
Name
7. Knowing ellipse and star forces diamond (13) in RRR, and knowing diamond, you can force square (11) in RRL.
6. Knowing pentagon and hexagon forces ellipse (2) in LRRF: and
star (9) in LRR.
5. Knowing pentagon forces circle (8) in LLLL and LRLR.
4. Knowing hexagon forces rectangle (6) in LLLR.
3. Using the possible hexagon weights from step 2, you list the
possible hexagon:pentagon combinations that would fit at
LLRL-9:1, 7:2, 5:3, 3:4, and 1:5. Only one of these com Ainations fits the clue given, forcing hexagon (5) and pentagon (3).
2. In LLLR, list possible rectangle: hexagon combinations that total
11-2:9, 4:7, 6:5, 8:3, and 10:1.
1. In LR, rectangle must be an even number because when sibtracted from the 44 that that arm of the mobile weighs, you get
a number that can be further divided into halves.
Solution
PUZZLE 13
0 J J U J J0 J J J J U U J 9 A I I I,
R = right arm
L = Left arm
C = center arm
Key
=2
• = 5
Values
Sampl
0
RRRL - (from t
right arm of th
left arm of that
9
9 U ! U U ^9 , 9, 1.9, ' 9, 9 ' t
Name
PUZZLE 14
4?
Extending step 3, list possible circle:square:triangle comt inations (at LL) and find 2:30:1, 6:28:3, 10:26:5, and 14:24:7.
and 14:24.
equals two diamonds (step 2). Using an If-Then table to list
possible circle:square combinations, you find 2:30, 6:28; 10:26,
RR is divided into two equal parts; it must then be an ev?n
number. Therefore, RL, the square, must also be even because it
forces the first combination and you learn circle (2), squire
(30), triangle (1), diamond (15), and trapezoid (13).
2:30:1:15:13, 6:28:3:14:8, and 10:26:5:13:3. (A triangle value
of 7 would force a negative value for trapezoid.) The cluf.
5. Building on the triangle weights in step 4, list circle:square:
triangle:diamond:trapezoid (at LRR) combinations and fir d
4.
3.
2. Diamond at RRL equals half of square.
4, 6, 8, 10, 12, or 14.
ber, it yields an even number. Circle cannot be more than 14
because it would be too heavy at LRR. Therefore circle ec uals 2,
1. Circle is even at R because, when subtracted from an eve i num-
Solution
PUZZLE 14
=2
R = right arm
L = left arm
C = center arm
Key
• = 13
0
Values
0
Sampl
Q
RRRL = (from t
right arm of th
left arm of that
= 14
[3 E3 0
t lIL)Li UUI)UU0U , U U U U U U JUUUUUUVUL)L)0L)l.
Name
Knowing the above forces diamond (27) at RRLR and stagy (20)
6.
7. Knowing diamond forces hexagon (1) at LRLL.
at RRR.
Knowing rectangle forces pentagon (5) at RRLLR.
5.
4. Knowing trapezoid forces circle (6) at RRRRR and triangh (11)
at LLLRL.
3. Knowing ellipse forces rectangle (10) at RLLRL.
2. Using the clue, you learn ellipse (4) and trapezoid (3) in
LLLRRR.
square (7) in LRLRL and LLLRRL.
1. Breaking down the weights of the arms into halves, you learn
Solution
PUZZLE 15
R= right arm
L = left arm
C = center arm
Key
6
=7
•=
0
Values
Sampl
Q
RRRL = (from
right arm of th
left arm of that
p= 11
O= 4
10 JUU0000 0 UUUUUU0 U1. !000U0 0 UUUUU0 U0
Name
Do=
J
PUZZLE 16
7. Knowing the above forces diamond (24) in RLR.
6. Knowing the above forces triangle (2) in RLLR.
5. Knowing trapezoid forces hexagon (6) in RLLLR.
4. That leaves you with four possible weights for the portion of LR
below the circle: 40, 48, 56, and 64. Subtracted from the weight
of LR (68), you would get these four differences: 28, 20, 12,
and 4. Given the second clue, you learn circle (28). Work- ng
backwards, you force rectangle (10) and trapezoid (3).
these four: 3:10, 7:12, 11:14, and 15:16.
3. Keeping in mind that (a) the weight of LR equals 68, (b) ellipse
equals 17, and (c) the sum of trapezoid plus rectangle is an odd
number, list possible trapezoid: rectangle combinations and find
2. In LRL, if ellipse is an odd number (17), then trapezoid must
also be odd because they add up to an even number.
R = right arm
L = Left arm,,,
C = center arm
Key
0=3
X34
Values
Sampt
A►
0
RRRL = (from t
right arm of th
left arm of that
®=6
,= 17
U U U U U U U.1 U U U 11 11 U U U 0 4 ► U I 0 U U U U U 1 U 1 U 1
1. Breaking down the weights of the arms, you learn square (34) in
LLL and ellipse (17) in LLR.
Solution
9 ^,.)
Name
00
41 .
A 4^
U)
U
V U
UU
2
L = left arm
C = center ar4n
RRRL.
zoid ;6) in
Knowing triangle and hexagon, you can force trape
RRLRR and then circle (10) in RRLRL.
R = right arm
Key
rz=15
0
q
e (2) in
Knowing triangle and pentagon, you can force squar
the ellipse (1) in
7. Knowing hourglass and other shapes forces
LRRL.
lass figure ('.5) in
6. Knowing the above shapes forces the hourg
(40) in RLLL.
5. Knowing the above shapes forces rectangle
4.
3.
ble by 3.
2. Given the clue, you know that triangle must be divisi
i even
By analyzing RLR, you learn that triangle also must be ai
at can
number. (When subtracted from 88, it yields a number tF
12, and
be divided in half-an even number.) You eliminate 6,
differ18 because, when subtracted from the 88 in RLR, their
le
ences cannot be further subdivided to fit RLRLL. Only triang
in
(24) can be subdivided to fit RLRLL, forcing diamond (8:
RLRLL and hexagon (4) in RLRR.
Samp
Q
'
left arm of th
RRRL = (from
right arm of t
D= 6
A=24
V110 V 11 ki U Al U ► U i
Values
10 11.1 1-1 U
the value of thf- pen1. Breaking down arms by halves, you learn
tagon (11) at RRLLR.
Solution
iU
t
PUZZLE 17
1 IL1 'U 'UU
Name
9. Knowing the above forces circle (30) at L.
8. Knowing the above forces square (36) at LR.
7. Knowing the above forces hexagon (18) at RRL.
6. Knowing the above forces diamond (5) at RRRR.
5. Knowing the above forces triangle (9) at CLLR.
4. Knowing the above forces trapezoid (11) at CLR.
3. Knowing the above forces rectangle (3) at CLRL.
ellipse (4).
le
2. At CR, triangle equals ellipse plus diamond. At RRR, trianc
equals two pentagons plus diamond. Therefore, ellipse eqt als
two pentagons. The second clue forces pentagon (2) and
subtracted from even weights, they each yield even weights.
Solution
1. Circle (at L) and square (at R) must be even because, when
PUZZLE 18
'^ `J
^)
.
► ► I ► ' l')
► ► ► ^► ?
. ,.
I
)
„► 1)
► kI
I k
R = right arm
L = left arm
C = center arm
Key
QQ = 30
Values
►
=5
®
I
P
RRRL = (from
right arm of th
left arm of tha
Sampl
I
Name
tr angle
le
5. Using an If-Then table to list possible diamond (even):triang
plus triangle (odd) combinations that equal 18, you find
4:(7 + 7) and 8:(5 + 5).
4. Using the circle possibilities of 9 and 7 from step 3, you now
that R must then weigh either 18 or 14.
list
3. Remembering that circle must be a single-digit, odd number,
possible C:circle possibilities and find 7:9, 15:7, 23:5, and 31:3.
But C weights of 23 and 31 are too large to be divided in:o two
single-digit numbers, so eliminate these combinations.
2. You know that L must be even because it is made up of two circles. (Two evens or two odds yield an even.) Since L is ev?n, so
is R. At R, diamond subtracted from the even total yields an
even number (it is halved), so diamond must be even. Using the
given clue, all other weights are odd.
1. The center arm (C) must be odd. When C is subtracted frori the
total (43), the remainder is split between L and R. If C is add,
you know that either diamond or rectangle is even. (An even
number plus an odd number equals an odd number.)
Solution
PUZZLE 19
i
an
4:3,
2:5,
find:
For a sum of 7, you
mond equals 4 fits the possibilities li
15, you find 6:9 and 8:7. Of these, o
the possibilities listed in step 6.
R = right arm
L = left arm
C = center arm
Key
♦=4
Values
=3
RRRL = (from
right arm of th
left arm of tha
Sampl
(3), and circle (7). Since the second
duplication of values for rectangle a
8. You are now left with two possible se
combinations. They are 4:3:7 and 8:
strategy, you find that the first comb
mond (4), rectangle (3), triangle (7),
combination would yield diamond (8
.+ J J J Jy ) A A I J UU) A I I I`.l",I-J J',11
m
Name
aF
5. Knowing the above shapes forces trapezoid (19) at L or CL.
4. Knowing square forces ellipse (6) in LRLR and hexagon (13) in
RLLR.
3. Knowing diamond allows you to learn the weight of CRRR and
CRRL, and by extension, CRL. Knowing circle at CRL farces square
(3), and then, moving back to CRRL, forces star (11).
in
2. Using the above weights at LLR, you can force pentacon (15)
LLLL and diamond (7) in LLLR.
the clues. You know that rectangle squared equals circle and
that five triangles equal one rectangle. If you list possible
triangle: rectangle: circle combinations, you see that at. numbers
in these puzzles must be one or two digits. Triangle ('l), rectangle (5), and circle (25) are the only values that allow you
to stay within the puzzle's limit. (If triangle equals 2, then by
extension, square equals 100.)
Solution
1. Since you cannot break down any arms into halves, start with
PUZZLE 20
0
C = center arm
R = right arm
I = left arm
Key
=13
0 -7
Q = 1
Values
Sa
RRRL = (f
right arm
left arm of
A =19
3
)')1)
11 J JUJU U J J U J'U UI'Lllu ) Q V 1 J ° U 01)111 VV1)MVU
Name
6. Knowing that RR would equal 2 ellipses and listing pos;ible RR
weights, you find the following: 60, 58, 56, 54, and 52. Since
you know that the weight of RR below the triangle is 40, you
5. Ellipse stands alone at RLR. You could think of RL (or RR), then,
as equaling 2 ellipses. Therefore, circle plus 4 ellipses would
equal the total weight of R. Listing possible circle:ellipse combinations, you find 1:30, 5:29, 9:28, 13:27, and 17:26. Other odd
weights for circle yield non-integer values.
4. Knowing square and rectangle, you learn that circle plu ; diamond at RRL equals 20. Therefore, circle has an odd-nu nbered
value of 19 or less.
3. A square value of 5 at RRRL would not allow RRRR to be halved,
so square (10) at RRRL and rectangle (5) at RRRRR and RRRRL
are forced.
2. The second clue tells you that square is a multiple of 5. But at
RRRL, a square value of 15 or higher would be too grea': for R.
odd number, they each yield an even number.
1. Diamond and circle are odd because, when subtracted from an
Solution
PUZZLE 21
C = center arm
R = right arm
L = left arm
Key
= 10
Values
.I
Sam
<^
,
RRRL = (fro
right arm of
left arm of t
=27
=5
11. Knowing the above forces hexagon
10. Knowing star forces trapezoid (6) a
9. Knowing diamond forces star (3) at
^'
,JJJJJJJJJJUUlL1 '0 `UVV'UUVUFvVUVUU
Name
4. You also know that hexagon plus 2 stars could equal i5. Using
an If-Then table to list hexagon:star possibilities when RR
equals 55, you find 7:24, 21:17, 35:10, and 49:3. (Even hexagon
weights yield non-integer star weights.) But if RR and RL are 55,
3. You know that hexagon plus 2 stars could equal 56. Wing an IfThen table to list hexagon:star possibilities when RR equals 56,
you find 14:21, 28:14, and 42:7. (Odd hexagon weights yield
non-integer weights.) But if RR and RL are 56, then s:ar must
be even to allow the further subdivision below the stir at RL.
This leaves only one possible hexagon star combination: 28:14.
Trying star equals 14 at RL, however, forces a weight Af 21 at
RLLL. Since RLLL must be halved, star equals 14 is eliminated.
2. Hexagon is limited to 7, 14, 21, 28, 35, 42, and 49. Larger
values would be too great at RR (which is either 55 o' 56).
tagon is 5, the puzzle weight (227) minus 5 equals 222. This
difference cannot be halved again as needed at R. If pentagon
is 3, then R and L would be 112 each. If pentagon is 7, then
they would be 110 each.
1. The first clue tells you that pentagon must be 3, 5, or 7. If pen-
Solution
PUZZLE 22
-1
R = right arm
L = left arm
C = center arm
Key
=6
=3
Values
Sa
RRRL = (fr
right arm
left arm of
=49
8. Using the third clue with the abo
force square (40) at L.R.
7. Knowing the above forces circle (1
6. Knowing the above forces diamon
rectangle (6) at RLLR.
5. Knowing star forces hexagon (49)
JJJJJUUUJJUUUl'UUU111) 0 0 U11 UUU
Name
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
(4 x 47) + 40 = 228
(4 x 48) + 41 = 233 (This one works!)
(4 x 49) + 42 = 238
(4 x 50) + 43 = 243
you find
of the
C = center arm
R = right arm
L = left arm
Key
e=_ 20 @ aZ
2b. Or you can guess and check. Applying your knowledge of averages, you find the mean of the five sections is 46.6. Testing
various combinations of numbers that are close to the m?an,
Samp
0
0
RRRL = (from
right arm of th
Left arm of th
0 =33
Q=6
2a. You can now state the problem as an algebraic equation, and
solve for a: 4a + (a - 7) = 233.
Values
8. Knowing triangle forces rectangle (4
7. Knowing trapezoid forces triangle (7)
6. In LRRL, hexagon plus pentagon equ
in LRL, then, equals four trapezoids.
weighs 45, is equal to nine trapezoid
and force diamond (20) and hexago
ilia' i`'UUUUUU^UUUUU
-
1. The clue tells you that the neutral central arm (C) is seven more
than the sum of rectangle and triangle (which happens to be
RR). Therefore, subtracting seven from C (square plus pertagon)
equals RR (rectangle plus triangle). RR must equal RL, and since
R is made up of RR and RL, then L must also be equal in weight
to RR plus RL. This means that C (square plus pentagon) plus
the right and left sections equals 233 (the total).
Solution
PUZZLE 23
L!JJJ UJ 0 0 0 J 0 L C V C
I
Name
PUZZLE 24
orces pentagon (13) at LL.
7. Knowing the above forces rectangle (10) at LRL.
6. Knowing diamond and trapezoid forces triangle (7) at RR L.
5. Knowing trapezoid forces ellipse (1) and diamond (5) at LRR.
plication.)
combi4. Using this information, you list possible eltipse :trapezoid
nations (at LRRR) and find 2:1, 3:1, or 1:2. The second clue
forces trapezoid (2). (If trapezoid equals 1, then rectang .e
would equal diamond because of the property of one for multi-
3. Diamond cannot equal 2 or 3 (at LRRR) because it is balanced
by three shapes. Therefore, diamond equals 4 or 5.
2. Square (at L and RL) can equal 4, 9, 16, or 25 given the first
clue. (Any higher a square would prove too heavy for the side at
RL.) This means that diamond can equal 2, 3, 4, or 5.
tracted from the odd puzzle total (143), the remainder is a number that can be halved (weights L and R). Also, notice that if
pentagon equals 1, the most R and L each could be equals 71.
Solution
1. Pentagon at central arm (C) is odd. When the pentagon i5 sub-
PUZZLE 24
R = right arm
L = left arm
C = center arm
Key
=10
Q= 2
Values
Sampl
0
RRRL = (from
right arm of th
left arm of tha
=13
1
11
.)jJi. jjJJ.11 11UL)0010UJ11j1)j 9J J'I'Ik1'I'I'I^
Name
6. Pentagon at RRR is odd because, when subtracted from ar odd
weight, it yields an even weight.
5. Knowing the above forces rectangle (6) at LRRL.
4. At LLL, triangle equals ellipse plus two diamonds. Only diilmond
equals 3 (see step 2) is small enough to work without making LL
too heavy, so you learn diamond (3), square (18), ellipse (9),
hexagon (30), and triangle (15).
3. Using the above square possibilities, list possible combinations
for square:hexagon:triangle that will work at LL and find
18:30:15, 14:32:16, 10:34:17, 6:36:18, and 2:38:19.
2. Using an If-Then table to list possible diamond:square:ell pse
combinations that will work at LRL, you find 3:18:9, 7:16:8,
11:14:7, 15:12:6, and so forth. (Diamond weights of 1, 5, 9, 13,
and so on form non-integer weights for ellipse.)
1. Diamond (at LRL) and ellipse (at LRR) must be odd because,
when subtracted from odd weights, they each yield even
weights. Square (at LL) and rectangle (at R) must be evert
evert
because, when subtracted from even weights, they each
even weights.
Solution
PUZZLE 25
C = center arm
R = right arm
L = left arm
Key
Q = 13
=6
Sampl
RRRL = (from t
right arm of that
left arm of that
0
n
= 15
13 = 18
O = 3
Values