Puzzles Archive
This is a list of the previous puzzles that have been sent out by
E-mail.
Don't forget to signup for "The MindBender" here.
To see the answer, click and hold your mouse button just to the right
of the red arrow 
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the answer and make it visible.
March 11, 2002
MindBender
Vacations
Three married couples from Santa Fe, New Mexico planned vacation
trips, each couple to a different location. The destinations were
Atlanta, Georgia; Tucson, Arizona; and Santa Barbara, California.
From the following statements, determine the first names (one is
James and one is Jean), the last names (one is Anderson), and their
destinations.
1. Joyce said they were going east to visit relatives.
2. Jack and his wife planned a golfing trip to Tucson.
3. John, his wife, and the Abernathys discussed their trips west
while playing bridge together.
4. The Adamses do not play bridge.
5. Joan is not an Abernathy.
___________________________
Mini-MindBender for Kids
Sequence of Four
What is the missing number in the following sequence of sets of four
numbers?
1 2 -- 3 4 -- 5 6
8 4 -- 16 12 -- ?? 20
___________________________
...Answer to MindBender
Vacations
From statements 1, 2, and 3, Joyce's husband is not Jack or John.
Therefore, it is James. From statements 3 and 4, their last name is
Adams. From statement 1, they are going to Atlanta. From statement 5,
Joan is not Abernathy. Therefore, her last name is Anderson. From
statements 2 and 3, her husband is John. Their destination is Santa
Barbara. Therefore, Jean and jack Abernathy's destination is Tucson.
The summary is:
Jack and Jean Abernathy went to Tucson.
James and Joyce Adams went to Atlanta.
John and Joan Anderson went to Santa Barbara.
This MindBender was modified from a puzzle in Norman D. Willis'
"The Little Giant Encyclopedia of Logic Puzzles."
___________________________
Answer to Mini-MindBender for Kids
Sequence of Four
The missing number is 24, 4 times its diagonal partner. Or the
numbers can be seen to go in the following sequence by position in
each set:
1 3 5 (up by two or odds)
2 4 6 (up by two or evens)
8 16 24 (up by 8)
4 12 20 (up by 8)
This MindBender was modified from a puzzle in Dr. Abbie F. Salny's
"The Mensa 365 Brain Puzzlers Calendar" for 2001.
March 14, 2002
Midweek MindBender
Driving Time
You want to arrive at a future appointment exactly on time. You are
restricted in how much time you will have to make the trip (but I
won't tell you how long that is). So you make some practice runs
driving to your destination. If you average 40 miles per hour, you
arrive an hour earlier than your allowed travel time. If you average
30 miles per hour, you arrive exactly on time. If you average 20
miles per hour, you arrive two hours late. How far away from your
appointment are you?
___________________________
...Answer to Midweek MindBender
Driving Time
120 miles. Solve any part of 40*(X-1)=30*X=20*(X+2) for the ideal
driving time X (4 hours) and then multiply by 30 miles per hour.
This MindBender was modified from a puzzle in Dr. Abbie F. Salny's
"365 Brain Puzzlers" Calendar for 2002.
March 18, 2002
MindBender
Level
How many ways can you read "LEVEL" from the below diagram, using
letters that touch each other? Any L or any E can be used twice in
the same spelling.
L
LEL
LEVEL
LEL
L
___________________________
Mini-MindBender for Kids
Time Division
Think about a normal clock face (numbers 1 to 12 in a circle). Can
you add two straight lines to the clock face, dividing the clock
face into three parts, with the sum of the numbers in each part
identical? HINT: The sum of the numbers in each part is 1/3 of the
total sum of 1+2+3+ ... +12.
___________________________
...Answer to MindBender
Level
Starting with the farthest top, farthest bottom, farthest left, or
farthest right L (4 possibilities), there is only 1 E that can be
reached, then only 1 V, then 4 possible E's, then from each of those,
3 possible L's. This leaves us with 4*1*1*4*3=48 ways to read "LEVEL."
Starting from either of the other 4 L's, there are 2 E's that can be
reached, then only 1 V, then 4 possible E's, then from each of those,
3 possible L's. This leaves us with 4*2*1*4*3=96 ways to read "LEVEL."
48+96=144 total ways to read "LEVEL."
This MindBender was modified from a puzzle in Pierre Berloquin's book,
"100 Numerical Games."
___________________________
Answer to Mini-MindBender for Kids
Time Division
The two straight lines go:
From between the 10 and 11 to between the 2 and 3
From between the 8 and 9 to between the 4 and 5
The sum in each section is 11+12+1+2=9+10+3+4=5+6+7+8=26.
This MindBender was modified from a puzzle in Raymond Blum's book,
"Math Tricks, Puzzles & Games."
March 21, 2002
Midweek MindBender
Letter Sequence
What are the next two letters in the following sequence?
B C D G J O P Q R ? ?
___________________________
...Answer to Midweek MindBender
Letter Sequence
S and U. These are the capital letters that are drawn with at least
one curved line.
This MindBender was modified from a puzzle in Terry Stickels'
"Mindbending Puzzles" calendar for 2001.
March 25, 2002
MindBender
Missing Symbols
What are the missing symbols in the following equation?
((5 ? 9) ? 4) ? 8 = 19.25
___________________________
Mini-MindBender for Kids
Jelly Beans
If a jelly bean weighs one gram plus a half a jelly bean, how much do
100 jelly beans weigh?
___________________________
...Answer to MindBender
Missing Symbols
((5 * 9) / 4) + 8 = 19.25
This MindBender was modified from a puzzle in Victor Serebriakoff's
book, "The Mammoth Book of Astounding Puzzles."
___________________________
Answer to Mini-MindBender for Kids
Jelly Beans
If one jelly bean weighs one gram plus a half a jelly bean, then two
jelly beans weigh two grams plus one jelly bean. So taking one jelly
bean away from each side, one jelly bean weighs two grams. Therefore,
100 jelly beans weigh 200 grams.
This MindBender was modified from a puzzle in Adam Hart-Davis' book,
"Amazing Math Puzzles."
March 27, 2002
Midweek MindBender
Sequence
What number completes the following sequence?
4 9 19 ?? 79 159
___________________________
...Answer to Midweek MindBender
Sequence
39. Each number is twice the previous number plus one. Or the differences keep
doubling: 5, 10, 20, 40, 80.
This MindBender was modified from a puzzle in Dr. Abbie F. Salny's
"365 Brain Puzzlers" Calendar for 2002.
April 1, 2002
MindBender
Cryptogram
Today's MindBender is a Cryptogram, or as some call them, a Cryptoquip.
Each letter stands for another. If you think Z=N, for example, Z
would equal N throughout the puzzle. Decode the following:
Cryptogram Clue: Z equals N
IJDYZX SUO RBSUORBSYLE LPZSOES, SUO MPEODE AODO XDOBSMH
PJSZJRTODOI TH SUO AYZZODE.
___________________________
Mini-MindBender for Kids
Jelly Beans
Abby has found all her jelly beans. She realizes that numbers are
important and with the help of her mom and dad, she counts the jelly
beans. Her gramma wants to know how many jelly beans she has. Abby
says that she could share her jelly beans equally with her parents (a
total of three people). She says that could also share them equally
with her four grandparents (a total of five people). She then says
that she could share them equally with six of her cousins (a total of
seven people). She tells her gramma that she has the smallest number
of jelly beans that satisfy the above sharing conditions. How many
jelly beans does Abby have?
___________________________
...Answer to MindBender
Cryptogram
DURING THE MATHEMATICS CONTEST, THE LOSERS WERE GREATLY
OUTNUMBERED BY THE WINNERS.
This Cryptogram came from the Minneapolis Star Tribune newspaper.
___________________________
Answer to Mini-MindBender for Kids
Jelly Beans
The number of jelly beans must be a multiple of 3 and 5 and 7. So
the smallest number possible is 3*5*7 = 105 jelly beans. So Abby
has 105 jelly beans. Don't eat them all in one day!
The MindBender moderator is the source for this Mini-MindBender.
April 4, 2002
Midweek MindBender
Number
What is the number that is five more than one-half of one-quarer
of one-quarter of one-half of 16,000?
___________________________
...Answer to Midweek MindBender
Number
255. ((((16000/2)/4)/4)/2)+5=255
This MindBender was modified from a puzzle in Dr. Abbie F. Salny's
"365 Brain Puzzlers" Calendar for 2002.
April 8, 2002
MindBender
Merged Series
Two series are merged together below. Which two letters should appear
next?
D V H Q L L P G T ? ?
___________________________
Mini-MindBender for Kids
Elizabeths
Three girls (ages 8, 9, and 10) each are named variations of the name
Elizabeth. These name variations are Bess, Liz, and Beth. The girls'
last names (in no particular order) are Jones, Smith, and Adams. From
the following statements, determine the full name and age of each girl.
1. Jones is younger than Adams, but older than Smith.
2. Beth is not the youngest or the oldest.
3. Liz's last name is Smith.
___________________________
...Answer to MindBender
Merged Series
B and X. Every fourth letter of the alphabet forms one series. Every
fifth letter from the end of the alphabet makes up the other series.
These two series alternate every other letter in the combined series.
This MindBender was modified from a puzzle in "Mensa, The Biggest
Puzzle Book Ever," a book edited by Robert Allen.
___________________________
Answer to Mini-MindBender for Kids
Elizabeths
Bess Adams is 10. Liz Smith is 8. Beth Jones is 9.
This MindBender was modified from a puzzle in "Brain Games," a book
by Dona Herweck Rice, et al.
April 11, 2002
Midweek MindBender
Partial Word
Fill in the question marks with letters to complete a mathematically
related word:
?X??N?N??A??Y
___________________________
...Answer to Midweek MindBender
Partial Word
EXPONENTIALLY
This MindBender was modified from a puzzle in Dr. Abbie F. Salny's
"365 Brain Puzzlers" Calendar for 2002.
April 15, 2002
MindBender
Multiply By 3
What is the smallest integer whose units digit is 3 and when the
integer is multiplied by 3, the product is the same as the original
number except that the units digit (3) is moved to the left end of
the answer? For example, if 26473 were the original number, the
product would be 32647.
___________________________
Mini-MindBender for Kids
Groceries
You go shopping for groceries and get vegetables, a dessert, and
some meat. The dessert costs twice as much as the vegetables and
the meat costs twice as much as the dessert. You spend a total of
$14. How much did each item cost?
___________________________
...Answer to MindBender
Multiply By 3
The answer is a very large number. The original number is:
1034482758620689655172413793
Start with 3*3. The result is 9, so the last two digits of the
original number are 93. 93*3=79 with a carry of 2, so the last
three digits of the original number are 793. 793*3=379 with a
carry of 2, so the last four digits of the original number are
3793. Continuing on in this manner yields the above answer.
The MindBender moderator is the source for this MindBender.
___________________________
Answer to Mini-MindBender for Kids
Groceries
Let the items be represented by V, D, and M. D=2*V. M=2*D=4*V.
V+2*V+4*V=14. So 7*V=14. V=2. D=4 and M=8. So the vegetables cost
$2, the dessert cost $4, and the meat cost $8.
This MindBender was modified from a puzzle in Victor Serebriakoff's
book, "The Mammoth Book of Astounding Puzzles."
April 22, 2002
MindBender
TEXAS
Complete the following square using the letters of TEXAS so that no
row, column, or diagonal contains the same letter more than once.
One horizontal line will spell out TEXAS correctly.
---------------------
| | | | T | |
---------------------
| | | E | X | |
---------------------
| | | A | | |
---------------------
| | S | | | |
---------------------
| | | X | | |
---------------------
___________________________
Mini-MindBender for Kids
Addition
In the following addition, A, B, C, and D represent the numbers 3, 4,
5, and 9 (but not necessarily in that order). Can you work the problem
to figure out which letters are represented by which numbers?
ABAC
+BCDD
_____
8ACB
___________________________
...Answer to MindBender
TEXAS
---------------------
| X | A | S | T | E |
---------------------
| S | T | E | X | A |
---------------------
| E | X | A | S | T |
---------------------
| A | S | T | E | X |
---------------------
| T | E | X | A | S |
---------------------
This MindBender was modified from a puzzle in "Mensa, The Biggest
Puzzle Book Ever," a book edited by Robert Allen.
___________________________
Answer to Mini-MindBender for Kids
Addition
From the 8-sum, (A B) must be (3 5), (5 3), (4 3), or (3 4). Trying
each of those combinations leads to only (3 4) as possible. This
then gives us the following answer:
3439
+4955
_____
8394
This MindBender was modified from a puzzle in Highlights
"Mathmania," a good thinking resource for young kids.
April 29, 2002
MindBender
Ship's Crew
A, B, and C are captain, first mate, and second mate on a ship, but
not necessarily in that order. From the following statements, determine
who is captain, who is first mate, and who is second mate?
1. If A is captain, then B is first mate.
2. If B is not captain, then C is first mate.
3. If A is first mate, then B is second mate.
___________________________
Mini-MindBender for Kids
SORT to PACK
Change one letter at a time to change SORT to PACK in four steps.
SORT
----
----
----
PACK
___________________________
...Answer to MindBender
Ship's Crew
Assume A is captain. From 1, B is first mate. From 2, C is first mate.
This means both B and C are first mate. So A is not captain.
Assume A is first mate. From 3, B is second mate. From 2, C is first
mate. This means both A and C are first mate. So A is not first mate.
Therefore, A must be second mate. If B is first mate, C must also be
first mate (from 2). So B must be captain and C must be first mate.
Summary: B is captain; C is first mate; A is second mate.
Note that even though statement 2's assumption is false, C can still
be first mate.
This MindBender was modified from a puzzle in Norman D. Willis'
"The Little Giant Encyclopedia of Logic Puzzles."
___________________________
Answer to Mini-MindBender for Kids
SORT to PACK
Here is one solution. There may be others.
SORT
PORT
PART
PACT
PACK
This MindBender was modified from a puzzle in Terry Stickels'
"Mindbending Puzzles" calendar for 2001.
May 2, 2002
Midweek MindBender
Strange Sequence
What would be the next line in this sequence of lines of ones and
zeroes?
10
011
10101
01101101
1010110101101
___________________________
...Answer to Midweek MindBender
Strange Sequence
Moving from one line to the next, each 1 is replaced by 01 and each 0
is replaced by 1. So the next line is:
011011010110110101101
This MindBender was modified from a puzzle in Cliff Pickover's
"Mindbending Puzzles" calendar for 2002.
May 6, 2002
MindBender
Four Words
What single three letter word can replace the ??? in all four words
below to make a new word in each case?
???BOARD
???STONE
???PUNCH
???HOLE
___________________________
Mini-MindBender for Kids
Frame Game
Find the phrase or title hidden in the frame:
-----------------
| |
| egsg |
| |
| segg |
| |
| gges |
| |
| sgge |
| |
| sgeg |
| |
-----------------
___________________________
...Answer to MindBender
Four Words
KEY to make KEYBOARD, KEYSTONE, KEYPUNCH, and KEYHOLE.
This MindBender was modified from a puzzle in Dr. Abbie F. Salny's
"365 Brain Puzzlers" Calendar for 2002.
___________________________
Answer to Mini-MindBender for Kids
Frame Game
Scrambled eggs.
This MindBender was modified from a puzzle in Terry Stickels'
"Mindbending Puzzles" calendar for 2000.
May 9, 2002
Midweek MindBender
Answer Rhymes
Fill in the blanks to find a rhyme for the questions:
Sample. Peter Pan Novel? HOOK BOOK
1. Male Mensa members? --S- --Y-
2. Code creator's equine? M---- ---S-
3. Wealthy Winnie the Pooh? M--------R- --N-- --A-
4. Unrefined guy? --U-- ---E
5. Boston Pops tryout? --S----- --D-----
6. Annoying invitee? -U--- P---
7. Seasonal percussionist? ---M-- -R---E-
8. A master's way with old furniture? A----U- T---N----
___________________________
...Answer to Midweek MindBender
Answer Rhymes
1. WISE GUYS
2. MORSE HORSE
3. MILLIONAIRE HONEY BEAR
4. CRUDE DUDE
5. MUSICIAN AUDITION
6. GUEST PEST
7. SUMMER DRUMMER
8. ANTIQUE TECHNIQUE
This MindBender was modified from a puzzle in the February 2002 issue
of the Reader's Digest.
May 13, 2002
MindBender
2442 Digits
In writing the numbers 1, 2, 3, and so on all the way up to N, a
total of 2442 digits are printed. What is the value of N?
___________________________
Mini-MindBender for Kids
Base B
If 52 (base B) is twice as much as 25 (base B), what is the base B?
___________________________
...Answer to MindBender
2442 Digits
There will be 9 one digit numbers (1 to 9), 90 two digit numbers (10
to 99) and N-99 three digit numbers (100 to N). If N is 999, there
will be 9+2*90+3*(900)=2889 digits printed. This is too many for
the question, so there are no four digit numbers required. To find
the value of N, solve 9+2*90+3*(N-99)=2442 or 189+3*N-297=2442 or
3*N=2550 or N=850.
This MindBender was modified from a puzzle in Summer 2002 issue of
"The Original Logic & Math Puzzles" by Ebb Publications.
___________________________
Answer to Mini-MindBender for Kids
Base B
8. 5*B+2=2*(2*B+5) or 5*B+2=4*B+10 or B=8.
This MindBender was modified from a puzzle in Summer 2002 issue of
"The Original Logic & Math Puzzles" by Ebb Publications.
May 16, 2002
Midweek MindBender
Math Signs
Insert mathematical signs between numbers below to make a true statement.
5 3 5 5 = 70
___________________________
...Answer to Midweek MindBender
Math Signs
5 X 3 X 5 - 5 = 70
This MindBender was modified from a puzzle in Cliff Pickover's
"Mindbending Puzzles" calendar for 2002.
May 20, 2002
MindBender
XYZ
X, Y, and Z satisfy:
X-Y-Z=0 and 2X+Y+3Z=0
If Z is nonzero, what is the ratio of X to Z?
___________________________
Mini-MindBender for Kids
Triangle Side
If the sides of a triangle are 5, 8, and X, what are the possible
values of X?
___________________________
...Answer to MindBender
XYZ
X-Y-Z=0 so Y=X-Z
2X+Y+3Z=0 so Y=-2X-3Z
Therefore, X-Z=-2X-3Z or 3X=-2Z or X/Z=-2/3
This MindBender was modified from a puzzle in Summer 2002 issue of
"The Original Logic & Math Puzzles" by Ebb Publications.
___________________________
Answer to Mini-MindBender for Kids
Triangle Side
The third side of a triangle has to be greater than the diffenrence
of the other two sides and less than the sum of the other two sides.
So X is greater than 3 and X is less than 13.
This MindBender was modified from a puzzle in Summer 2002 issue of
"The Original Logic & Math Puzzles" by Ebb Publications.
May 23, 2002
Midweek MindBender
Ages
Jim's age is the same as his mother's with the digits reversed. A
year ago, Jim was exactly one-half of his mother's age. How old
are they now?
___________________________
...Answer to Midweek MindBender
Ages
Jim is 37. His mother is 73. The following is one method to solve this
age puzzle. Let Jim's age be TU where T is the tens digit and U is the
units digit. Then we have:
10T+U-1=(10U+T-1)/2 or
20T+2U-2=10U+T-1 or
19T=8U+1
If T is greater than 3, no value of U works. Trying T=1, T=2, and
T=3 results in T=3 and U=7 as the only solution.
This MindBender was modified from a puzzle in Dr. Abbie F. Salny's
"365 Brain Puzzlers" Calendar for 2002.
June 3, 2002
MindBender
Logical Numbers
What number will replace the "??" below to complete the logic of the
following set of numbers?
8 1 12 10 14 11 ?? 3 7 5 16 9
___________________________
Mini-MindBender for Kids
Double Word Change
Change one letter at a time to change MILD and COLD both into FARE
in four steps. At each step, you should have a good English word.
M I L D
- - - -
- - - -
- - - -
F A R E
C O L D
- - - -
- - - -
- - - -
F A R E
___________________________
...Answer to MindBender
Logical Numbers
6. Starting with the two ends (8 and 9) and moving toward the middle,
each pair of numbers adds to 17.
This MindBender was modified from a puzzle in Terry Stickels'
"Are You As Smart As You Think?"
___________________________
Answer to Mini-MindBender for Kids
Double Word Change
One way is shown below. There may be other ways.
M I L D
M I L E
F I L E
F I R E
F A R E
C O L D
C O R D
C O R E
C A R E
F A R E
This MindBender was modified from a puzzle in Terry Stickels'
"Are You As Smart As You Think?"
June 6, 2002
Midweek MindBender
Letter Sequence
What is the next letter in the following letter sequence?
O T F S N E ??
___________________________
...Answer to Midweek MindBender
Letter Sequence
T for Thirteen. The sequence is the initial letters of odd numbers
starting at one.
This MindBender was modified from a puzzle in Terry Stickels'
"Mindbending Puzzles" calendar for 2000.
June 17, 2002
MindBender
Single Number
Find the single five-digit number Y, where Y is a multiple of 7, Y+1 is a
multiple of 13, Y+2 is a multiple of 23, and Y+3 is a multiple of 53.
___________________________
Mini-MindBender for Kids
Missing Sum
Can you determine the value of the missing vertical sum (??) in the
square below where the numbers to the right of the square are
horizontal or row sums and the numbers below the square are vertical
or column sums?
_________________
| A | B | C | C | 48
_________________
| C | A | A | A | 46
_________________
| B | B | B | C | 70
_________________
| A | C | A | B | 54
_________________
54 ?? 54 48
___________________________
...Answer to MindBender
Single Number
This is a specific example of a general type problem:
Find the smallest number X that when divided by A leaves a remainder of R;
when divided by B leaves a remainder of S; when divided by C leaves a
remainder of T; and when divided by D leaves a remainder of U.
General approach to this type of problem:
Assume that A, B, C, and D are prime to each other. If not some slight
adjustment is needed. Determine a multiplier M such that the product B*C*D*M
results in a remainder of R when divided by A. Similarly, determine a
multiplier N such that the product A*C*D*N results in a remainder of S when
divided by B. Similarly, determine a multiplier P such that the product
A*B*D*P results in a remainder of T when divided by C. Similarly, determine
a multiplier Q such that the product A*B*C*Q results in a remainder of U when
divided by D. Now add these four products together. The sum,
B*C*D*M+A*C*D*N+A*B*D*P+A*B*C*Q will have the desired remainders when divided
by A, B, C, and D. This is because, for example, when dividing by A, the
second through fourth terms in the sum divide by A exactly. Therefore, the
total sum has the same remainder as the first term in the sum and we made
that remainder R by choosing the multiplier M. The same holds when dividing
the sum by B or C or D, by similar reasoning. Now reduce the above sum by
A*B*C*D as many times as necessary to arrive at the lowest number X. This
reduction by A*B*C*D will not affect the remainder properties.
In this particular example, A, B, C, and D are 7, 13, 23, and 53. R, S, T,
and U are 0, 12, 21, and 50. M, N, P, and Q are 0, 5, 20, and 6. So,
A*B*C*D = 7*13*23*53 = 110929 and
B*C*D*M+A*C*D*N+A*B*D*P+A*B*C*Q = 13*23*53*0+7*23*53*5+7*13*53*20+7*13*23*6 =
0+42665+96460+12558 = 151683 and 151683-110929 = 40754
So the final answer is 40754.
This MindBender was modified from a puzzle in the June-July 2002
issue of "The Actuarial Digest." Thanks also to my daughter Terri
and her husband Matt for telling me about it.
___________________________
Answer to Mini-MindBender for Kids
Missing Sum
The missing sum is 62. The values of A, B, and C are 13, 21, and 7.
However, there is no need to determine their values since the sum of
the horizontal sums must equal the sum of the vertical sums:
48+46+70+54=54+??+54+48 or ??=62
This MindBender was modified from a puzzle in "Mensa, The Biggest
Puzzle Book Ever," a book edited by Robert Allen.
June 20, 2002
Midweek MindBender
A Number
A certain number plus ten is two less than five times that number.
What is the number?
___________________________
...Answer to Midweek MindBender
A Number
Three. The Algebra equation to use is: N+10=5*N-2.
This MindBender was modified from a puzzle in Cliff Pickover's
"Mindbending Puzzles" calendar for 2002.
June 24, 2002
MindBender
Number Class
Five of the following six numbers belong to the same class. One of
them does not! What is the class of numbers and which number does
not belong?
40320 5040 56478 6 362880 2
___________________________
Mini-MindBender for Kids
Math Scores
The teacher passed back the math exams. Five students anxiously
awaited their grades. The grades were A, B, C, C-, and D. Using the
clues below, determine each student's grade.
1. Chelsea, who did not get an A, scored higher than Morgan and Linda.
2. Marcia and Linda both scored higher than Casie.
3. Morgan received a C on her test.
___________________________
...Answer to MindBender
Number Class
The number that does not belong is 56478. All the rest of the numbers
are factorials (the result of multiplying an integer by all the
integers below it). 2=2! 6=3! 5040=7!=7*6*5*4*3*2*1 40320=8! 362880=9!
This MindBender was modified from a puzzle in Victor Serebriakoff's
book, "The Mammoth Book of Mindbending Puzzles."
___________________________
Answer to Mini-MindBender for Kids
Math Scores
The grades were: Chelsea: B; Linda: C-; Marcia: A; Morgan: C; Casie: D.
This MindBender was modified from a puzzle in "Brain Games," a book
by Dona Herweck Rice, et al.
June 27, 2002
Midweek MindBender
Interesting Number
There is an interesting two-digit number that has an interesting
property. If the two digits are added together and then that result
is squared, the reverse of the original number is the result. What
is the original number?
___________________________
...Answer to Midweek MindBender
Interesting Number
18.
This MindBender was modified from a puzzle in Dr. Abbie F. Salny's
"The Mensa 365 Brain Puzzlers Calendar" for 2001.
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