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 and drag down. This will highlight the answer and make it visible.


July 1, 2002
MindBender
Alphametic
Solve the following addition alphametic if B=1:
ABCD
ABCD
ABCD
______
EBCEF
___________________________

Mini-MindBender for Kids
Saving Fuel
The inventor had discovered three separate ways to save fuel while driving. Each one saved 30% on fuel consumption. The inventor thought he would save 90% by using all three. However, that was not the case. How much fuel did he save, assuming that all the ways were independent and cumulative?
___________________________
...Answer to MindBender
Alphametic
7143
7143
7143
______
21429
There can be no carry from the hundreds column. So, A must be 7 to have a two-digit product of 3 that ends in 1. Thus, E is 2. Then C must be 4 with no carry from the units column. Since there is no carry in the units column, D must be 3 since 1 and 2 are used and any larger D would produce a carry. Thus, F must be 9.
The MindBender moderator is the source for this MindBender.
___________________________

Answer to Mini-MindBender for Kids
Saving Fuel
He saved 1- (.7*.7*.7) = 1 - .343 = .657 or 65.7%.
This MindBender was modified from a puzzle in Dr. Abbie F. Salny's "The MENSA Genius Quiz-A-Day Book."


July 4, 2002
Midweek MindBender
A Number
What number should logically replace the question mark below?
200 390 1170 8920
140 273 819 ?
___________________________

...Answer to Midweek MindBender
A Number

6244. Each number in the second line is 0.7 times the number above it.
This MindBender was modified from a puzzle in Terry Stickels' "Mindbending Puzzles" calendar for 2001.


July 8, 2002
MindBender
Missing Number
What is the missing number in the following grid:
144 9 4
175 7 5
392 8 ?
___________________________

Mini-MindBender for Kids
Bicycle Mile
How many revolutions are made by a bicycle wheel (that is 28 inches in diameter) to travel one mile?
___________________________
...Answer to MindBender
Missing Number
7. The first number in each row is the product of the second number times the square of the third number.
The MindBender moderator is the source for this MindBender.
___________________________

Answer to Mini-MindBender for Kids
Bicycle Mile
If you use 22/7 as a value for Pi, then the circumference of the wheel is 22/7 * 28 = 88 inches. Then one mile in inches divided by 88 inches is the number of revolutions made: 5280 * 12 / 88 = 720 or 720 revolutions.
This MindBender was modified from a puzzle in Philip Carter and Ken Russell's "Power Puzzles 2."


July 11, 2002
Midweek MindBender
Sequence
What number completes the following sequence?
60 60 24 7 ?? 10
___________________________

...Answer to Midweek MindBender
Sequence

52. 60 seconds in a minute, 60 minutes in an hour, 24 hours in a day, 7 days in a week, 52 weeks in a year, 10 years in a decade.
This MindBender was modified from a puzzle in Dr. Abbie F. Salny's "365 Brain Puzzlers" Calendar for 2002.


July 15, 2002
MindBender
No Seven
Imagine that there is no digit 7 in our number system. Thus, counting goes like this: 1, 2, 3, 4, 5, 6, 8, 9, 10, 11, 12, 13, 14, 15, 16, 18, 19, etc. The seventh number is now 8 and the 16th number is now 18. Complete the following: The ___th number is written as 4685.
___________________________

Mini-MindBender for Kids
Pennies
Alex has been saving a penny every day for quite a while (but less than one year). His mom wanted to know how many pennies he had saved. Alex said that if he counted his pennies by 2s or 3s or 4s or 5s or 6s, there is always one penny left over. But if he counts them by 7s, there are no pennies left over. How many pennies does Alex have?
___________________________
...Answer to MindBender
No Seven
There are at least two ways to solve this MindBender. The first:
In every hundred numbers, 19 are not used because they have at least one digit 7 in them. These numbers would end in 07, 17, 27, 37, 47, 57, 67, 70 through 79, 87, and 97. In addition, 81 additional numbers are not used for the 700s (700 to 799), 1700s, 2700s, and 3700s. From 4600 to 4685, 17 numbers are not used. Thus, the number written as 4685 is the (4685-46*19-4*81-17=4685-874-324-17=3470)th or 3470th number.
The second method:
4685 is a base 9 number where the digit 8 is the same as a 7 in a normal base 9 number. 4685 or really 4675(base 9)=4*9**3 + 6*9**2 + 7*9 + 5 (remember that the 8 is the same as the 7 in a normal base 9 number) Therefore, 4685=2916+486+63+5=3470 or the number written as 4685 is the 3470th number.
This MindBender was modified from a puzzle in David J. Bodycombe's "The IQ Obstacle Course."
___________________________

Answer to Mini-MindBender for Kids
Pennies
301 pennies. The smallest number divisible by 2, 3, 4, 5, and 6 is 60. So, Alex must have 61, 121, 181, 241, 301, or 361 pennies. The only one of those numbers divisible evenly by 7 is 301.
This MindBender was modified from a puzzle in Highlights "Mathmania," a good thinking resource for young kids.


July 18, 2002
Midweek MindBender
Radical Sequence
What comes next in the following sequence?
1 1.7321 2.2361 2.6458 3 3.3166 ??
___________________________

...Answer to Midweek MindBender
Radical Sequence

3.6056 which is the square root of 13. The numbers in the sequence are the square roots of 1, 3, 5, 7, 9, 11, and 13.
This MindBender was modified from a puzzle in Terry Stickels' "Mindbending Puzzles" calendar for 2000.


July 22, 2002
MindBender
Multiply Arithmetic
Solve the following multiplication arithmetic. A given letter always represents the same digit.

ABCDE
X 4
______
EDCBA
___________________________

Mini-MindBender for Kids
Remote Boats
At a carnival, remote controlled boat races are held. Each heat has 4 boats racing and the heat winner goes on to the next round of the competition. If there are 82 entrants, how many heats are required to determine an overall champion?
___________________________
...Answer to MindBender
Multiply Arithmetic
A must be 1 or 2 to avoid a two digit product. Since E*4 (or A) must be even, A must be 2. The only numbers that, when multiplied by 4, result in a number ending in 2, are 3 and 8. So E must be 3 or 8. E cannot be 3 since 4*A plus a possible carry cannot be 13 since 13 is two digits. Therefore E is 8. 4*B must produce no carry. So B must be 1 or 2. A is 2. Therefore, B is 1. So 4*D+3 must end in 1. Only D=2 or D=7 do this. A is 2, so D is 7. 4*C+3 must be 30 plus C. Only C=9 will do this. So the final answer is:

21978
X 4
______
87912
This MindBender was modified from a puzzle in James Fixx's "Games for the Superintelligent."
___________________________

Answer to Mini-MindBender for Kids
Remote Boats
Since there will be 81 losers and each heat knocks 3 contestants out of the competition, there needs to be 81/3 or 27 heats.
This MindBender was modified from a puzzle in David J. Bodycombe's book, "The Mammoth Puzzle Carnival."


July 25, 2002
Midweek MindBender
Pronouns
"She" contains two personal pronouns: she and he. Can you name a five letter word starting with "u" that contains five personal pronouns?
___________________________

...Answer to Midweek MindBender
Pronouns

Ushers. It contains: us, she, he, her, hers.
This MindBender was modified from a puzzle in Cliff Pickover's "Mindbending Puzzles" calendar for 2002.


July 29, 2002
MindBender
Seventh Term
What is the seventh term of the following series?
6, -4, 2 2/3, __, __, __, ??
___________________________

Mini-MindBender for Kids
New Words
Find a 5-letter word that, when placed in front of the following six words, produces six new words: BOAT MAID KEEPER WIFE HOLDER COAT.
___________________________
...Answer to MindBender
Seventh Term
128/243. Each term is the previous term multiplied by -2/3. So the seventh term is 6 * (-2/3)**6 or 128/243. This is an example of a geometric series, where each term (after the first) is multiplied by a constant to get the next term.
This MindBender was modified from a puzzle in Philip Carter and Ken Russell's "Power Puzzles 2."
___________________________

Answer to Mini-MindBender for Kids
New Words
HOUSE. It gives us HOUSEBOAT HOUSEMAID HOUSEKEEPER HOUSEWIFE HOUSEHOLDER HOUSECOAT.
This MindBender was modified from a puzzle in Philip Carter and Ken Russell's "Power Puzzles 2."


August 1, 2002
Midweek MindBender
Ages
Emily is now as old as Megan was two years ago. In eight years, Emily will be eight-ninths as old as Megan. How old are they now?
___________________________

...Answer to Midweek MindBender
Ages

Emily is 8. Megan is 10. Use the following equatios to solve:
E=M-2
E+8=8/9*(M+8)
M-2+8=8/9*(M+8)
M+6=8/9*(M+8)
9*(M+6)=8*(M+8)
9M+54=8M+64
M=10
This MindBender was modified from a puzzle in Dr. Abbie F. Salny's "The Mensa 365 Brain Puzzlers Calendar" for 2001.


August 5, 2002
MindBender
Neighbors
Five neighbors, the Quigleys, the Rodneys, the Smiths, the Taylors, and the Ungers, live either adjacent to or directly across the street from each other. Two of their houses are white, one is gray, one is green, and one is blue. On which side of the street is each neighbor's house and what is its color?
1. The two white houses are directly across the street from each other, at the west end of the street.
2. The Quigleys' house is not adjacent to either of the white houses.
3. The Rodneys' house is on the same side of the street as the Quigleys' house.
4. The blue house is immediately to the east of the Rodneys' house on the same side of the street.
5. The Smiths' house, which is not gray or white, is next door to the Ungers' house, which is on the south side of the street.
6. The Taylors' house is directly across the street from the Smiths' house.
___________________________

Mini-MindBender for Kids
Riddle Word
Using the rhyme below, find the five letter word that is described.
My first is in see but not in look.
My second in hope but not in book.
My third in pin but not in pen.
My fourth in red but not in den.
My last in tree but not in bear.
My whole an object you can wear.
___________________________
...Answer to MindBender
Neighbors
From statements 1 and 2, we can conclude that the Quigley house is one of at least three houses on its side of the street. From statement 5, the Unger and the Smith houses are the two on the south side of the street. The Unger house is at the west end of the street, and is white. The Smith house is either green or blue. From statement 3, the Rodneys and the Quigleys are two of the three neighbors that live on the north side of the street. From statements 4 and 6, the Taylor house is between the Rodney and the Quigley house, and is blue. The Rodneys live in the second white house. The Smith house is green and the Quigley house is gray. The summary solution is: 
          East 
  ___________________ 
  | Quigley |       | 
  | gray    |       | 
N ___________________ S 
o | Taylor  | Smith | o 
r | blue    | green | u 
t ___________________ t 
h | Rodney  | Unger | h 
  | white   | white | 
  ___________________ 
          West
This MindBender was modified from a puzzle in Norman D. Willis' "The Little Giant Encyclopedia of Logic Puzzles."
___________________________

Answer to Mini-MindBender for Kids
Riddle Word
Shirt.
This MindBender was modified from a puzzle in Dr. Abbie F. Salny's "The Mensa 365 Brain Puzzlers Calendar" for 2001.


August 8, 2002
Midweek MindBender
Interesting Numbers
The numbers 11, 22, 26, 101, 111, and 836 share an interesting characteristic. What is it? Hint: Think palindrome.
___________________________

...Answer to Midweek MindBender
Interesting Numbers

The square of each number is a palindrome (it reads the same backward and forward). 11*11=121; 22*22=484; 26*26=676; 101*101=10201; 111*111=12321; and 836*836=698896.
This MindBender was modified from a puzzle in Terry Stickels' "Mindbending Puzzles" calendar for 2001.


August 12, 2002
MindBender
Four-letter Word
Find a four-letter word with the following connections. You can do it many ways: you can do it on, you can do it in, you can do it out, you can do it back, and you can do it by a fiber. It appears shamefaced with a hound after it.
___________________________

Mini-MindBender for Kids
N's
If the numbers from 1 to 20 are written out in words (one, two, ..., twenty), how many N's will there be?
___________________________
...Answer to MindBender
Four-letter Word
HANG. Hang on, hang in, hang out, hang back, hang by a thread, hangdog.
This MindBender was modified from a puzzle in Victor Serebriakoff's book, "The Mammoth Book of Mindbending Puzzles."
___________________________

Answer to Mini-MindBender for Kids
N's
17.
This MindBender was modified from a puzzle in Victor Serebriakoff's book, "The Mammoth Book of Mindbending Puzzles."


August 15, 2002
Midweek MindBender
Six-digit Number
There is a six-digit number in which the first digit is one more than the second; the second digit is one more than the third; the fourth digit is one less than the third; and the fifth and sixth digits, taken together as a two-digit number, are the sum of the first and second. The sum of all the digits is 32. What is the six-digit number?
___________________________

...Answer to Midweek MindBender
Six-digit Number

876515.
This MindBender was modified from a puzzle in Dr. Abbie F. Salny's "365 Brain Puzzlers" Calendar for 2002.


August 19, 2002
MindBender
Balance
Consider three weight types: A-weights, B-weights, and C-weights. Weights of the same type consistently weigh exactly the same amount. On a balance scale, 3 A-weights balance with the combination of 4 B-weights and 4 C-weights. 2 A-weights balance with the combination of 3 B-weights and 2 C-weights. How many C-weights combined with 1 B-weight balance with 2 A-weights?
___________________________

Mini-MindBender for Kids
Tricycle Tires
If you ride a tricycle for 100 miles and want to equally use the three tires on the tricycle plus two spare tires that you have, how many miles will each tire travel assuming you change tires to give each tire equal use?
___________________________
...Answer to MindBender
Balance
6 C-weights. Consider the weight amounts as A, B, and C. Then we have: 3A=4B+4C and 2A=3B+2C or A=4B/3+4C/3 and A=3B/2+C. Therefore, 4B/3+4C/3=3B/2+C or 4C/3-C=3B/2-4B/3 or C/3=9B/6-8B/6 or C/3=B/6 or 2C=B. To balance the 2 A-weights, we need the equivalent of 3 B-weights and 2 C-weights. Since we want to have only 1 B-weight, we need 2 C-weights plus the 4 C-weights that equal the 2 B-weights we don't have. Therefore, we need a total of 6 C-weights.

An alternate solution with no division from one of the Mindbender Solvers:

Subtract 1 formula from the other giving:
(4B + 4C = 3A)
-(3B + 2C = 2A)
________________
B + 2C = A
Next substitute:
3B + 2C = 2(B + 2C)
3B + 2C = 2B + 4C
B = 2C
The rest becomes simple:
2A = 2B + 4C substite 2C for a B
2A = B + 6C
This MindBender was modified from a puzzle in Victor Serebriakoff's book, "The Mammoth Book of Astounding Puzzles."
___________________________

Answer to Mini-MindBender for Kids
Tricycle Tires
60 miles. Total tire travel distance is 3*100 or 300 miles. Since you have 5 tires, each tire travels 300/5 or 60 miles.
This MindBender was modified from a puzzle in Victor Serebriakoff's book, "The Mammoth Book of Astounding Puzzles."


August 22, 2002
Midweek MindBender
Repeater
What is the fraction equivalent of the following repeating decimal?
.980198019801...
___________________________

...Answer to Midweek MindBender
Repeater

99/101. Simply place the repeating part over the same number of 9s as the number of digits that are in the repeating portion and reduce. 9801/9999=99/101.
This MindBender was modified from a puzzle in Terry Stickels' "Mindbending Puzzles" calendar for 2000.


August 26, 2002
MindBender
Old/Older/Oldest
If Tom is twice as old as Howard will be when Jack is as old as Tom is now, who is oldest, next oldest, and youngest?
___________________________

Mini-MindBender for Kids
TicTacToe Twist
Here is a twist on the very old game of TicTacToe. Using a 3X3 TicTacToe game board and three markers for each palyer (for example, three pennies and three dimes), follow the following rules of play. Flip a coin to see who goes first. The players take turns placing their markers on the game board. The first player is not allowed to place his/her first marker in the center square. The object of the game is to be the first player to get three markers in a row, horizonally, vertically, or diagonally. After all six markers are on the game board, the players take turns sliding any one of their markers horizontally, vertically, or diagonally (only along either of the two main diagonals) to an open space.
___________________________
...Answer to MindBender
Old/Older/Oldest
Tom is oldest, then Jack, then Howard. From the problem statement, we have the following:
T=2(H+y) where y is positive.
J+y=T so Tom is older than Jack (T>J).
From above we have T=2H+2y or T-y=2H+y and T-y=J.
Substituting gives us J=2H+y so Jack is older than Howard (J>H).
And in summary, we have T>J>H.
This MindBender was modified from a puzzle in C. R. Wylie Jr.'s book, "101 Puzzles in Thought & Logic."
___________________________

Answer to Mini-MindBender for Kids
TicTacToe Twist
There is no answer to this Mini-MindBender for Kids. Just play the game and enjoy it.
This MindBender was modified from a puzzle in Raymond Blum's book, "Math Tricks, Puzzles & Games."


August 29, 2002
Midweek MindBender
Mathematical Anagram
Rearrange the letters of "eleven plus two" to make the following mathematically true:
eleven plus two = an anagram of "eleven plus two"
___________________________

...Answer to Midweek MindBender
Mathematical Anagram

eleven plus two = twelve plus one
This MindBender was modified from a puzzle in Cliff Pickover's "Mindbending Puzzles" calendar for 2002.


September 2, 2002
MindBender
Word Feature
The following six words share a common feature. What is it? EVIL BATS REWARD POTS SPOOL TIME
___________________________

Mini-MindBender for Kids
New Words
Can you find a four-letter word that makes four new words when it (the four-letter word) is placed where the four question marks are? FIRST???? ????SHAKE BACK???? ????STAND
___________________________
...Answer to MindBender
Word Feature
Each word is a different word when spelled backwards: LIVE STAB DRAWER STOP LOOPS EMIT
This MindBender was modified from a puzzle in Terry Stickels' "Are You As Smart As You Think?"
___________________________

Answer to Mini-MindBender for Kids
New Words
HAND. It will make the words: FIRSTHAND HANDSHAKE BACKHAND HANDSTAND.
This MindBender was modified from a puzzle in Terry Stickels' "Are You As Smart As You Think?"


September 5, 2002
Midweek MindBender
Sequence
Find the missing number in the following sequence:
1 2 4 7 11 ?? 22
___________________________

...Answer to Midweek MindBender
Sequence

16. Each number in the sequence adds 1, 2, 3, 4, 5, and 6, respectively, to the preceding number.
This MindBender was modified from a puzzle in Dr. Abbie F. Salny's "The Mensa 365 Brain Puzzlers Calendar" for 2001.


September 9, 2002
MindBender
Swimming Pool
A rectangular swimming pool of constant depth is twice as long as it is wide. If its length is reduced by 12 feet and its width is increased by 10 feet, the pool will hold exactly the same amount of water. What are the original length and width of the swimming pool?
___________________________

Mini-MindBender for Kids
Three Boys
Three boys, Andy, Ben, and Carl are 8, 9, and 10, but not necessarily respectively. Their last names are Smith Jones, and Rogers, but not necessarily respectively. 1) Jones is younger than Carl, but older than Smith. 2) Ben is not the youngest. Determine each boy's full name and age.
___________________________
...Answer to MindBender
Swimming Pool
Algebra to the puzzle rescue again. Let W be the width and 2*W be the length. Then we have:
2*W*W=(2*W-12)*(W+10) or
2*W*W=2*W*W+8*W-120 or
8*W=120 or W=15 so the width is 15 feet and the length is 30 feet.
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
Three Boys
Andy Smith is 8. Ben Jones is 9. Carl Rogers is 10. The first statement means that Jones is 9, Smith is 8, and Carl Rogers is 10. Statement 2 requires that Ben Jones is 9. So Andy Smith is 8.
This MindBender was modified from a puzzle in "Brain Games," a book by Dona Herweck Rice, et al.


September 12, 2002
Midweek MindBender
Marbles
A bag of marbles can be divided equally among 2, 11, or 13 kids. What is the minimum number of marbles that can be in the bag?
___________________________

...Answer to Midweek MindBender
Marbles

286. This is the least common multiple of 2, 11, and 13. 286 = 2*11*13.
This MindBender was modified from a puzzle in Terry Stickels' "Mindbending Puzzles" calendar for 2001.


September 23, 2002
MindBender
15 Lines
Four straight lines can cross each other in a maximum of 6 intersections (assuming that no lines are parallel). If there are 15 straight lines, what is the maximum number of intersections?
___________________________

Mini-MindBender for Kids
The Dog
Alex went to school and told his teacher that the dog ate his math homework. He turned in the following combination addition and subtraction problem where "?" indicates a spot where the dog ate a hole in the paper. Can you figure out what Alex's correctly solved problem was? 
 ??35 
+524? 
_____ 
 66?? 
-?374 
_____ 
 4??2
___________________________
...Answer to MindBender
15 Lines
105. The answer can be found in two ways. For the four line case, the first line can cross 3 other lines; the second line can cross 2 lines (other than the first line's crossing which has already been counted); the third line can cross 1 line (other than the first line's and the second line's crossings which have already been counted); the fourth line's crossings have all been counted. This leaves us with 3+2+1=6 intersections. This answer can also be calculated by the combinations of 4 things taken 2 at a time, or C(4,2)=4!/((4-2)!*2!)= 4*3*2*1/(2*1*2*1)=6. For the 15 line case, the answer is 14+13+12+...+1=105 or C(15,2)=15!/((15-2)!*2!)=15*14/2=105.
This MindBender was modified from a puzzle in David J. Bodycombe's "The IQ Obstacle Course."
___________________________

Answer to Mini-MindBender for Kids
The Dog
Start with the units part of the subtraction and just keep working to find all the missing digits. 
 1435 
+5241 
_____ 
 6676 
-2374 
_____ 
 4302
This MindBender was modified from a puzzle in Highlights "Mathmania," a good thinking resource for young kids.


September 26, 2002
Midweek MindBender
Ages
John's age is now two-thirds Mary's age. In four years, John will be three-fourths Mary's age. How old are they now?
___________________________

...Answer to Midweek MindBender
Ages

John is 8. Mary is 12.
J=2/3*M
J+4=3/4*(M+4)
2/3*M+4=3/4*M+3
1=3/4*M-2/3*M
1=1/12*M
M=12
J=2/3*12=8
This MindBender was modified from a puzzle in Dr. Abbie F. Salny's "365 Brain Puzzlers" Calendar for 2002.


September 30, 2002
MindBender
Anagram
The following 15 letters can be anagrammed into a four-word message that is commonly used on postcards:
EEEEIOUHHRRSWWY
Can you determine what that four-word message is?
___________________________

Mini-MindBender for Kids
The Dog Again
Max went to school and told his teacher that the dog ate his math homework. He turned in the following combination addition and subtraction problem where "?" indicates a spot where the dog ate a hole in the paper. Can you figure out what Max's correctly solved problem was? 
  7?8 
+ ?1? 
_____ 
 ?107 
- 42? 
_____ 
  ??1
___________________________
...Answer to MindBender
Anagram
WISH YOU WERE HERE.
This MindBender was modified from a puzzle in Dr. Abbie F. Salny's "The Mensa 365 Brain Puzzlers Calendar" for 2001.
___________________________

Answer to Mini-MindBender for Kids
The Dog Again
Start with the units part of the addition and just keep working to find all the missing digits. 
  788 
+ 319 
_____ 
 1107 
- 426 
_____ 
  681
This MindBender was modified from a puzzle in Highlights "Mathmania," a good thinking resource for young kids.


October 3, 2002
Midweek MindBender
Another Sequence
What are the next two numbers in the following sequence?
15 30 20 27 25 24 30 21 35 18 ?? ??
___________________________

...Answer to Midweek MindBender
Another Sequence

40 and 15. There are two different sequences alternating every other number. The first sequence begins with 15 and increases by 5 each time. The second sequence begins at 30 and decreases by 3 each time.
This MindBender was modified from a puzzle in Terry Stickels' "Mindbending Puzzles" calendar for 2000.


October 7, 2002
MindBender
Ten-digit Numbers
Find two ten-digit numbers, each containing the digits from 0 to 9 once and only once, with the property that successive pairs of digits, from left to right, are evenly divisible in turn by 2, 3, 4, 5, 6, 7, 8, 9, and 10. For example, if the ten-digit number is ABCDEFGHIJ, then AB is evenly divisible by 2, BC by 3, and so on, until IJ is evenly divisible by 10.
___________________________

Mini-MindBender for Kids
Important Numbers
Abby knows for absolutely sure that "numbers are important" and so she is bothered by the following incomplete combination addition and subtraction problem. Occurrences of "?" indicate spots where the problem is incomplete. Can you figure out what Abby's missing important numbers are? 
  ?1?? 
+ 5?13 
______ 
  ?677 
- 5?3? 
______ 
  22?5
___________________________
...Answer to MindBender
Ten-digit Numbers
The solutions are 1,872,549,630 and 7,812,549,630, and are derived as follows. Every second digit starting at the second digit must be even and every second digit starting at the first must be odd. The 0 and 5 can be placed immediately. The sixth digit must be 4. The seventh digit is odd (since every second digit must be even), so it must be 9. The eighth digit must be 6. The ninth digit must be 3. The third digit is 1 or 7, so the fourth digit must be 2. The three first digits are therefore 187 or 781.
This MindBender was modified from a puzzle on Humor Shack's Daily Riddle email list. Subscribe by visiting
http://www.humorshack.com/subscribe
___________________________

Answer to Mini-MindBender for Kids
Important Numbers
Start with the units part of the addition and just keep working to find all the missing digits and you will then know all the important numbers in this puzzle. 
  2164 
+ 5513 
______ 
  7677 
- 5432 
______ 
  2245
This MindBender was modified from a puzzle in Highlights "Mathmania," a good thinking resource for young kids.


October 10, 2002
Midweek MindBender
Mathematical Signs
Insert mathematical signs between the numbers before the equal sign to make the following statement true.
22 3 4 5 = 15
___________________________

...Answer to Midweek MindBender
Mathematical Signs

22 - 3 X 4 + 5 = 15
This MindBender was modified from a puzzle in Cliff Pickover's "Mindbending Puzzles" calendar for 2002.


Next Page


Other Pages




If you have any riddles or MindBenders to add please e-mail me or e-mail Mike Avery
PaintSaint helped make this page. check out his Riddles page.