Take any two odd numbers, like 5 and 7. Multiply them together. 5 X 7 = 35

Now some more odd numbers and multiply them together.

What can you say about multiplying two odd numbers?

Answer: When you multiply two odd numbers the answer is always an odd number.

Now take two even numbers, like 4 and 8. Multiply them together. 4 X 8 = 32

Now take some more even numbers and multiply them together.

What can you say about multiplying two even numbers?

Answer: When you multiply two even numbers the answer is always an even number.

Now take an even and an odd number, like 5 and 8. Multiply them together. 5 X 8 = 40

Now take some more even and odd pairs of numbers and multiply them together.

What can you say about multiplying an odd and even number?

Answer: The answer is always even.

If multiplication works like this with odd and even pairs, what can you say about multiplying more that two numbers together?

Look at the following multiplications and see if you can see a pattern:

2 X 3 X 4 X 5 = 120 E O E O = E

2 X 3 X 4 X 5 X 7 = 840 E O E O O = E

2 X 3 X 2 X 3 X 5 X 3 = 540 E O E O O O = E

2 X 3 X 3 X 5 X 3 = 270 E O O O O = E

3 X 5 X 3 = 45 O O O = O

Answer: Any even number in the multiplication will make the answer even.

## Friday, June 19, 2015

## Thursday, June 18, 2015

### Sums of Consecutive Numbers

Alice has been thinking about sums of consecutive numbers. Here is part of her working out:

9 = 4 + 5 and 2 + 3+ 4

10 = 1 + 2 + 3 + 4

11 = 5 + 6

12 = 3 + 4 + 5

13 = 6 + 7

14 = 2 + 3 + 4 + 5

15 = 7 + 8 and 4 + 5 + 6 and 1 + 2 + 3 + 4 + 5

Rice looked over Alice's shoulder:

"I wonder if we could write

"Some numbers can be written in more than one way! I wonder which ones?"

"9
, 12
and 15
can all be written using three consecutive numbers. I wonder if all multiples of 3
can be written in this way?"

Let's see if you can answer some of the questions.

Getting Started:

Start by trying some simple cases; perhaps start by exploring what happens when you add two or three consecutive numbers.

1 + 2 + 3 = 6

2 + 3 + 4 = 9

3 + 4 + 5 = 12

Can you explain why the total has gone up by 3
?

Let us know what you find out. I'll put your explanations with your name on the bottom as answers.

9 = 4 + 5 and 2 + 3+ 4

10 = 1 + 2 + 3 + 4

11 = 5 + 6

12 = 3 + 4 + 5

13 = 6 + 7

14 = 2 + 3 + 4 + 5

15 = 7 + 8 and 4 + 5 + 6 and 1 + 2 + 3 + 4 + 5

Rice looked over Alice's shoulder:

"I wonder if we could write

**every**number as the sum of consecutive numbers?""Some numbers can be written in more than one way! I wonder which ones?"

"9

Let's see if you can answer some of the questions.

Getting Started:

Start by trying some simple cases; perhaps start by exploring what happens when you add two or three consecutive numbers.

1 + 2 + 3 = 6

2 + 3 + 4 = 9

3 + 4 + 5 = 12

Let us know what you find out. I'll put your explanations with your name on the bottom as answers.

## Wednesday, June 17, 2015

### Classic Number Guessing

This trick will impress even your math teacher (but she may already know it). |

- Think of a number.
- Double it.
- Add 10
. - Halve it.
- Take away you original number.
- Is your answer
5?

Why does this happen?

Write the answer on a piece of paper without letting anybody see it and seal it in an envelope. Have somebody hold the envelope and at the end ask them to open it and reveal the number you wrote at the beginning. Wow, Magic!

You can also try this with a different number than 10 to be added.

What is your result then?

Answer: You can add any number instead of 10 and the final answer will be 1/2 of that number.

### Magic 123

Start with a number and count the number E of even and the number O of odd
digits. Let's say you pick the number 7397418. E=2 and O=5,

Write them down next to each other following by their sumE +
O. So the new number is 257.

Treat the result as a new number and continue the process. So 257 becomes 123. 123 becomes 123.

In this example, iterations converge very rapidly. Moreover, in just a few steps they reach the number 123 which has exactly 3 digits of which 1 is even and two are odd. Therefore, applying the computations to 123 produces the number 123 itself, such that further iterations become really mindless. Of course we can always start with another number.

Write them down next to each other following by their sum

Treat the result as a new number and continue the process. So 257 becomes 123. 123 becomes 123.

In this example, iterations converge very rapidly. Moreover, in just a few steps they reach the number 123 which has exactly 3 digits of which 1 is even and two are odd. Therefore, applying the computations to 123 produces the number 123 itself, such that further iterations become really mindless. Of course we can always start with another number.

### The Month With 19 Days

If you had a really old calendar, you would run across a month that really looks weird.

The strange appearance reflects the

The

By 1582, the cumulative effect of 11 minutes error has shifted the dates of the seasons by 13 days from Caesar's time. Pope Gregory XIII's reform reclaimed only 11 of the lost 13 days so that the date of the

The new dating system was proclaimed in 1582 by Pope Gregory XIII. Those countries where the change has been accepted immediately (the Italian states, Portugal, Spain, 1582; the catholic German states, 1583), the day following October 4, was reckoned as October 15. (As a matter of fact, St. Theresa of Ă€vila died during the night between the 4th and 15th of October 1582.) Protestant German states adopted the new calendar in 1699;

**September 1752**

**Su Mo Tu We Th Fr Sa
1 2 14 15 16
17 18 19 20 21 22 23
24 25 26 27 28 29 30**

The strange appearance reflects the

**we use today.***change from the Julian to the Gregorian calendar*The

*old style*Julian calendar was established by Julius Caesar in 45 B.C. to fix the Roman republican calendar (reckoned from the founding of Rome). Caesar, advised by Sosigenes, made the new calendar solar, not lunar. The length of the solar year was estimated at 365.25 days which was off by about 11 minutes. It's this 11 minutes that caused the problems with the Julian calendar.By 1582, the cumulative effect of 11 minutes error has shifted the dates of the seasons by 13 days from Caesar's time. Pope Gregory XIII's reform reclaimed only 11 of the lost 13 days so that the date of the

*vernal equinox*was restored to March 21, the date it had at the time of the Council of Nicaea, 325 A.D. With this reform we have fewer leap years too. In general, as before, every fourth year is leap except for those that are divisible by 100 but not by 400. So, for example, 1996 and 2000 are leap whereas 1900 and 2100 are not. The fact is, however, that, if all years divisible by 4,000 are denied their exceptional leap status, we would get even better conformity with actual time measurements. So far, no decision has been made to this effect. There is time yet.The new dating system was proclaimed in 1582 by Pope Gregory XIII. Those countries where the change has been accepted immediately (the Italian states, Portugal, Spain, 1582; the catholic German states, 1583), the day following October 4, was reckoned as October 15. (As a matter of fact, St. Theresa of Ă€vila died during the night between the 4th and 15th of October 1582.) Protestant German states adopted the new calendar in 1699;

**England and its colonies**in 1752 (which explains the September 1752's anomaly); Sweden in 1753; Japan in 1783; China in 1912; Russia in 1918, Greece in 1923.If ever asked: What's better, the Sun or the Moon? -
reply: The Moon. For the Sun only shines during daytime when it's light anyway whereas the Moon shines at night. |

Kozma Prutkov

### Baffling Prediction

Following is an simplified excerpt from

Now deal twelve cards to the table, face down. You ask the spectator to touch any four. The touched cards are turned face up. The remaining cards are gathered and returned to the bottom of the pack.

We will assume the four face-up cards to be a three, six, ten, and king. You state that you will deal cards on top of each of the four, dealing enough cards to bring the total of each pile up to ten. For example, you deal seven cards on the three, counting "4, 5, 6, 7, 8, 9, 10." Four cards are dealt on the six. No cards are dealt on the ten. Each court card counts as ten, so no cards are placed on the king.

The values of the four cards are now added: 3, 6, 10, and 10 equals 29. The spectator is handed the pack and asked to count to the 29th card. This card is turned over. Your prediction is now read. It is, of course, the name of the chosen card.

It is interesting to note that in this trick, as well as in others based on the same principle, you may permit the spectator to assign any value, from 1 to 10, to the jacks, kings, and queens. For example, he may decide to call each jack a 3, each queen a 7, and each king a 4. This has no effect whatever on the working of the trick, but it serves to make it more mysterious. Actually, the trick requires only that the deck consist of 52 cards - it matters not in the least what these cards are. If they were all deuces the trick would work just as well. This means that a spectator can arbitrarily assign a new value to any card he wishes without affecting the success of the trick!

Further mystification may be added by stealing two cards from the pack before showing the trick. In this case ten cards are dealt on the table instead of twelve. After the trick is over, the two cards are secretly returned to the pack. Now if a spectator tries to repeat the trick exactly as he saw it, it will not work.

*M.Gardner,**Mathematics, Magic and Mystery**The Trick:*You, as the magician, ask a spectator to shuffle a deck of cards and place it on the table. You then write the name of a card on a piece of paper and place it face down without letting anyone see what has been written.Now deal twelve cards to the table, face down. You ask the spectator to touch any four. The touched cards are turned face up. The remaining cards are gathered and returned to the bottom of the pack.

We will assume the four face-up cards to be a three, six, ten, and king. You state that you will deal cards on top of each of the four, dealing enough cards to bring the total of each pile up to ten. For example, you deal seven cards on the three, counting "4, 5, 6, 7, 8, 9, 10." Four cards are dealt on the six. No cards are dealt on the ten. Each court card counts as ten, so no cards are placed on the king.

The values of the four cards are now added: 3, 6, 10, and 10 equals 29. The spectator is handed the pack and asked to count to the 29th card. This card is turned over. Your prediction is now read. It is, of course, the name of the chosen card.

*Method:*After the deck is shuffled you casually note the bottom card of the pack. It is the name of this card that you write as your prediction. The rest works automatically. Gathering the eight cards and placing them on the bottom of the pack places the glimpsed card at the 40th position. After the cards are properly dealt, and the four face-up cards totaled, the count will invariably fall on this card. The fact that the deck is shuffled at the outset makes the trick particularly baffling.It is interesting to note that in this trick, as well as in others based on the same principle, you may permit the spectator to assign any value, from 1 to 10, to the jacks, kings, and queens. For example, he may decide to call each jack a 3, each queen a 7, and each king a 4. This has no effect whatever on the working of the trick, but it serves to make it more mysterious. Actually, the trick requires only that the deck consist of 52 cards - it matters not in the least what these cards are. If they were all deuces the trick would work just as well. This means that a spectator can arbitrarily assign a new value to any card he wishes without affecting the success of the trick!

Further mystification may be added by stealing two cards from the pack before showing the trick. In this case ten cards are dealt on the table instead of twelve. After the trick is over, the two cards are secretly returned to the pack. Now if a spectator tries to repeat the trick exactly as he saw it, it will not work.

### Brain Teasers #1

1. You begin reading a book 240 pages long. If you read half of the remaining book each day how long would it take you to finish the book?

Answer: You will never finish it.

2. Add a single line to the equation in order to make it true.

105 + 2 + 5 = 350

Answer: Add a line to the second plus sign in order to make it into a four.

105 + 245 = 350

Or: Using the same small diagonal line and placing it on the equals sign also makes the statement true, since 105+2+5 does not equal 350.

3. What is the product of multiplying all the numbers from 20 to 0 in order? (20 x 19 x 18.....)

Answer: Zero

4. What number when multiplied against itself will result in a number which includes the numbers (1 - 9) in forward order, and then descending in order from the 9? (i.e., to get the number: 12,345,678,987,654,321)

Answer: 111,111,111 X 111,111,111 = 12,345,678,987,654,321

5. Which number should replace the question mark?

A. 4

B. 5

C. 6

D. 7

Answer: A, the sum of the first two numbers in a row is the product of the last two numbers in the row.

6. At the end of a banquet 10 people shake hands with each other. How many handshakes will there be in total?

A. 100

B. 20

C. 45

D. 50

E. 90

Answer: C, 9+8+7+6+5+4+3+2+1=45

7. 165135 is to peace as 1215225 is to

A. lead

B. love

C. loop

D. castle

Answer: love, letters are encoded with ASCII type code

Answer: You will never finish it.

2. Add a single line to the equation in order to make it true.

105 + 2 + 5 = 350

Answer: Add a line to the second plus sign in order to make it into a four.

105 + 245 = 350

Or: Using the same small diagonal line and placing it on the equals sign also makes the statement true, since 105+2+5 does not equal 350.

3. What is the product of multiplying all the numbers from 20 to 0 in order? (20 x 19 x 18.....)

Answer: Zero

4. What number when multiplied against itself will result in a number which includes the numbers (1 - 9) in forward order, and then descending in order from the 9? (i.e., to get the number: 12,345,678,987,654,321)

Answer: 111,111,111 X 111,111,111 = 12,345,678,987,654,321

5. Which number should replace the question mark?

17 | 8 | 5 | 5 |

13 | 7 | 5 | 4 |

6 | 12 | 6 | 3 |

10 | 6 | 4 | ? |

A. 4

B. 5

C. 6

D. 7

Answer: A, the sum of the first two numbers in a row is the product of the last two numbers in the row.

6. At the end of a banquet 10 people shake hands with each other. How many handshakes will there be in total?

A. 100

B. 20

C. 45

D. 50

E. 90

Answer: C, 9+8+7+6+5+4+3+2+1=45

7. 165135 is to peace as 1215225 is to

A. lead

B. love

C. loop

D. castle

Answer: love, letters are encoded with ASCII type code

## Tuesday, June 16, 2015

### Classic Puzles - Party Time

Puzzle:

Several kids of varying ages brought things for the end of year party. Sheldon, who is 11, brought POTATO CHIPS. Margaret, who is 12, brought PINK LEMONADE. Peter, who is 9, brought PAPER CUPS. Sheila, who is the same age as Margaret, brought PARTY POPPERS. Newt, who is the same age as Sheldon, brought PAPER PLATES. Young Henry brought POPCORN. How old is Henry?

Henry is nine if he copied everyone else and brought something with the same number of letters as their age.

Several kids of varying ages brought things for the end of year party. Sheldon, who is 11, brought POTATO CHIPS. Margaret, who is 12, brought PINK LEMONADE. Peter, who is 9, brought PAPER CUPS. Sheila, who is the same age as Margaret, brought PARTY POPPERS. Newt, who is the same age as Sheldon, brought PAPER PLATES. Young Henry brought POPCORN. How old is Henry?

Henry is nine if he copied everyone else and brought something with the same number of letters as their age.

### Classic Puzzles - Calendar Confusion

### The Puzzle:

If I said that in three days' time it would be a Thursday, I am sure that most of you would have no difficulty telling me that today was a Monday.

Try this one then. Yesterday was two days before Monday. What day is it today? Yes, you're right again. It's Sunday. Do you get the idea?

Now let's tackle a similar question from The National Mathematics Contest (1991) Paper:

Three days ago, yesterday was the day before Sunday. What day will it be tomorrow?

Try this one then. Yesterday was two days before Monday. What day is it today? Yes, you're right again. It's Sunday. Do you get the idea?

Now let's tackle a similar question from The National Mathematics Contest (1991) Paper:

Three days ago, yesterday was the day before Sunday. What day will it be tomorrow?

When this was tested, the order of popularity of answers handed in were: Sunday (9), Thursday (4), Monday (3), Friday (2), Wednesday (1), Saturday (1). THE RIGHT ANSWER WAS THURSDAY. Why?

Three days ago, yesterday was the day before Sunday, so three days ago was itself Sunday.

That means today is Wednesday, so tomorrow is Thursday

### Classic Puzzles - Alphabet Numbers

### The Puzzle:

Given only one of each letter in the alphabet, what are

the smallest and largest numbers that you could write down?

the smallest and largest numbers that you could write down?

Help: You can't use the same letter twice, so million, trillion, and googol are all out. Answer below.

Using only one of each letter in the alphabet, you can spell:

ZERO or NOUGHT

MINUS FORTY (allowing negative numbers)

FIVE THOUSAND

### Classic Puzzles - 12 Days of Christmas

Traditional Cross Stitch |

## The Puzzle:

According to the traditional song, on the first day of Christmas (25th December), my true love sent to me:

. A partridge in a pear tree

On the second day of Christmas (26th December), my true love sent to me THREE presents:

. Two turtle doves

. A partridge in a pear tree

On the third day of Christmas (27th December and so on) my true love sent to me SIX presents:

. Three French hens

. Two turtle doves

. A partridge in a pear tree

This carries on until the twelfth day of Christmas, when my true love sends me:

Twelve drummers drumming

Eleven pipers piping

Ten lords a-leaping

Nine ladies dancing

Eight maids a-milking

Seven swans a-swimming

Six geese a-laying

Five gold rings

Four calling birds

Three French hens

Two turtle doves

A partridge in a pear tree

After the twelve days of Christmas are over, how many presents has my true love sent me altogether?

. A partridge in a pear tree

On the second day of Christmas (26th December), my true love sent to me THREE presents:

. Two turtle doves

. A partridge in a pear tree

On the third day of Christmas (27th December and so on) my true love sent to me SIX presents:

. Three French hens

. Two turtle doves

. A partridge in a pear tree

This carries on until the twelfth day of Christmas, when my true love sends me:

Twelve drummers drumming

Eleven pipers piping

Ten lords a-leaping

Nine ladies dancing

Eight maids a-milking

Seven swans a-swimming

Six geese a-laying

Five gold rings

Four calling birds

Three French hens

Two turtle doves

A partridge in a pear tree

After the twelve days of Christmas are over, how many presents has my true love sent me altogether?

As usual, you can check the answer below by highlighting the yellow.

Day by Day:

1 + 3 + 6 + 10 + 15 + 21 + 28 + 36 + 45 + 55 + 66 + 78 = 364 presents

Which is really interesting when you think there are 365 days in a typical year!

1 + 3 + 6 + 10 + 15 + 21 + 28 + 36 + 45 + 55 + 66 + 78 = 364 presents

Which is really interesting when you think there are 365 days in a typical year!

It's also interesting that if you count up the name of each present, you come up with this symmetrical list:

12 Partridges (12 days x 1)

22 Turtle Doves (11 days x 2)

30 French Hens (10 days x 3)

36 Calling Birds (9 days x 4)

40 Gold Rings (8 days x 5)

42 Geese Laying (7 days x 6)

42 Swans swimming (6 days x 7)

40 Maids a Milking (5 days x 8)

36 Ladies dancing (4 days x 9)

30 Lords a Leaping (3 days x 10)

22 Pipers Piping (2 days x 11)

12 Drummers drumming (1 day x 12)

22 Turtle Doves (11 days x 2)

30 French Hens (10 days x 3)

36 Calling Birds (9 days x 4)

40 Gold Rings (8 days x 5)

42 Geese Laying (7 days x 6)

42 Swans swimming (6 days x 7)

40 Maids a Milking (5 days x 8)

36 Ladies dancing (4 days x 9)

30 Lords a Leaping (3 days x 10)

22 Pipers Piping (2 days x 11)

12 Drummers drumming (1 day x 12)

### McNugget Numbers

With only 2 pence and 5 pence coins, one cannot make 3 pence, but one can make any higher amount. |

**Frobenius problem**, after the mathematician Ferdinand Frobenius). The problem that asks for the largest monetary amount that cannot be obtained using only coins of specified denominations. For example, the largest amount that cannot be obtained using only coins of 3 and 5 units is 7 units. The solution to this problem for a given set of coin denominations is called the

**Frobenius number.**

**"Various British Pennys". Licensed under CC BY-SA 3.0 via Wikipedia**

By Fritz Saalfeld (Own work) [CC BY-SA 2.5

The McNuggets version of the coin problem was introduced by Henri Picciotto, who included it in his algebra textbook. Picciotto thought of the application in the 1980s while dining with his son at McDonald's, working the problem out on a napkin. A McNugget number is the total number of McDonald's Chicken McNuggets in any number of boxes. The original boxes (prior to the introduction of the Happy Meal-sized nugget boxes) were of 6, 9, and 20 nuggets.

If you look at the numbers you can reach with boxes of 6, 9, and 20 nuggets, you notice that you can only reach multiples of 3 and multiples of 3 plus multiples of 20. So starting with 1, you can come up the the following list of numbers you cannot reach (non-McNugget numbers):

- 1, 2, 3, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19, 22, 23, 25, 28, 31, 34, 37, and 43

44 = 6 + 9 + 9 + 20

45 = 9 + 9 + 9 + 9 + 9

46 = 6 + 20 + 20

47 = 9 + 9 + 9 + 20

48 = 6 + 6 + 9 + 9 + 9 + 9

49 = 9 + 20 + 20

Now you know that

**all**the numbers above these can be reached. You can add 6 to each of these six numbers and get the next six numbers. You can do this to infinity, so 43 is the largest non-McNugget Number. Anytime you can get the smallest number that number times in a row, you've got the largest non-McNugget number.

Bonus Round: Since the introduction of the 4 piece Happy Meal, what is the largest non-McNugget number. Use 4, 6, 9 and 20. Answer Below.

Highlight to read answer

The answer is 11, did you get it?

## Sunday, June 14, 2015

### What's A Crytogram and How Do I Solve Them?

When people started using codes for messages, a simple code was good enough. Check out substitution codes and Augusta's code in the blog. As people got familiar with codes, they realized that they could be "broken", even if you didn't have a decoder ring. People got so good at breaking substitution codes that they started using them as puzzles called cryptograms. Cryptograms are still printed in some newspapers and you can buy books full of them at the supermarket, but the internet allows you to enjoy solving cryptograms anytime you want to.

A cryptogram is a block of text which has been made unreadable through the use of what is called a "substitution cypher"(Check it out in the blog list). This means that each letter used in the original text has been substituted with another (G becomes A, F becomes P, etc.). Letter/word positions, spaces and punctuation remain unchanged.

Instead of making up some cryptograms, I have put a link to a site that specializes in cryptograms below. I'm also going to list some of their hints on how to solve them.

Many, if not most puzzles, will have one or more words which are composed of only a single letter. In the English language, the only two commonly used one-letter words are

Frequency analysis is a fancy term for a simple idea - certain letters appear far more often in the English language than others. That's where

How does this help? Well, you'll notice in our (Cryptograms.org) cryptograms, we provide a number below each letter. That number tells you how often that particular letter appears in the puzzle (i.e. that letter's "frequency analysis"). If, for example, a letter appears twelve times in a puzzle, much more often than any other letter, then it is a very good bet (though by no means certain) that that letter can be decoded to one of the ETAOIN group. More often than not, it will decode to 'E' or 'T'.

####

You may have hated learning about contractions, but here in crypto-land, contractions are extremely useful! Contractions are simply words that combine two words into a shorter, single word by replacing certain internal letters with an apostrophe. Some examples are

The reason contractions and possessives are so useful in decoding cryptograms is that only a small number of letters can be used in them immediately after the apostrophe. Possessives will only ever use 'S' - contractions have more options, however:

#4

By now you maybe have placed an 'A' or an 'I' on the board, if there were any one-letter words available, and maybe you've even placed an 'E' or a 'T' via frequency analysis. At this point you may start to see some two- and three-letter words which now have a single letter decoded in them. There are only a handful of common two-letter words, and not very many more three-letter words, so you can start analyzing each to see where they may and may not fit.

Be especially sure to search for appearances of 'THE' and 'AND' - two of the most commonly used words in the english language. Even if no letters have yet been decoded you can often use frequency analysis (remember ETAOIN?) to find one or both of these words. Look for three letter words with a frequency analysis pattern of HIGH-MEDIUM-HIGH (for 'THE') and HIGH-HIGH-MEDIUM (for 'AND').

####

Certain less-common letters in the english language tend to "pair up" with other letters in two-letter sequences commonly referred to as "digraphs." 'H' is one example - particularly when it is the last letter of a word. A partially-decoded word like ----H, for example, will probably end in -CH, -PH, -SH or -TH, just because there are very few other letters that can pair up with H near the end of a word.

It is also extremely useful to look for double-letter digraphs, i.e. letters which appear in duplicate (one directly after the other) in the same word. These can often be a dead giveaway, and especially so in 3- and 4-letter words. Only two vowels, 'E' and 'O', are commonly used as double-letter vowel digraphs.

####

Longer words with more than 5 or 6 letters will often contain prefixes and/or suffixes, both of which can be a big help in decoding a puzzle. Try to keep some of the more common prefixes and suffixes in mind for these longer words, and see if any of them might fit the bill.

Some of those suffixes also have frequently appearing, longer variants which can sometimes decode additional letters:

####

We've already covered common words with one, two and three letters, but there are a handful of other, longer words which also appear frequently in the english language.

I like cryptograms, but they are hard to make up and take a lot of effort to solve. One of my favorite places for cryptograms is here:

Cryptograms.org - online and printable cryptograms.

You can get your cryptograms with the letters already counted and a "used letter" list that updates as you fill in the chart. The hints above are from this site.

A cryptogram is a block of text which has been made unreadable through the use of what is called a "substitution cypher"(Check it out in the blog list). This means that each letter used in the original text has been substituted with another (G becomes A, F becomes P, etc.). Letter/word positions, spaces and punctuation remain unchanged.

Instead of making up some cryptograms, I have put a link to a site that specializes in cryptograms below. I'm also going to list some of their hints on how to solve them.

Solving Cryptograms

#1. Start with the one-letter words. Many, if not most puzzles, will have one or more words which are composed of only a single letter. In the English language, the only two commonly used one-letter words are

*I*and*a*, so it's usually a safe bet that any single-letter word in your puzzle can be decoded to one of those two.#2. Frequency analysis and the importance of ETAOIN. Frequency analysis is a fancy term for a simple idea - certain letters appear far more often in the English language than others. That's where

*ETAOIN*comes in handy. No, that's not the name of an exotic tribe or an extinct tongue. ETAOIN is simply a mnemonic device combining the six letters which appear most frequently in the English language. The letter 'E' appears much more frequently than any other letter in the alphabet, with 'T' the most common after that, 'A' the third most common, and so on.How does this help? Well, you'll notice in our (Cryptograms.org) cryptograms, we provide a number below each letter. That number tells you how often that particular letter appears in the puzzle (i.e. that letter's "frequency analysis"). If, for example, a letter appears twelve times in a puzzle, much more often than any other letter, then it is a very good bet (though by no means certain) that that letter can be decoded to one of the ETAOIN group. More often than not, it will decode to 'E' or 'T'.

####
#3. Contractions and possessives are your friend.

You may have hated learning about contractions, but here in crypto-land, contractions are extremely useful! Contractions are simply words that combine two words into a shorter, single word by replacing certain internal letters with an apostrophe. Some examples are *don't*,*they've*,*he'll*,*he's*,*I'm*,*she'd*, etc. Possessives also use apostrophes in a similar way, to show ownership - i.e.*woman's*,*child's*,*dog's*, etc.The reason contractions and possessives are so useful in decoding cryptograms is that only a small number of letters can be used in them immediately after the apostrophe. Possessives will only ever use 'S' - contractions have more options, however:

Common Endings for Contractions (With Examples)
| |

'T | won't don't isn't aren't weren't shouldn't didn't can't |

'S | he's she's it's |

'D | I'd he'd she'd they'd |

'M | I'm |

'RE | they're you're |

'VE | they've you've |

'LL | I'll he'll she'll they'll it'll |

**.**

Move on to the two- and three-letter words. By now you maybe have placed an 'A' or an 'I' on the board, if there were any one-letter words available, and maybe you've even placed an 'E' or a 'T' via frequency analysis. At this point you may start to see some two- and three-letter words which now have a single letter decoded in them. There are only a handful of common two-letter words, and not very many more three-letter words, so you can start analyzing each to see where they may and may not fit.

Most Common Two- and Three-Letter Words
| |

Two Letters: | of to in it is be as at so we he by or on do if me my up an go no us am |

Three Letters: | the and for are but not you all any can had her was one our out day get has him his how man |

Be especially sure to search for appearances of 'THE' and 'AND' - two of the most commonly used words in the english language. Even if no letters have yet been decoded you can often use frequency analysis (remember ETAOIN?) to find one or both of these words. Look for three letter words with a frequency analysis pattern of HIGH-MEDIUM-HIGH (for 'THE') and HIGH-HIGH-MEDIUM (for 'AND').

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#5. Look for digraph patterns.

Useful Letters with Commonly Appearing Digraphs
| |

H | CH SH TH PH WH |

K | CK SK LK KE |

Q | QU |

X | EX |

It is also extremely useful to look for double-letter digraphs, i.e. letters which appear in duplicate (one directly after the other) in the same word. These can often be a dead giveaway, and especially so in 3- and 4-letter words. Only two vowels, 'E' and 'O', are commonly used as double-letter vowel digraphs.

Common Words with Double-Letter Digraphs
| |

3 Letters | all add bee boo ell ebb egg fee goo too tee see |

4 Letters | ball been beer beet beep bell boom boot book bull butt call cell coon dell doll door doom fall fell feel feet foot food fool fuss full gull gall hall hell heed heel hill hull hoop hood hoof hoot jeep keen keel keep less lees mall need peel pall pool poof poll poor peek pass root reel reef reed roll room rood sass sell seen seem seed seek seer seep soon soot sill tall tell teen teem teed tool wall well watt weed week weep |

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#6. Consider common prefixes and suffixes

Longer words with more than 5 or 6 letters will often contain prefixes and/or suffixes, both of which can be a big help in decoding a puzzle. Try to keep some of the more common prefixes and suffixes in mind for these longer words, and see if any of them might fit the bill.Common Prefixes and Suffixes
| |

Prefixes | DE- DIS- EN- EM- IN- IM- MIS- OVER- PRE- RE- UN- |

Suffixes | -ABLE -AL -ED -EN -ER -EST -FUL -IBLE -IC -ING -ION -IVE -LESS -LY -MENT -NESS -OUS |

Some of those suffixes also have frequently appearing, longer variants which can sometimes decode additional letters:

Common Suffix Variants
| |

-ION | -TION -ATION -ITION |

-OUS | -IOUS -EOUS -ATIOUS -ITIOUS |

-IVE | -ATIVE -ITIVE |

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#7. Remember the more common words.

We've already covered common words with one, two and three letters, but there are a handful of other, longer words which also appear frequently in the english language.Common English Language Words
| |

4 letters | that with have this will your from they know want been good much some time very when come here just like long make many more only over such take than them well were |

5 letters | about where which their there today every would after other being first great these since under where while after |

6+ letters | through people between before |

I like cryptograms, but they are hard to make up and take a lot of effort to solve. One of my favorite places for cryptograms is here:

Cryptograms.org - online and printable cryptograms.

You can get your cryptograms with the letters already counted and a "used letter" list that updates as you fill in the chart. The hints above are from this site.

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