# Expanded Form

The **expanded form** of a number writes it as a sum, with each digit makes an individual term multiplied by its place value. For example $523$ has an expanded form of $5 \times 100 + 2 \times 10 + 3 ,$ and $6203$ has an expanded form of $6 \times 1000 + 2 \times 100 + 0 \times 10 + 3 .$

In general expanded form helps understand the meaning of place value. It can be useful in thinking of different number bases and also help in the solving of number cryptogram puzzles.

## Definition and examples

The expanded form of a number gives the number as a sum where each digit is separate term multiplied by its place value.

$943 = 9 \times 100 + 4 \times 10 + 3 \times 1$ because there are 9 hundreds, 4 tens, and 3 ones. (The $\times 1$ part in front of the $3$ is optional.)

$41000 = 4 \times 10000 + 1 \times 1000$ because there 4 ten thousands and 1 thousand. We can also write alternately write it as $41000 = 4 \times 10000 + 1 \times 1000 + 0 \times 100 + 0 \times 10 + 0 \times 1 .$

## Number bases

Expanded form can be used to think about bases other than 10.

For example, while $62 = 6(10) + 2(1) ,$ it also can be written in base 2 as 111110 since $62 = 1(32)+1(16)+1(8)+1(4)+1(2)+0(1) .$ (The number 111110 is then sometimes written as ${111110}_2$ to indicate base 2.)

In general, if we're in base $b,$ we can think of the expanded form as

$\ldots + (\text{third to last digit})(b^2) + (\text{second to last digit})(b^1) + (\text{last digit})(b^0)$

where the dots extend as far as necessary to make a complete sum. For example, when $b = 2$ (binary, as is used in computers):

$\ldots + (\text{third to last digit})(2^2) + (\text{second to last digit})(2^1) + (\text{last digit})(2^0)$

or alternately

$\ldots + (\text{third to last digit})(4) + (\text{second to last digit})(2) + (\text{last digit})(1) .$

Write the base 10 number 92 in base 3.

Base 3 will be of the format:

$\ldots + (\text{third to last digit})(3^2) + (\text{second to last digit})(3^1) + (\text{last digit})(3^0)$

The places in base 10 that would be "ones", "tens", "hundreds", "thousands", and "ten thousands" are now "ones" $(3^0=1),$ "threes" $(3^1=3),$ "nines" $(3^2=9),$ "twenty-sevens" $(3^3=27),$ "eighty-ones" $(3^4=81).$

Note that $3^5 = 243$ which is larger than our target number, so the highest we need to go is the "eighty-ones" place. Also, just like how in base 10 the ten digits we can use are $0, 1, 2, 3, 4, 5, 6, 7, 8,$ and $9,$ in base 3 the three digits we can use are $0, 1,$ and $2 .$

We can have at most one 81 from 92, leaving $92 - 81 = 11 .$ We can't remove 27 from 11, but we can remove one 9, leaving $11 - 9 = 2 .$ The 2 can the be included in the ones place. So we have

as our number. In expanded form the number ${10102}_3$ is $92 = 1(81) + 0(27) + 1(9) + 0(3) + 2(1) .$

More information can be found at the wiki page on number bases.

## Cryptogram solving

A cryptogram is a mathematical puzzle where various symbols are used to represent digits, and a given system has to be true. An example puzzle is below.

$\large{\begin{array}{ccccccc} && & & S& E & N&D\\ +&& & & M& O & R&E\\ \hline & & & M & O& N & E&Y \end{array}}$

Note we consider $SEND$ to be the four-digit number made of the letters $S,$ $E,$ $N,$ and $D.$ We can alternately write this in expanded form as $1000S + 100E + 10N + D .$ This is a common technique for cryptogram solving. (More detail about the above problem is at the wiki page dedicated to cryptograms.)

When a two digit number $AB$ is added to its reverse $BA,$ the sum is 55. Neither $A$ nor $B$ is zero. Find the largest possible value of $AB .$

We can write $AB$ in expanded form as $10A + B .$

We can write $BA$ in expanded form as $10B + A .$

The sum is 55, so $10A + B + 10B + A = 55 .$ Combining like terms, $11A + 11B = 55 ,$ which also implies $A + B = 5 .$

Now it's not too hard to list the possibilities for $A$ and $B:$ $1 + 4 = 5, 2 + 3 = 5, 3 + 2 = 5, 4 + 1 = 5 .$ $41$ is the largest number that can be made from this set, so is the desired answer.

$\begin{array} {ccccc} \huge & & & & \huge \color{#69047E}{X}& \huge \color{#69047E}{X}\\ \huge & & & & \huge \color{#D61F06}{Y}& \huge \color{#D61F06}{Y}\\ \huge + & & & & \huge \color{#3D99F6}{Z}& \huge \color{#3D99F6}{Z}\\ \hline \huge& & & \huge \color{#69047E}{X} & \huge \color{#D61F06}{Y} & \huge \color{#3D99F6}{Z} \end{array}$

**If each letter represents a different nonzero digit, what must $\large \color{#3D99F6}{Z}$ be?**