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.

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 $41000 = 4 \times 10000 + 1 \times 1000 + 0 \times 100 + 0 \times 10 + 0 \times 1 .$

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.

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.