Place Value Games for 4th Graders

In early childhood education, kids deal with only smaller numbers that they use in their everyday life. In 4th grade curriculum, they get introduced to multi-digit whole numbers, which they just imagine to exist. So, the introduction should be done using a lot of educational resources like math manipulatives, place value charts, fun math activities, and learning games for kids.  Many math games for 4th graders are available online.

Place Value Progression

In grade 4, children learn about multi-digit place values and the connection between the place values of the digits of a number.

Consider the number 35269.

The number at the right most, 9 is at one's place, the number 6 at ten’s place, 2 is the hundred’s place, 5 is at thousand’s place, and 3 is at ten thousand’s place.

 

That is, there are 3 ten thousands, 5 thousands, 2 hundreds, 6 tens, and 9 ones in the number 35269.

The number can be expanded as   or  

In whole numbers, as you go from right to left, the place value gets multiplied by 10.

In other words, as you go from left to right, the place value gets divided by 10.

SplashLearn has a variety of place value activities for kids on understanding the multi-digit whole numbers place value definition. Practice place value up to hundred thousands to solve math problems involving place values.

Understanding Multi-digit Whole Numbers

In this level, children associate the place value strategies with reading and writing multi-digit whole numbers. Whole numbers are written in the groups of 3 separated by commas. Each set separated by commas should read as ones, tens, and hundreds respectively from right to left followed by base thousand unit.

A place value chart would make things clearer.

The place value worksheets for reading and writing multi-digit whole numbers available at SplashLearn covers the topic well.

 

Rounding Multi-digit Whole Numbers using Place Values

Once the kids have a deeper understanding of what is place value system, they can explain and reason about rounding a number to a specific place value.

For example, to round the number 77 to the nearest ten, first, the child needs to understand the number of tens in 77, that is 7. Now, the tens before and after 77 are 70 and 80 respectively. But 77 is closer to 80 than 70. Thus, rounding off the number 77 to the nearest ten results in 80.

The same logic applies for multi-digit whole numbers. To round 3578 to the nearest hundred, first, consider the hundreds before and after the number. The digit at the hundred’s place is 5 and there are 35 hundreds in 3578. Thus, the hundred before and after the number are 3500 and 3600 respectively. Plot the number on a number line.

Since 3578 is closer to 3600 than 3500, the rounded to the nearest hundred is 3600.

The child basically follows the place value strategies than the standard algorithm for rounding here.

The child also applies the similar logic in rounding numbers before and after operations like addition, multiplication etc in math word problems. Many math worksheets for rounding numbers are available at round numbers.

Addition of Multi-digit Whole Numbers using Standard Algorithm

In lower grades, kids add numbers using a number line or by decomposing it to smaller place values etc. By grade 5, they learn to use the standard algorithm of addition by lining-up the same place value numbers together.

For example, to find the sum . The first step is to line up the digits that have the same place value. That is, here 4 and 1 have the same place value, hence they need to be lined up.

So, the correct alignment would be:

 

Now, the corresponding place values can be added to find the sum. There are 4 ones in the first number and 1 one in the second, adding them together there are 5 ones. Similarly, there are 2 tens and 7 tens that add up to 9 tens. There are 3 hundreds in the first number and no hundreds in the second number, making 3 hundreds in total.

Thus, .

The procedure is smoother when there are no compositions of place values.

Children already know that a bundle of ten ones make-up one “ten”, a bundle of ten tens, make a “hundred” etc.

Consider the sum  .

First, line up at the right-hand side with the one’s places.

   

Start with adding the ones together, . But, 10 ones will add up to one ten and would be moved to the next place value on the left, tens leaving no more ones at the one’s place. Thus, the one’s place in the sum has a zero and one ten need to be added with the tens of the addends.

Continuing the process by adding the tens, there are  tens in the sum. Again, 10 tens make a hundred and it gets carried to the next place value, hundreds.

In a similar way, hundreds, thousands, and ten thousands can also be added.

Thus, the sum is 60,100.

Using place value chart, this can be shown as:

Addition of multi-digit numbers is explained in two different sections in SplashMath.

 

Subtraction of Multi-digit Whole Numbers using Standard Algorithm

Subtraction by taking away the base ten blocks from the minuend is something that kids are familiar with. In this level, kids get introduced to standard algorithm for subtraction.

Consider  .

Like addition, the standard algorithm for subtraction starts with aligning the numbers on the right-hand side with one’s places. The digits at the corresponding place values need to be subtracted from the other.

 

Here, 6 ones need to be subtracted from 3 ones which are not possible. Children have experience of decomposing one of the nearby tens into 10 ones while subtracting two digit numbers in grade 2. A similar procedure is followed here. But, instead of showing the base ten block numerals are used here. So, one of the 5 tens from the minuend is decomposed into 10 ones, thus, having 13 ones to subtract 6 ones from.

Now, the 2 tens have to be subtracted from 4 tens resulting in 2 tens.

At the hundred’s place, 8 hundreds have to be taken away from 1. This requires a decomposition of one thousand from the left.

Now, the 5 ten thousands have to be taken away from 5 ten thousands resulting in 0 ten thousands. Also, taking away zero hundred thousands from 8 hundred thousands resulting in 8 hundred thousands.

Thus, 856153 – 52826 = 803327 .

 

Multiplication using Place Value Strategies

Two-digit by One-digit: The place value strategy to multiply a two-digit number by a one-digit number is to decompose the two-digit number into tens and ones. That is, write the two-digit number in expanded form.

Consider the product  . The number 34 can be written in expanded form as 30 + 4.

A rectangle of dimensions 34 and 5 would have the area equal to the product. If the rectangle is divided lengthwise into two, each of length 30 and 4 respectively, the total area is the sum of the areas of the smaller rectangles.

This can be represented using array area model as:

That is, the total area is  .

 

Three-digit or Four-digit by One-digit: When the two-digit number is replaced by three-digit or four-digit numbers, the mathematics behind the process remains the same. The number is written in the expanded form and added them after multiplying each part separately.

For example, consider the product  .

The number 2843 can be written in expanded form as .

This can be represented using array area model as:

Again, the area of the rectangle of sides 4 and 2843 is the sum of the areas of the smaller rectangles.

That is,

Simplifying:  

Two-digit by Two-digit:

The mathematical techniques used for multiplying a two-digit number by another one is the same. Since both the numbers have ones’ and tens’ places, both have to be written in the expanded form before multiplying.

Consider . The two numbers can be expanded as  and  respectively. So, the array area model rectangle of the product would have dimensions  and . The product would be equivalent to the area of the rectangle which is divided into 4 smaller rectangles.

 

Find each product.

 

Add them.

 

Thus, .

A formative assessment of the concept learned can be done using free printable worksheets for kids available on the internet.

Division by One-digit Numbers

At this level, kids are not expected to follow the standard algorithm for dividing numbers. Instead, they use the connections between division and multiplication. And the logical question of how many divisors are “contained” in the dividend.

Consider the example  .

The child knows that  and . So, the quotient would be in between 10 and 20. Write the number 55 as . The number 30 contains 10 number of 3s. Now, consider 25. The child already knows that . So, the number 25 contains 8 number of 3s with a remainder 1.

Thus, the number 55 can be written as  and .

Since 30 has ten 3s and 24 has eight 3s,  has 10 + 8  or eighteen 3s. Thus, 55 can be written as . That is, when 55 is divided by 3, the quotient is 18 and the divisor is 1.

When the dividend is 3- or 4-digit numbers the calculations and estimation of the contained divisors become more tricky.

Consider  .

The dividend 2786 can be written in expanded form as  .

Consider a few multiples of 6 to estimate the number of 6s contained in 2786.

 and  are way too bigger than the dividend and  is less than the dividend. So, the quotient would be between 400 and 500. Decompose 2786 with one part equal to 2400. That is, .

In the next level, the child needs to estimate the number of 6s in 386.

The number 300 is less than 386 and there are fifty 6s in 300.

Thus,  .

Now, the next estimation would be the number of 6s in 86.

So, the number of 6s in 86 is between 10 and 15. But since 6 times 15 is just 4 greater than 86, try

Thus, the expanded form can be .

Now, the last digit in the expanded form is 2 and 2 < 6. So, the procedure can be stopped.

The total number of 6s in 2786 would be the number of 6s in each of these addends in the expanded form.

 

So, the total number is  or .

Thus, the quotient is 464 and the remainder is 2.

SplashLearn provides well-organized 4th grade worksheets from easy to difficult math problems for both multiplication and division.