digit number problem solving

Digit Word Problems

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Digit Word Problems

• The tens digit of a certain number is 3 less than the units digit. The sum of the digits is 11. What is the number?

Let x = units digit

x – 3 = tens digit

x + (x – 3) = 11

2x – 3 = 11

x = 7 (units digit)

x – 3 = 4 (tens digit)

Using the values above for units and tens, we find the number 47.

• The tens digit of a number is twice the units digit. If the digits are reversed, the new number is 27 less than the original. Find the original number.

Let x = units digit 2x = tens digits

Then the original number is 10(2x) + x, the reserved number is 10(x) + 2x, and the new number is the original number less than 27.

10(x) + 2x = 10(2x) + x – 27

12x = 21x – 27

x = 3  (units digit)

2x = 6 (tens digit)

The number is (6 X 10) + 3 or 63.

• The sum of the digits in a two-digit number is 12. If the digits are reversed, the number is 18 greater than the original number. What is the number?

Let x = units digit 12 – x = tens digit

Then the original number is 10(12 – x) + x and the reserved  number is 10(x) + (12 – x).

Equation:  The reserved number is the original number plus 18.

10(x) + (12 – x) = 10(12 – x) + (x) + 18

10x + 12 – x = 120 – 10x + x + 18

9x + 12 = 138 – 9x

12 – x = 5 (tens digit)

The number is ( 5 x 10) + 7 or 57

10(7) + (12 – 7) = 10(12- 7) + (7) + 18

70 + 5 = 50 + 7 + 18

• The tens digit of a certain number is 5 more than the units digit. The sum of the digits is 9. Find the number.

x + 5 = tens digit

x + (x + 5)  = 9

x = 2         (unit digit)

x + 5 = 7   (tens digit)

The number is (7 X 10) + 2 or 72.

• The tens digit of a two-digit number is twice the units digit. If the digits are reversed, the new number is 36 less than the original number. Find the number.

2x = tens digit

Then the number is 10(2x) + x and the reversed number is 10(x)+ 2x.

10(x) + 2x= 10(2x) + x – 36

12x  = 21x – 36

x = 4 (units digit)

2x = 8 (tens digit)

The number is (8 X 10) + 4 or 84.

• The sum of the digits of a two-digit number is 13. The units digit is 1 more than twice the tens digit. Find the number.

13 – x = tens digit

The units digit is twice the tens digit plus 1.

x = 2(13 – x)  + 1

x = 26 – 2x + 1

x = 27 – 2x

x = 9  (units digit)

13 – x = 4 (tens digit)

The number is (4 x 10)  + 9 or 49.

• The sum of the digits of a three-digit number is 6. The hundreds digit is twice the units digit, and the tens digit equals the sum of the other two. Find the number.

2x = hundreds digit

x + 2x = tens digit

x + 2x + (x + 2x) = 6

x = 1 (units digit)

2x = 2 (hundreds digit)

x + 2x = 3 (tens digit)

The number is (2 x 100) + (3 x 10) + 1 or 231.

• The units digit is twice the tens digit. If the number is doubled, it will be 12 more than the reversed number. Find the number.

Let x = tens digit.

2x = units digit.

Then the number is 10(x) + 2x and the reversed number is 10(2) = x.

Two times the number equals 12  more than the reversed number.

2[10(x) + (2x)] = 10(2x) + x + 12

2(12x) = 21x + 12

24x = 21x + 12

x = 4 (tens digit)

2x = 8 (units digit)

The number is (4 x 10) + 8 or 48.

• Eight times the sum of the digits of a certain two-digit number exceeds the number by 19.The tens digit is 3 more than the units digit. Find the number.

Let x = units digit (smaller)

x + 3 = tens digit

then the number is 10(x + 3) + x.

Eight times the sum of the digits exceeds the number by 19.

8[ x + (x + 3) ] – [ 10(x + 3) + x ] = 19

8[ 2x + 3 ] – [ 10x+30 + x ] = 19

16x + 24 – [ 9x + 30 ] = 19

16x + 24 – 9x – 30 = 19

5x – 6 = 19

x = 5 (units digit)

x + 3 = 8 (tens digit)

The number is (8 x 10) + 5 or 85.

• The ratio of the units digit to the tens digit of a two-digit number one-half. The tens digit is 2 more than the units digit. Find the number.

x + 2 = tens digit

The ratio is a fractional relationship.

x/(x + 2) = 1/2

Multiply by the LCD, 2(x + 2)

x = 2 (units digit)

x + 2 = 4 (tens digit)

The number is (4 x 10) + 2 or 42.

• There is a two-digit number whose units digit is 6 less than the tens digit. Four times the tens digit plus five times the units digit equals 51. Find the digits.

x + 6 = tens digit

Four times the tens digit plus five times the units digit equal 51.

4(x + 6) + 5x = 51

4x + 24 + 5x = 51

9x + 24 = 51

x = 3 (units digit)

x + 6 = 9 (tens digit)

The number is (9 x 10) + 3 or 93.

• The tens digit is 2 less than the units digit. If the digits are reversed, the sum of the reversed number and the original number is 154. Find the original number.

Let  x = units digit

x – 2 = tens digit

Then the number is 10(x – 2) + x and the reversed number is 10(x) + (x – 2)

The reversed number plus the original number equal 154.

10(x) + (x – 2) + 10(x – 2) + x = 154

10x + x – 2 + 10x -20 + x = 154

22x – 22 = 154

x = 8 (units digit)

x – 2 = 6 (tens digit)

The number is (6 X 10) + 8 or 68.

• A three-digit number has a tens digit 2 greater than the units digit and a hundreds digit 1 greater than the tens digit. The sum of the tens and hundreds digits is three times the units digit. What is the number?

(x + 2) + 1 = hundreds digit

The sum of the tens and hundreds digits is three times the units digit.

(x + 2) + (x + 2) + 1 = 3x

2x + 5 = 3x

x + 2 = 7 (tens digit)

(x + 2) + 1 = 8 (hundreds digit)

The number is (8 x 100) + (7 x 10) + 5 or 875.

• The sum of the digits of a two-digit number is 9. The value of the number is 12 times the tens digit. Find the number.

Let u = units digit

Let t = tens digit

Let u + 10t = the number itself

u + t = 9  sum of the digits is 9

u + 10t = 12t   value of the number is 12 times the tens digit

u = 2t   sub this into u + t = 9

So the tens digit is 3 and the units must be 6

So the number is 36

• The sum of the digits of a two-digit number is 12. If 15 is added to the number, the result is 6 times the units digit. Find the number.

u + t = 12      sum of the digits is 12

u + 10t + 15 = 6u      if 15 is added to the number, the result is 6 times the units digit

10t + 15 = 5u   Simplify

2t + 3 = u   divide through by 5

u = 2t + 3   sub this into u + t = 12

2t + 3 + t = 12

3t + 3 = 12

So the tens digit is 3 and the units must be 9

So the number is 39

39 + 15 = 6(9)

• The sum of the digits of a two-digit number is 8. If the digits of the number are reversed, the new number is 18 less than the original number. Find the number.

Let u + 10t = the original number

10u + t = the number with the digits reversed

u + t = 8      sum of the digits is 8.

10u + t + 18 = u + 10t     if the digits of the number are reversed, the new number is 18 less than the original number

9u + 18 = 9t    simplify

u + 2 = t      divide through by 9

t = u + 2     sub this into u + t = 8

u + (u + 2) = 8

So the units digit is 3 and the tens must be 5

So the number is 53

35 + 18 = 53

10u + t + 36 = u + 10t

9u + 36 = 9t    simplify

u + 4 = t      divide through by 9

t = u + 4     sub this into t = 2u

So the units digit is 4 and the tens must be 8

So the number is 48 and the number with the digits reversed is 84

48 + 36 = 84

• The units digit of a two-digit number is 4 times the tens digit. If the digits are reversed, the new number is 54 more than the original number. Find the number.

10u + t – 54 = u + 10t

9u – 54 = 9t    simplify

u – 6 = t      divide through by 9

u = t + 6     solve for u

u = t + 6     sub this into u = 4t

So the tens digit is 2 and the units digit must be 8

So the number is 28 and the number with the digits reversed is 82

82 = 28 + 54

• The sum of the digits of a two-digit number is 11. If 27 is added to the number, the digits will be reversed. Find the number.

u + 10t + 27 = 10u + t

9t + 27 = 9u    simplify

t + 3 = u      divide through by 9

u = t + 3     sub this into u + t = 11

(t + 3) + t = 11

2t + 3 = 11

So the tens digit is 4 and the units digit must be 7

So the number is 47 and the number with the digits reversed is 74

47 + 27 = 74

• The units digit of a two-digit number is 1 less than 3 times the tens digit. It the digits are reversed, the new number is 45 more than the original number. Find the number.

10u + t – 45 = u + 10t

9u – 45 = 9t    simplify

u – 5 = t      divide through by 9

u = t + 5     solve for u

u = t + 5     sub this into u + 1 = 3t

(t + 5) + 1 = 3t

So the tens digit is 3 and the units digit must be 8

So the number is 38 and the number with the digits reversed is 83

8 + 1 = 3(3)

38 + 45 = 83

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"Number" Word Problems

What are "number" word problems.

Number word problems involve relationships between different numbers; these exercises ask you to find some number (or numbers) based on those relationships.

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Algebra Word Problems

How do you solve number word problems?

To set up and solve number word problems, it is important clearly to label variables and expressions, using your translation skills to convert the words into algebra. The process of clear labelling will often end up doing nearly all of the work for you.

Number word problems are usually fairly contrived, but they're also fairly standard. Keep in mind that the point of these exercises isn't their relation to "real life", but rather the growth of your ability to extract the mathematics from the English. These exercises are a great way to stretch your mental muscles, use what you know already, apply your logic (and common sense), and then hippity-hop your way to the answer.

What is an example of solving a number word problem?

• The sum of two consecutive integers is 15 . Find the numbers.

They've given me many pieces of information here.

• I'm adding (that is, summing) two things
• the numbers are integers (like −3 and 6 )
• the second number is 1 more than the first
• the result of the addition will be 15

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How do I know that the second number will be larger than the first by 1 ? Because the two integers are "consecutive", which means "one right after the other, not skipping over anything between". (Examples of consecutive integers would be −12 and −11 , 1 and 2 , and 99 and 100 .)

The "integers" are the number zero, the whole numbers, and the negatives of the whole numbers. In going from one integer to the next consecutive integer, I'll have gone up by one unit.

I need to figure out what are the two numbers that I'm adding. The second number is defined in terms of the first number, so I'll pick a variable to stand for this number that I don't yet know:

1st number: n

The second number is one more than the first, so my expression for the second number is:

2nd number: n + 1

I know that I'm supposed to add these two numbers, and that the result will be (in other words, I should set the sum equal to) 15 . This, along with my translation skills, allows me to create an equation, being the algebraic equivalent to "(this number) added to (the next number) is (fifteen)":

n + ( n + 1) = 15

This is a linear equation that I can solve :

2 n + 1 = 15

The exercise did not ask me for the value of the variable n ; it asked for the identity of two numbers. So my answer is not " n  = 7 "; the actual answer, taking into account the second number, too, is:

The numbers are 7 and 8 .

It usually isn't required that you write your answer out like this; sometimes a very minimal " 7, 8 " is regarded as acceptible form. But the exercise asked me, in complete sentences, a question about two numbers; I feel like it's good form to answer that question in the form of a complete sentence.

What do they mean when they say "consecutive even (or odd) integers"?

Some number word problems will refer to "consecutive even (or odd) integers". This means that they're talking about two whole numbers (or their negatives) that are both even or else both odd; in particular, the two numbers are 2 units apart.

• The product of two consecutive negative even integers is 24 . Find the numbers.

I'll start with extracting the information they've given me.

• I'm multiplying (that is, finding the product of) two things
• those two things are numbers
• those two numbers are integers
• those two integers are even
• those two even integers are negative
• the second even integer is 2 units more than the first
• when I multiply, I'll get 24

How do I know that one number will be 2 more than the other? Because these numbers are consecutive even integers; the "consecutive" part means "the one right after the other", and the "even" part means that the numbers are two units apart. (Examples of consecutive even integers are 10 and 12 , −14 and −16 , and 0 and 2 .)

The second number is defined in terms of the first number, so I'll pick a variable for the first number. Then the second number will be two units more than this.

1st number: n 2nd number: n + 2

When I multiply these two numbers, I'm supposed to get 24 . This gives me my equation:

( n )( n + 2) = 24

This is a quadratic equation that I can solve :

( n )( n + 2) = 24 n 2 + 2 n = 24 n 2 + 2 n − 24 = 0 ( n + 6)( n − 4) = 0

This equation clearly has two solutions, being n  = −6 and n  = 4 . Since the numbers I am looking for are negative, I can ignore the " 4 " solution value and instead use the n  = −6 solution.

Then the next number, being larger than the first number by 2 , must be n  + 2 = −4 , and my answer is:

The numbers are −6 and −4 .

In the exercise above, one of the solutions to the exercise — namely, n  = −6 — was one of the solutions to the equation; the other solution to the equation — namely, n  = 4 — had the sign opposite to the other answer to the exercise.

You will encounter this pattern often in solving this type of word problem. However, do not assume that you can use both solutions if you just change the signs to be whatever you think they ought to be. While this often works, it does not always work, and it's sure to annoy your grader. Instead, throw out invalid results, and solve properly for the valid ones.

• Twice the larger of two numbers is three more than five times the smaller, and the sum of four times the larger and three times the smaller is 71 . What are the numbers?

The point of exercises like this is to give me practice in unwrapping and unwinding these words, somehow turning the words into algebraic expressions and equations. The point is in the setting-up and solving, not in the relative "reality" of the exercise. That said, how do I solve this? The best first step is to start labelling.

I need to find two numbers and, this time, they haven't given me any relationship between the two, like "two consecutive even integers". Since neither number is defined by the other, I'll need two letters to stand for the two unknowns. I'll need to remember to label the variables with their definitions.

the larger number:  x

the smaller number:  y

Now I can create expressions and then an equation for the first relationship they give me:

twice the larger:  2 x

three more than five times the smaller:  5 y + 3

relationship between ("is"):  2 x = 5 y + 3

And now for the other relationship they gave me:

four times the larger:  4 x

three times the smaller:  3 y

relationship between ("sum of"):  4 x + 3 y = 71

Now I have two equations in two variables:

2 x = 5 y + 3

4 x + 3 y = 71

I will solve, say, the first equation for x = :

x = (5/2) y + (3/2)

(There's no right or wrong in this choice; it's just what I happened to choose while I was writing up this page.)

Then I'll plug the right-hand side of this into the second equation in place of the x :

4[ (5/2) y + (3/2) ] + 3 y = 71

10 y + 6 + 3 y = 71

13 y + 6 = 71

y = 65/13 = 5

Now that I have the value for y , I can back-solve for x :

x = (5/2)(5) + (3/2)

x = (25/2) + (3/2)

x = 28/2 = 14

As always, I need to remember to answer the question that was actually asked. The solution here is not " x  = 14 ", but is instead the following:

larger number: 14

smaller number: 5

What are the steps for solving "number" word problems?

The steps for solving "number" word problems are these:

• Read the exercise through once; don't try to start solving it before you even know what it says.
• Figure out what you know (for instance, are you adding or multiplying?).
• Figure out what you don't know; this will probably be the value(s) of number(s).
• Pick one or more useful variables for unknown(s) that you need to find.
• Use the variable(s) and the known information to create expressions.
• Use these expressions and the known information to create one or more equations.
• Solve the equation(s) for the unknown(s).
• Check your definition(s) for your variable(s).
• Use this/these definition(s) to state your answer in clear terms.

But more than any list, the trick to doing this type of problem is to label everything very explicitly. Until you become used to doing these, do not attempt to keep track of things in your head. Do as I did in this last example: clearly label every single step; make your meaning clear not only to the grader but to yourself. When you do this, these problems generally work out rather easily.

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Digit Problems

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• Subtraction

3 Digit Math Problems

Three-digit numbers have three digits ranging from 100 to 999. These numbers are composed of hundreds, tens, and unit digits, which give them a unique value and place in the number system. Understanding and working with 3-digit numbers is an essential skill in mathematics.

They play a crucial role in developing math skills as they provide real-life contexts for applying mathematical concepts. Solving word problems helps students understand how to use mathematical operations such as addition and subtraction in practical situations. It enhances critical thinking and problem-solving abilities and promotes a deeper understanding of mathematical concepts.

Addition of 3 Digit Math Problems

Problem 1: adding two 3-digit numbers.

Example problem: 342 + 178 Solution steps and calculations:

• Line up the numbers vertically, aligning the units, tens, and hundreds of digits.
• Start by adding the units digits: 2 + 8 = 10. Write down 0 and carry over 1 to the tens column.
• Add the tens digits: 4 + 7 + 1 (carried over) = 12. Write down 2 and carry over 1 to the hundreds column.
• Add the hundreds of digits: 3 + 1 (carried over) = 4.
• Therefore, 342 + 178 = 520.

To add two 3-digit numbers, we start from the rightmost digit and move to the left, carrying over any excess values to the next column. By summing the corresponding digits in each column, we obtain the result. In the given example, the units column gives us 0, the tens column gives us 2, and the hundreds column gives us 4, resulting in 520.

Problem 2: Adding a 2-digit and a 3-digit number

Example problem: 97 + 534 Solution steps and calculations:

• Add the units digits: 7 + 4 = 11. Write down 1 and carry over 1 to the tens column.
• Add the tens digits: 9 + 3 + 1 (carried over) = 13. Write down 3 and carry over 1 to the hundreds column.
• Add the hundreds of digits: 5 + 1 (carried over) = 6.
• Therefore, 97 + 534 = 631.

Adding a 2-digit and a 3-digit number is similar to adding two 3-digit numbers. We start from the rightmost digit and move to the left, carrying over any excess values. In the given example, the units column gives us 1, the tens column gives us 3, and the hundreds column gives us 6, resulting in 631.

Subtraction 3 Digit Math Problems

Problem 1: subtracting a 3-digit number from another 3-digit number.

Example problem: 572 – 294 Solution steps and calculations:

• Start by subtracting the units digits: 2 – 4. Since 4 is greater than 2, borrow 1 from the tens column, making it 12 – 4 = 8.
• Subtract the tens digits: 7 – 9 (borrowed 1) = -2. Since we cannot have a negative digit, borrow 1 from the hundreds column, making it 17 – 9 = 8.
• Subtract the hundreds of digits: 5 – 2 = 3.
• Therefore, 572 – 294 = 278.

To subtract a 3-digit number from another 3-digit number, we start from the rightmost digit and move to the left. If the digit being subtracted is greater than the corresponding digit in the minuend, we borrow from the next column. In the given example, the units column gives us 8, the tens column gives us -2, and the hundreds column gives us 3, resulting in a difference of 278.

Problem 2: Subtracting a 2-digit number from a 3-digit number

Example problem: 765 – 87 Solution steps and calculations:

• Subtract the units digits: 5 – 7. Since 7 is greater than 5, borrow 1 from the tens column, making it 15 – 7 = 8.
• Subtract the tens digits: 6 – 8 (borrowed 1) = -2. Since we cannot have a negative digit, borrow 1 from the hundreds column, making it 16 – 8 = 8.
• Subtract the hundreds of digits: 7 – 0 = 7.
• Therefore, 765 – 87 = 678.

When subtracting a 2-digit number from a 3-digit number, the process is similar to subtracting two 3-digit numbers. We start from the rightmost digit and move to the left, borrowing from the next column if needed. In the given example, the units column gives us 8, the tens column gives us -2, and the hundreds column gives us 7, resulting in a difference of 678.

Mixed Addition and Subtraction Word Problems

Problem 1: mixed addition and subtraction involving 3-digit numbers.

Example problem: 432 + 275 – 167 Solution steps and calculations:

• Perform addition first: 432 + 275 = 707.
• Then subtract 707 – 167 = 540.

When faced with mixed addition and subtraction word problems involving 3-digit numbers, it is important to follow the order of operations (BIDMAS/BODMAS). In this case, addition is performed first, resulting in 707. Then, the subtraction is carried out, resulting in the final answer of 540.

Problem 2: Mixed addition and subtraction with multiple 3-digit numbers

Example problem: 846 – 219 + 374 Solution steps and calculations:

• Perform the subtraction first: 846 – 219 = 627.
• Then, perform the addition: 627 + 374 = 1001.

In this example of mixed addition and subtraction with multiple 3-digit numbers, we start with subtraction to obtain 627. Then, we perform the addition operation, resulting in the final answer, 1001.

Practicing 3-digit addition and subtraction, especially through word problems, helps develop essential math skills such as critical thinking, problem-solving, and applying mathematical concepts in real-life scenarios. It enhances numerical fluency and provides a solid foundation for more advanced math topics.

To strengthen math skills further, it is crucial to continue practicing and solving word problems involving 3-digit addition and subtraction. Regular practice will improve computational abilities and enhance logical reasoning and analytical thinking, empowering individuals to tackle more complex math problems in the future.

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Digit-Related Problems

For any three digit number, let h = the hundreds digit t = the tens digit and u = the units digit  

The number = 100 h + 10 t + u The number with digits reversed = 100 u + 10 t + h The sum of digits = h + t + u The product of digits = htu  

Example In a three digit number, the hundreds digit is twice the units digit. If 396 be subtracted from the number, the order of the digits will be reversed. Find the number if the sum of the digits is 17.  

The hundreds digit is twice the units digit $h = 2u$       ← equation (1)  

The sum of the digits is 17 $h + t + u = 17$       ← equation (2)  

396 be subtracted from the number $(100h + 10t + u) - 396 = 100u + 10t + h$

$99h - 99u = 396$

$h - u = 4$       ← equation (3)  

Substitute h = 2u to equation (3) $2u - u = 4$

From equation (1) $h = 2(4)$

From equation (2) $8 + t + 4 = 17$

The number is $100h + 10t + u = 854$           answer

• Example 01 | Digit-related problem
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Lesson Word problems on reversing digits of numbers

Problems on Digits and Numbers

We will learn how to solve different types of problems on digits and numbers.

1.  Numbers 1, 2, 3, 4, ........., 98, 99, 100 are multiplied together. The number of zeros at the end of the product on the right will be equal to

The value of the ‘n’ = 100.

Therefore, number of zeros at the end of 1 × 2 × 3 × 4 × 5 × .............. × 99 × 100

= (100 ÷ 5) + (100 ÷ 5^2)

Answer: (a)

Note: Number of zeros at the end of the product of natural numbers = n/5 + n/5^2 + n/5^3 + ........... (upto n terms)

2. How many three digit natural numbers are possible?

Number of three digit numbers = 9 × 10^(3 - 1) = 9 × 10^2 = 900

Answer: (d)

Note: Number of numbers with specific number of digits = 9 × 10^(d - 1), where ‘d’ = number of digits.

3. The difference of the squares of two nos. is 135 & their difference is 5. The product of the numbers is

 (b) 178

Solution: Let, two numbers be ‘a’ and ‘b’

According to the problem,

a - b = 5 and a^2 - b^2 = 135

Therefore, a + b = 135 ÷ 5 = 27

Since, a = (27 + 5) ÷ 2 = 16 and b = 16 - 5 = 11

Therefore, required value of ab = 16 × 11 = 176

4. The sum of x and y is three times their difference. Find the ratio of x and y:

a + b = 3(a - b)

or, 2a = 4b

Therefore, a : b = 4 : 2 = 2 : 1

5. The sum of the two numbers m and n is 5760, & the difference is one-third of the greater number. Which is the greater number?

Let, two numbers be x and y.

Now according to the problem,

x/3 = x - y

 or, 3x - 3y = x

or, 2y = 3y

or, x :  y = 3 : 2

Therefore, the greatest number = 5760 × 3/(3 + 2) =5760 × 3/5 = 3456

Answer: (b)

Math Employment Test Samples From Problems on Digits and Numbers to HOME PAGE

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Course: 3rd grade   >   Unit 3

• Subtraction by breaking apart

Break apart 3-digit subtraction problems

• Adding and subtracting on number line
• Subtract on a number line
• Methods for subtracting 3-digit numbers
• Select strategies for subtracting within 1000

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3-Digit Subtraction Worksheets

Equal parts exciting and challenging, our printable 3-digit subtraction worksheets help children enhance their subtraction skills with regrouping (or borrowing) and without regrouping. Awaiting to be explored here are an enthralling ensemble of activities like cross-number and picture puzzles, matching 3-digit numbers, finding the missing digits, and solving easy subtraction word problems. These pdf subtraction within 1000 worksheets are ideal for 2nd grade, 3rd grade, and 4th grade kids. Experience the nuance with our free worksheets!

3-Digit Minus 2-Digit Subtraction | No Regrouping

Say hello to cutting-edge subtraction on the back of our pdf 3-digit subtraction worksheets without borrowing! Young students are required to subtract 2-digit numbers from 3-digit numbers.

• Download the set

3-Digit Minus 2-Digit Subtraction | Regrouping

Compelling and trailblazing, our printable worksheets help kids find their feet working with 2-digit subtrahends and 3-digit minuends by regrouping. The answer keys are a big relief!

3-Digit Subtraction | Fill in the Missing Digits

Darting between easy and moderate levels, these exercises for grade 3 and grade 4 practice finding missing digits by adding or subtracting the given digits in each subtraction problem.

3-Digit Subtraction | Circle the Numbers

Feel always-at-it with a wealth of 10 problems in each pdf subtraction within 1000 worksheet! Circle the pairs of numbers that arrive at the given difference when subtracted.

3-Digit Subtraction | No Regrouping

No longer will subtracting 3-digit numbers feel weary and stressful! Kids can use these printables to practice subtracting two 3-digit numbers with no regrouping or borrowing.

3-Digit Subtraction | Regrouping

Renew children's enthusiasm for performing regrouping from hundreds to tens or tens to ones! Escorted by subtraction word problems, the learners will morph into true-blue subtraction fans!

3-Digit Subtraction | Cut-and-Glue Picture Puzzles

Upskill the fledgling mathematicians with printable 3-digit subtraction worksheets! The task is to match the answers to the numbers on the picture cards to complete the mystery picture.

3-Digit Subtraction | Cross-Number Puzzles

Ring in a subtraction leisure full of fun and frolic for grade 2 and grade 3 kids with our exclusive cross-number puzzles involving 3-digit minus 3-digit and 3-digit minus 2-digit subtraction!

3-Digit Subtraction Riddles

Watch a refreshing spell of amusement aboard as students solve riddles in printable worksheets. Work out 3-digit subtraction problems; map out the letters to reveal the answers.

3-Digit Minus 2-Digit or 3-Digit Line up Subtraction

This section of our pdf 3-digit subtraction worksheets focuses on line-up subtraction. Rearrange the numbers in vertical (column) form as per their place values and perform subtraction.

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4-Digit Addition Word Problems Worksheets

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Our pdf 4-digit addition word problems worksheets boast a wealth of real-life scenarios so children can practice adding numbers in the thousands to their heart's content. Students practice regrouping – or carrying – tens, hundreds, or thousands while performing the column addition or horizontal addition of four-digit numbers using real-life story problems. These printable word problems on addition worksheets, included with answer keys, are an effective tool to get your students adding numbers within 1000 with oodles of enthusiasm!

Our 4-digit addition word problem worksheets are suitable for students belonging to grade 3 and grade 4.

Related Printable Worksheets

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SOLVING WORD PROBLEMS BASED ON TWO DIGIT NUMBER

Key Concept :

Let us consider a  two digit number as xy , where x is the digit at tens place and y is the digit at ones place .

Then, the two digit number can be written in expanded form as shown below.

xy = 10x + 1y

Problem 1 :

A two digit number is four times the sum of its digits and twice the product of the digits. Find the number.

Let xy be the required two digit number.

Given : The  two digit number is four times the sum of its digits.

xy = 4(x + y)

10x + y = 4x + 4y

10x - 4x + y - 4y = 0

6x - 3y = 0

y = 2x ----(1)

Given : The two digit number is four times the sum of its digits.

xy = 2  ⋅ x ⋅ y

10x + 1y = 2xy  ----(2)

Substitute y = 2x.

10x + 1(2x) = 2x(2x)

10x + 2x = 4x 2

x =  ¹²⁄₄

Substitute x = 3 into (1).

Therefore, the two digit number is 36.

Problem 2 :

A two digit number such that the product of its digits is 21. When 36 is subtracted from the number the digits are interchanged. Find the number.

Let xy be the two digit number.

Given : The two digit number such that the product of its digits is 21.

x  ⋅  y = 21  ----(1)

Given : When 36 is subtracted from the number the digits are interchanged.

xy - 36 = yx

10x + y - 36 = 10y + x

10x - x + y - 10 y = 36

 9x - 9y = 36

Divide both sides by 9.

x = y + 4 ----(2)

Substitute x = y + 4 in to (1).

(y + 4)  ⋅  y = 21

y 2  + 4y  = 21

y 2  + 4y - 21 = 0

y 2  - 3y + 7y - 21 = 0

y(y - 3) + 7(y - 3) = 0

(y - 3)(y + 7) = 0

y - 3 = 0  or  y + 7 = 0

y = 3  or  y = 7

y represents the ones place of the two digit number and it can not be negative.

Substitute y = 3 into (2).

Therefore, the two digit number is 73.

Problem 3 :

A two digit number is such that the product of its digits is 12. When 36 is added to this number the digits are interchanged. Find the numbers.

Let xy be the required two digit number

A two digit number such that the product of its digits is 12.

x   ⋅  y = 12  -----(1)

When 36 is added to the number the digits are interchanged

xy + 36 = yx

10x + y + 36 = 10y + x

9x - 9y = -36

x = y - 4 ----(2)

Substitute x = y - 4 into (1).

(y - 4)  ⋅  y = 12

y 2 - 4y = 12

y 2 - 4y - 12 = 0

Solve by factoring.

y 2 - 6y + 2y - 12 = 0

y(y - 6) + 2(y - 6) = 0

(y - 6)(y + 2) = 0

y - 6 = 0  or  y + 2 = 0

y = 6  or  y = -2

Substitute y = 6 into (2).

Therefore, the two digit number is 26.

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Gizmodo Monday Puzzle: The Only Number That Describes Itself

Only one 10-digit number meets the criteria to solve this brainteaser..

Here’s a classic brainteaser that I don’t like. What’s the next number in this sequence: 1, 11, 21, 1211, 111221, …? The answer is 312211, because each number describes the digits in the number that precedes it. We open with 1, an arbitrary choice, but the next number describes 1 as “a single one,” i.e. “one one,” i.e. 11. The next entry describes 11 as “two ones,” or 21. This, in turn, is “one two followed by one one,” or 1211, and so on.

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Legendary mathematician John Conway studied this so-called “look-and-say” sequence and actually proved some interesting results about it. It clearly goes on forever, and the numbers grow to infinity, but surprisingly no digits other than 1, 2, and 3 ever appear. If you keep describing bigger and bigger numbers in this way, you’ll never generate a string of four ones (or twos or threes) in a row. Conway also studied the sequences that spring from different starting numbers other than 1. He proved that no matter what whole number you open with, the resulting sequence will diverge to infinity… except for one. Determining which one is your bonus puzzle this week.

I like the idea of numbers describing other numbers, but I’d prefer it not cast as a puzzle to solve. My gripe with sequence puzzles is that they’re open to multiple possible solutions. You could surely cook up some strange mathematical operation that produces the same first five numbers as the look-and-say sequence but then deviates from there. Your main puzzle this week concerns a number that describes itself . And rest assured it has only one solution.

Did you miss last week’s puzzle? Check it out here , and find its solution at the bottom of today’s article. Be careful not to read too far ahead if you haven’t solved last week’s yet!

Puzzle #39: A Self-Referential Number

Only one 10-digit number has the following property. Its left-most digit is the number of 0s in the number, the next digit is the number of 1s in the number, the next is the number of 2s, and so on until the right-most digit, which is the number of 9s in the number. Find the number. Numbers can’t begin with a zero.

An example of a four-digit number with this property is 2020. The first digit indicates that the number contains two 0s, the next indicates zero 1s, the next indicates two 2s, and the final indicates zero 3s.

Bonus: you can seed the look-and-say sequence with any whole number. For example, if you started with 39, then the next entry would be 1319 (one three, one nine). Conway proved that all seeds yield a sequence whose entries grow to infinity, with only one exception. Find the exception.

I’ll be back next Monday with the solutions and a new puzzle. Do you know a cool puzzle that you think should be featured here? Message me on X @JackPMurtagh or email me at [email protected]

Solution to Puzzle #38: Tax Evasion

Shout-out to 8x10 for a swift answer to last week’s tax evasion puzzle. I hope the IRS doesn’t monitor these...

You can win a maximum of $50 in The Taxman Game. See the turns below:

• You take $11 and the Tax Collector takes $1 (1 is the only available factor of 11)
• You take $10 and the Tax Collector takes $2 and $5
• You take $9 and the Tax Collector takes $3
• You take $8 and the Tax Collector takes $4 ($2 was already taken on move 2)
• You take $12 and the Tax Collector takes $6
• You’re out of legal moves so the Tax Collector takes the final check of $7

Your winnings total $8 + $9 + $10 + $11 + $12 = $50.

To save you time trying to hoard even more of Uncle Sam’s due, here’s a little argument that proves the above strategy is optimal. You can only ever take at most one prime numbered paycheck throughout the whole game. Because once you do, the Tax Collector takes the $1 paycheck and all other primes become off limits (the Tax Collector wouldn’t get paid). To avoid the $1 paycheck being wasted on another turn, you must start with a prime number and you might as well make it as big as possible, hence opening with $11.

Now the very end of the game will involve the Tax Collector taking the $7 paycheck no matter what happens, because you can never take it for yourself and no multiples of 7 are available to make the Tax Collector take it earlier. So effectively three paychecks are out of play ($11, $1, and $7), leaving nine remaining. You cannot get more than four of these nine because the Tax Collector must get paid on every turn. The strategy we gave gets you $12, $10, $9, and $8, the four largest remaining paychecks. So our approach cannot be improved.

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