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The digits of positive integer, having three digit , are in AP and their sum is 15. The number obtained by ,reversing the digit is 594 less than the original number. find the number.
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The digits of positive integer, having three digit , are in AP and the...
Let a - d, a, a + d be the three digits of a three digit number respectively at unit place, ten's place, hundreth place.
∴ The number = 100 (a + d) + 10 + a + a - d
Since the sum of digits is 15
∴ a -d + a + a + d = 15 or 3a = 15 or a = 5
The number obtained by reversing the digits = 100 (a - d) + 10a + (a + d) ...(ii)
According to the given condition
100 (a - d) + 10 a + a + d = 100 (a + d) + 10a + a - d - 594
or 100a - 100d + 10a + a + d = 100a + 100d + 10a + a - d - 594
or 198d = 594 or d = 3
∴ Original number = 100 (5 + 3) + 10 (5) + 5 - 3 = 800 + 50 + 2 = 852
Verification: On reversing the digits, the number is 258
852 - 258 = 594
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The digits of positive integer, having three digit , are in AP and the...
Given information:
- The digits of a positive integer, with three digits, are in an arithmetic progression (AP).
- The sum of the digits is 15.
- The number obtained by reversing the digits is 594 less than the original number.
Let's solve the problem step by step:
Step 1: Understanding the arithmetic progression
- Let's assume the three digits of the number to be a-d, a, and a+d.
- The sum of these digits would be (a-d) + a + (a+d) = 3a, which is given to be 15.
- So, 3a = 15, or a = 15/3 = 5.
Step 2: Forming the number with the digits
- The number with digits a-d, a, and a+d can be represented as (a-d)*100 + a*10 + (a+d).
- Substituting the value of a as 5, we get the number as (5-d)*100 + 5*10 + (5+d).
Step 3: Reversing the digits and forming the second number
- Reversing the digits gives us a new number with digits (a+d), a, and (a-d).
- This number can be represented as (a+d)*100 + a*10 + (a-d).
- Substituting the value of a as 5, we get the reversed number as (5+d)*100 + 5*10 + (5-d).
Step 4: Finding the relationship between the numbers
- The problem states that the reversed number is 594 less than the original number.
- Mathematically, we can write this as:
(5+d)*100 + 5*10 + (5-d) = (5-d)*100 + 5*10 + (5+d) - 594.
Step 5: Solving for the value of d
- Simplifying the equation from step 4, we get:
600 + 50d - 50 = 500 - 50d + 594.
- Combining like terms, we have:
100d - 50d = 500 + 594 - 600 + 50.
50d = 544.
d = 544/50 = 10.88.
- Since d needs to be an integer, we can conclude that there is no solution to the problem.
Conclusion:
- There is no three-digit number that satisfies the given conditions.
- The digits of a positive integer, with three digits, are in an arithmetic progression (AP).
- The sum of the digits is 15.
- The number obtained by reversing the digits is 594 less than the original number.
Let's solve the problem step by step:
Step 1: Understanding the arithmetic progression
- Let's assume the three digits of the number to be a-d, a, and a+d.
- The sum of these digits would be (a-d) + a + (a+d) = 3a, which is given to be 15.
- So, 3a = 15, or a = 15/3 = 5.
Step 2: Forming the number with the digits
- The number with digits a-d, a, and a+d can be represented as (a-d)*100 + a*10 + (a+d).
- Substituting the value of a as 5, we get the number as (5-d)*100 + 5*10 + (5+d).
Step 3: Reversing the digits and forming the second number
- Reversing the digits gives us a new number with digits (a+d), a, and (a-d).
- This number can be represented as (a+d)*100 + a*10 + (a-d).
- Substituting the value of a as 5, we get the reversed number as (5+d)*100 + 5*10 + (5-d).
Step 4: Finding the relationship between the numbers
- The problem states that the reversed number is 594 less than the original number.
- Mathematically, we can write this as:
(5+d)*100 + 5*10 + (5-d) = (5-d)*100 + 5*10 + (5+d) - 594.
Step 5: Solving for the value of d
- Simplifying the equation from step 4, we get:
600 + 50d - 50 = 500 - 50d + 594.
- Combining like terms, we have:
100d - 50d = 500 + 594 - 600 + 50.
50d = 544.
d = 544/50 = 10.88.
- Since d needs to be an integer, we can conclude that there is no solution to the problem.
Conclusion:
- There is no three-digit number that satisfies the given conditions.
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The digits of positive integer, having three digit , are in AP and their sum is 15. The number obtained by ,reversing the digit is 594 less than the original number. find the number. Related: Exercise 2 - Chapter 5 - Arithmetic Progressions, Class 10, Mathematics?
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The digits of positive integer, having three digit , are in AP and their sum is 15. The number obtained by ,reversing the digit is 594 less than the original number. find the number. Related: Exercise 2 - Chapter 5 - Arithmetic Progressions, Class 10, Mathematics? for Class 10 2024 is part of Class 10 preparation. The Question and answers have been prepared according to the Class 10 exam syllabus. Information about The digits of positive integer, having three digit , are in AP and their sum is 15. The number obtained by ,reversing the digit is 594 less than the original number. find the number. Related: Exercise 2 - Chapter 5 - Arithmetic Progressions, Class 10, Mathematics? covers all topics & solutions for Class 10 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The digits of positive integer, having three digit , are in AP and their sum is 15. The number obtained by ,reversing the digit is 594 less than the original number. find the number. Related: Exercise 2 - Chapter 5 - Arithmetic Progressions, Class 10, Mathematics?.
The digits of positive integer, having three digit , are in AP and their sum is 15. The number obtained by ,reversing the digit is 594 less than the original number. find the number. Related: Exercise 2 - Chapter 5 - Arithmetic Progressions, Class 10, Mathematics? for Class 10 2024 is part of Class 10 preparation. The Question and answers have been prepared according to the Class 10 exam syllabus. Information about The digits of positive integer, having three digit , are in AP and their sum is 15. The number obtained by ,reversing the digit is 594 less than the original number. find the number. Related: Exercise 2 - Chapter 5 - Arithmetic Progressions, Class 10, Mathematics? covers all topics & solutions for Class 10 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The digits of positive integer, having three digit , are in AP and their sum is 15. The number obtained by ,reversing the digit is 594 less than the original number. find the number. Related: Exercise 2 - Chapter 5 - Arithmetic Progressions, Class 10, Mathematics?.
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The digits of positive integer, having three digit , are in AP and their sum is 15. The number obtained by ,reversing the digit is 594 less than the original number. find the number. Related: Exercise 2 - Chapter 5 - Arithmetic Progressions, Class 10, Mathematics?, a detailed solution for The digits of positive integer, having three digit , are in AP and their sum is 15. The number obtained by ,reversing the digit is 594 less than the original number. find the number. Related: Exercise 2 - Chapter 5 - Arithmetic Progressions, Class 10, Mathematics? has been provided alongside types of The digits of positive integer, having three digit , are in AP and their sum is 15. The number obtained by ,reversing the digit is 594 less than the original number. find the number. Related: Exercise 2 - Chapter 5 - Arithmetic Progressions, Class 10, Mathematics? theory, EduRev gives you an
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