Math Problem Statement

Déterminer un nombre de trois chiffres sachant que la somme de ces chiffres est égale à 17. Si on permute le chiffre des dizaines et celui des centaines, le nombre augmente de 360. Si on permute le chiffre des unités et celui des centaines, le nombre diminue de 198.

Solution

Let's analyze the problem step by step.

We are tasked with finding a three-digit number ABCABC such that:

  1. The sum of the digits is equal to 17.
  2. If we swap the tens and hundreds digits, the number increases by 360.
  3. If we swap the units and hundreds digits, the number decreases by 198.

Let's break down each condition:

Let:

  • AA be the hundreds digit.
  • BB be the tens digit.
  • CC be the units digit.

Thus, the original number can be expressed as: 100A+10B+C100A + 10B + C

Step 1: Translating Each Condition

Condition 1: Sum of Digits

The sum of the digits is given by: A+B+C=17A + B + C = 17

Condition 2: Swap Tens and Hundreds

Swapping the tens and hundreds digits results in the number 100B+10A+C100B + 10A + C. According to the problem, this swap increases the number by 360: 100B+10A+C=(100A+10B+C)+360100B + 10A + C = (100A + 10B + C) + 360 Simplifying, we get: 100B+10A+C100A10BC=360100B + 10A + C - 100A - 10B - C = 360 90B90A=36090B - 90A = 360 Dividing by 90: BA=4B - A = 4

Condition 3: Swap Units and Hundreds

Swapping the units and hundreds digits results in the number 100C+10B+A100C + 10B + A. According to the problem, this swap decreases the number by 198: 100C+10B+A=(100A+10B+C)198100C + 10B + A = (100A + 10B + C) - 198 Simplifying, we get: 100C+10B+A100A10BC=198100C + 10B + A - 100A - 10B - C = -198 99C99A=19899C - 99A = -198 Dividing by 99: CA=2C - A = -2

Step 2: Solving the System of Equations

We now have a system of three equations:

  1. A+B+C=17A + B + C = 17
  2. BA=4B - A = 4
  3. CA=2C - A = -2

Let's solve these equations systematically:

From equation (2), we find: B=A+4B = A + 4

From equation (3), we find: C=A2C = A - 2

Substitute B=A+4B = A + 4 and C=A2C = A - 2 into equation (1): A+(A+4)+(A2)=17A + (A + 4) + (A - 2) = 17 3A+2=173A + 2 = 17 3A=153A = 15 A=5A = 5

Using A=5A = 5 in the expressions for BB and CC: B=A+4=5+4=9B = A + 4 = 5 + 4 = 9 C=A2=52=3C = A - 2 = 5 - 2 = 3

Solution

The three-digit number is: 100A+10B+C=1005+109+3=500+90+3=593100A + 10B + C = 100 \cdot 5 + 10 \cdot 9 + 3 = 500 + 90 + 3 = 593

Verification

  1. Sum of digits: 5+9+3=175 + 9 + 3 = 17 (satisfied).
  2. Swapping tens and hundreds: 953593=360953 - 593 = 360 (satisfied).
  3. Swapping units and hundreds: 395593=198395 - 593 = -198 (satisfied).

Thus, the number is indeed 593.

Would you like further details on any step?


Related Questions:

  1. How would the solution change if the sum of the digits was different?
  2. What other methods could we use to solve this problem?
  3. Can we generalize this method for any three-digit number with similar conditions?
  4. What happens if we swap other combinations of digits in a three-digit number?
  5. How can systems of linear equations be used in real-world problems?

Tip:

When solving digit-based problems, always translate conditions into equations based on place values (hundreds, tens, and units) to simplify the solution.

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Math Problem Analysis

Mathematical Concepts

Algebra
Systems of Equations
Place Value

Formulas

A + B + C = 17
100B + 10A + C = (100A + 10B + C) + 360
100C + 10B + A = (100A + 10B + C) - 198

Theorems

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Suitable Grade Level

Grades 8-10