Math Problem Statement

A two-digit number is such that the sum of its digits is = of the number. When the digits of the number are reversed and the number is subtracted from the original number, the result obtained is 45. By letting the tens digit of the original number be x and the ones digit of the original number be y, form two simultaneous equations in terms of x and y in its simplest form and solve them to find the original number.

Solution

Let the two-digit number be represented as 10x+y10x + y, where xx is the tens digit and yy is the ones digit. Based on the problem:

Step 1: Form the equations

  1. Sum of the digits equals the number: x+y=10x+yx + y = 10x + y Simplify: x=9x    8x=0x = 9x \implies 8x = 0

ThuslyLet's correctly frame the problem:

Original Problem Restated

Let the two-digit number be 10x+y10x + y, where:

  • xx is the tens digit, and
  • yy is the ones digit.

1. First Condition

The sum of its digits equals one-third of the number: x+y=13(10x+y)x + y = \frac{1}{3}(10x + y) Simplify this equation: 3(x+y)=10x+y3(x + y) = 10x + y 3x+3y=10x+y3x + 3y = 10x + y 7x - 2y = 0 \tag{1}

2. Second Condition

When the digits are reversed, the resulting number subtracted from the original number is 45: (10x+y)(10y+x)=45(10x + y) - (10y + x) = 45 Simplify: 10x+y10yx=4510x + y - 10y - x = 45 9x9y=459x - 9y = 45 x - y = 5 \tag{2}

Step 2: Solve the Equations

We now solve the simultaneous equations:

  1. 7x2y=07x - 2y = 0
  2. xy=5x - y = 5

From (2):

x = y + 5 \tag{3}

Substitute x=y+5x = y + 5 into (1): 7(y+5)2y=07(y + 5) - 2y = 0 7y+352y=07y + 35 - 2y = 0 5y+35=05y + 35 = 0 5y=355y = -35 [ y = -7

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Math Problem Analysis

Mathematical Concepts

Algebra
Simultaneous Equations
Linear Equations

Formulas

Sum of digits equals one-third of the number: x + y = (1/3)(10x + y)
Difference after reversing digits: (10x + y) - (10y + x) = 45

Theorems

Basic principles of linear equations and simultaneous equations

Suitable Grade Level

Grades 8-10