The sum of a 2-digit number and the number obtained by reversing the order of the digits is 165. If the digits differ by 3, find the number.

To solve this problem, let’s denote the two-digit number as 10a + b, where a is the tens digit and b is the units digit. Based on the information provided, we can establish the following equations:

  • The sum of the number and its reverse is 165:

(10a + b) + (10b + a) = 165

This simplifies to:

11a + 11b = 165

Dividing the entire equation by 11 gives:

a + b = 15

  • Additionally, we also know that the digits differ by 3:

|a – b| = 3

Now, we have a system of two equations:

  1. a + b = 15
  2. |a – b| = 3

From the second equation, we can derive two cases:

Case 1: a – b = 3

In this case, we can express a as:

a = b + 3

Substituting this into the first equation:

(b + 3) + b = 15

Which simplifies to:

2b + 3 = 15

Thus:

2b = 12

b = 6

Substituting back to find a:

a = 6 + 3 = 9

So in this case, we have a = 9 and b = 6, which gives us the number 96.

Case 2: b – a = 3

In this case, we express b as:

b = a + 3

Substituting into the first equation:

a + (a + 3) = 15

This simplifies to:

2a + 3 = 15

So:

2a = 12

a = 6

Then substituting back to find b:

b = 6 + 3 = 9

Thus, we end up with the digits as a = 6 and b = 9, leading to the number 69.

Therefore, the two-digit number that satisfies the conditions of the problem is either 96 or 69. Since both numbers meet the original requirements, you can confirm that:

96 + 69 = 165 and the digits differ by 3.

More Related Questions