What is the two-digit number such that the sum of the number and its reverse is 165, and the digits differ by 3?

To find the two-digit number based on the given conditions, let’s denote the two-digit number as 10a + b, where a is the tens digit and b is the units digit.

1. The first condition states that the sum of the number and the number obtained by reversing its digits equals 165. So we have:

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

This simplifies to:

11a + 11b = 165

Dividing the entire equation by 11 gives:

a + b = 15

2. The second condition tells us that the digits differ by 3:

|a – b| = 3

This gives us two possibilities:

  • a – b = 3
  • b – a = 3 (which implies a – b = -3)

We will solve each case:

Case 1: a – b = 3

From the equation a – b = 3, we can express a as:

a = b + 3

Substituting this into the first equation a + b = 15 gives:

(b + 3) + b = 15

Which simplifies to:

2b + 3 = 15

Solving for b:

2b = 12

b = 6

Then substituting back to find a:

a = 6 + 3 = 9

Thus, the two-digit number is:

10a + b = 10(9) + 6 = 96

Case 2: a – b = -3 (b – a = 3)

From this equation, we get:

b = a + 3

Using this in the first equation a + b = 15 gives:

a + (a + 3) = 15

This simplifies to:

2a + 3 = 15

Thus:

2a = 12

a = 6

Now substituting back to find b:

b = 6 + 3 = 9

Therefore, in this case, the two-digit number is:

10a + b = 10(6) + 9 = 69

Final Result:

So, the two two-digit numbers that satisfy the conditions are 96 and 69.

To conclude:

  • If the number is 96, the reverse is 69.
  • If the number is 69, the reverse is 96.

Both pairs satisfy the original criteria where their sum equals 165 and their digits differ by 3.

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