Question

The difference between the digits of a twodigit number is 1. The number itself is 1 more than 5 times the sum of its digits. If the units digit is greater than the tens digit. Find the number




2. A two digit number is such that the sum of its digit is 11. The number is increased by 27 when the digits are interchanged. Find the number

Answer 1

Let's solve each problem step by step.

### Problem 1:

We have a two-digit number represented as:

- Let the tens digit be \(x\).
- Let the units digit be \(y\).

The conditions given are:

1. The difference between the digits is 1:
\[ y - x = 1 \quad \Rightarrow \quad y = x + 1 \]

2. The number itself is 1 more than 5 times the sum of its digits:
\[ 10x + y = 5(x + y) + 1 \]

Now, substituting \(y\) from the first equation into the second equation:

\[
10x + (x + 1) = 5(x + (x + 1)) + 1
\]

This simplifies to:

\[
10x + x + 1 = 5(2x + 1) + 1
\]
\[
11x + 1 = 10x + 5 + 1
\]
\[
11x + 1 = 10x + 6
\]
\[
11x - 10x = 6 - 1
\]
\[
x = 5
\]

Now, substituting \(x = 5\) back to find \(y\):

\[
y = x + 1 = 5 + 1 = 6
\]

Thus, the two-digit number is:

\[
10x + y = 10(5) + 6 = 50 + 6 = 56
\]

So, the number is **56**.

### Problem 2:

We have a two-digit number:

- Let the tens digit be \(a\).
- Let the units digit be \(b\).

The conditions given are:

1. The sum of its digits is 11:
\[ a + b = 11 \]

2. The number is increased by 27 when the digits are interchanged:
\[ 10b + a = 10a + b + 27 \]

Now, we'll simplify the second equation:

\[
10b + a - b - 10a = 27
\]
\[
9b - 9a = 27
\]
\[
b - a = 3 \quad \Rightarrow \quad b = a + 3
\]

Now, substitute \(b\) from this equation into the first equation:

\[
a + (a + 3) = 11
\]
\[
2a + 3 = 11
\]
\[
2a = 8
\]
\[
a = 4
\]

Now find \(b\):

\[
b = a + 3 = 4 + 3 = 7
\]

Thus, the two-digit number is:

\[
10a + b = 10(4) + 7 = 40 + 7 = 47
\]

So, the number is **47**.

### Summary:
1. The number from problem 1 is **56**.
2. The number from problem 2 is **47**.