If a two digit number is divided by the sum of the digits the quotient is 9. But the number obtained by reducing the digits at the tens place by four times the digits at the units place leaves the remainder 5 when divided by the sum of the digits of the original number find the original number 

Let the two-digit number be represented as \(10x + y\), where:

- \(x\) is the tens digit
- \(y\) is the units digit.

From the problem, we have two conditions:

### Condition 1:
When the number is divided by the sum of its digits, the quotient is 9. This gives us the equation:
\[
\frac{10x + y}{x + y} = 9
\]
Multiplying both sides by \(x + y\), we get:
\[
10x + y = 9(x + y)
\]
Expanding the right-hand side:
\[
10x + y = 9x + 9y
\]
Simplifying:
\[
10x + y - 9x - 9y = 0
\]
\[
x - 8y = 0
\]
This implies:
\[
x = 8y
\]

### Condition 2:
If the digit at the tens place is reduced by four times the digit at the units place, the remainder when dividing by the sum of the digits is 5. So, the modified number is:
\[
10(x - 4y) + y = 10x - 40y + y = 10x - 39y
\]
The sum of the digits of the original number is \(x + y\), and dividing the modified number by \(x + y\) gives a remainder of 5. Therefore, we have:
\[
(10x - 39y) \mod (x + y) = 5
\]
Substituting \(x = 8y\) from Condition 1 into this equation:
\[
(10(8y) - 39y) \mod (8y + y) = 5
\]
\[
(80y - 39y) \mod 9y = 5
\]
\[
41y \mod 9y = 5
\]
Simplifying:
\[
41 \mod 9 = 5
\]
This is true because \(41 \div 9 = 4\) with a remainder of 5.

Thus, the units digit \(y = 1\).

### Finding \(x\):
Since \(x = 8y\), we have:
\[
x = 8 \times 1 = 8
\]

### Conclusion:
The original number is \(10x + y = 10(8) + 1 = 81\).

So, the original number is **81**.

Would you like to dive into more details, or have any questions? 

Here are 5 related questions you can try:

1. What if the quotient was 8 instead of 9, how would the solution change?
2. If the original number was a three-digit number, how would you approach the problem?
3. How would you solve if the sum of the digits and quotient relation changed?
4. What is the remainder if we modify the digit in the units place instead of the tens place?
5. Can you find a similar problem where the difference of digits is involved?

**Tip:** Always keep track of conditions carefully; reducing complex problems into simpler equations often leads to a straightforward solution!

If a two digit number is divided by the sum of the digits the quotient is 9. But the number obtained by reducing the digits at the tens place by four times the digits at the units place leaves the remainder 5 when divided by the sum of the digits of the original number find the original number

Learn how to solve a challenging algebra problem involving a two-digit number where the quotient is 9 when divided by the sum of its digits. The solution uses key concepts like modular arithmetic and algebraic manipulation. Suitable for students in Grades 8-10.

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