how many positive integres less than 100 have the property that the square of the sum of its digits is greater than the number itself?


We are tasked with finding how many positive integers less than 100 have the property that the square of the sum of their digits is greater than the number itself.

### Let’s break this problem down:
- The integers we are considering are all numbers from 1 to 99.
- For each number \( n \), if \( n = 10a + b \), where \( a \) is the tens digit and \( b \) is the ones digit, the condition we are checking is whether:
\[
(s(a, b))^2 > n,
\]
where \( s(a, b) = a + b \) is the sum of the digits of \( n \).

### Step-by-step approach:
1. **Compute the sum of digits** for each \( n \) from 1 to 99.
2. **Square the sum of digits** and check whether it exceeds \( n \).
3. Count how many such \( n \) satisfy this condition.

Let’s calculate this systematically.

We’ll create a list of the integers less than 100 that satisfy this condition.

1. \( n = 1 \): \( s(0, 1) = 1 \), \( 1^2 = 1 \), \( 1^2 \not> 1 \) (False)
2. \( n = 2 \): \( s(0, 2) = 2 \), \( 2^2 = 4 \), \( 4 > 2 \) (True)
3. \( n = 3 \): \( s(0, 3) = 3 \), \( 3^2 = 9 \), \( 9 > 3 \) (True)
4. \( n = 4 \): \( s(0, 4) = 4 \), \( 4^2 = 16 \), \( 16 > 4 \) (True)
5. \( n = 5 \): \( s(0, 5) = 5 \), \( 5^2 = 25 \), \( 25 > 5 \) (True)
6. \( n = 6 \): \( s(0, 6) = 6 \), \( 6^2 = 36 \), \( 36 > 6 \) (True)
7. \( n = 7 \): \( s(0, 7) = 7 \), \( 7^2 = 49 \), \( 49 > 7 \) (True)
8. \( n = 8 \): \( s(0, 8) = 8 \), \( 8^2 = 64 \), \( 64 > 8 \) (True)
9. \( n = 9 \): \( s(0, 9) = 9 \), \( 9^2 = 81 \), \( 81 > 9 \) (True)
10. \( n = 10 \): \( s(1, 0) = 1 \), \( 1^2 = 1 \), \( 1^2 \not> 10 \) (False)
11. \( n = 11 \): \( s(1, 1) = 2 \), \( 2^2 = 4 \), \( 4 \not> 11 \) (False)
12. \( n = 12 \): \( s(1, 2) = 3 \), \( 3^2 = 9 \), \( 9 \not> 12 \) (False)
13. \( n = 13 \): \( s(1, 3) = 4 \), \( 4^2 = 16 \), \( 16 > 13 \) (True)
...

### Conclusion
By applying this logic for all numbers up to 99, we’ll compute the total number of integers where the square of the sum of the digits is greater than the integer itself.

Allow me to compute the complete list and provide the count.

There are 76 positive integers less than 100 for which the square of the sum of their digits is greater than the number itself.

Would you like more details or have any questions about the method?

Here are five related questions you could explore:
1. How does the result change for numbers less than 1000?
2. What if we look for numbers where the cube of the sum of the digits exceeds the number?
3. Can we generalize this problem for any base system other than base 10?
4. How does the distribution of satisfying numbers look across different ranges of \( n \)?
5. Can this problem be approached using a different mathematical technique?

**Tip**: Digit-based properties often reveal interesting number patterns, so exploring them for larger ranges might uncover hidden relationships.

How many positive integers less than 100 have the property that the square of the sum of its digits is greater than the number itself?

Discover how many positive integers less than 100 have the square of the sum of their digits greater than the number itself. This problem involves number theory concepts and inequalities. Suitable for students in Grades 9-11.

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