Question

1. In a positive integer, each digit after the leftmost one is larger than the digit to its left. Its square, which is a six-digit number, also has the same property. Find the original positive integer.

Answer 1

Let the original positive integer be of the form ABCDE, where A is the leftmost digit and B, C, D, E are its subsequent digits, with B > A, C > B, D > C, and E > D.

The largest possible value for A is 9, as A cannot be less than any of its subsequent digits.

The smallest possible value for B is A+1, and the largest possible value for B is 9, to maintain the relationship B > A.

Similarly, the smallest possible value for C is B+1, and the largest possible value is 9.

For D and E, the smallest values are C+1 and D+1 respectively, and the largest values are 9.

Therefore, we have the range of values for ABCDE as:

* 91234 <= ABCDE <= 98765

Squaring the largest possible value of ABCDE yields 975308641, which has seven digits instead of the required six digits. So the largest possible value of ABCDE is 98765.

The smallest possible value of ABCDE is 12345, and its square is 152399025, which satisfies the conditions.

Therefore, the original positive integer is 12345.

Answer 2

To find the original positive integer, let's start by considering the possible values for the leftmost digit.

1. The leftmost digit cannot be 0 because it would make the integer not positive.

2. The leftmost digit also cannot be 1, as 1 squared is still 1, and this doesn't satisfy the condition that the square is a six-digit number.

So, we start with the leftmost digit being 2. Now, we need to find the other digits such that each digit to the right is larger than the one to its left, and the square of the entire number is a six-digit number.

Let's consider the options for the second digit:

• If the second digit is 0, then the square of the number will have at least three trailing zeros, making it more than a six-digit number.
• If the second digit is 1, then the square will start with 4 (2 squared), and we need four more digits that are larger than 1, but it's impossible to satisfy this condition while keeping the square a six-digit number.
• If the second digit is 2, then the square will start with 4 (2 squared), and we need four more digits that are larger than 2. The only option is to use 3, 4, 5, and 6 as the remaining digits in some order.

So, the original positive integer must be 23, and its square is 529, which is indeed a six-digit number.