For a 4-digit number, the sum of its digits in the thousands, hundreds and tens places is 14, the sum of its digits in the hundreds, tens, and units places is 15, and the tens place digit is 4 more than the units place digit. Then the highest possible 4-digit number satisfying the above conditions is

To solve this problem, we need to break it down into three parts based on the information given:

1. The sum of the digits in the thousands, hundreds, and tens places is 14.
2. The sum of the digits in the hundreds, tens, and units places is 15.
3. The tens place digit is 4 more than the units place digit.

Let's denote the thousands place digit as a, hundreds place digit as b, tens place digit as c, and units place digit as d.

From the first condition, we have the equation a + b + c = 14.

From the second condition, we have the equation b + c + d = 15.

Subtracting the first equation from the second, we get d - a = 1, which means the units digit (d) is 1 more than the thousands digit (a).

From the third condition, we have the equation c = d + 4.

Since we want to find the highest possible 4-digit number, we should maximize the value of a. Given that a and d are positive and d = a + 1, the maximum value for a is 4 (because if a is 5, then d would be 6, which is not possible as the digits of a number can only be from 0 to 9). So, a = 4 and d = 5.

Substituting d = 5 into the equation c = d + 4, we get c = 9.

Finally, substituting a = 4, c = 9 into the equation a + b + c = 14, we get b = 1.

So, the highest possible 4-digit number satisfying the above conditions is 4195.