Question

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

Answer 1

We can create four placeholders to represent the four digits of a 4-digit number.
We now know that the place digits for thousands, hundreds, and tens add up to 14, while the place digits for hundreds, tens, and units add up to 15.

Thus, the unit's digit is one unit higher than the thousandth place, as we can conclude.
\(\frac{a}{}-\frac{}{}\frac{a+5}{}\frac{a+1}{}\)

In order to maximize the 4-digit number, a should have the largest possible value. Additionally, the tens digit can have a maximum value of 9. Thus, four is the maximum value of a.
\(\frac{4}{}-\frac{}{}\frac{4+5}{}\frac{4+1}{}\)

\(\frac{4}{}\frac{1}{}\frac{9}{}\frac{5}{}\)

To ensure that the sum of the thousands, hundreds, and tens place digits is satisfied, the value of the hundreds place should indeed be equal to 1. Therefore, the largest possible number meeting this criterion is 4195.