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

Correct Answer: 4195

For a 4-digit number we can make 4 placeholders to represent the four digits
Now, we know that the sum of the thousands, hundreds and the tens place digit is 14, and the sum of the hundreds, tens and the units place digit is 15.

So we can conclude that the unit's digit is one unit greater than the thousandth place.

\(\frac{a}{}\frac{}{}\frac{a+5}{}\frac{a+1}{}\)

To maximize the 4-digit number the value of a should be as large as possible. And the value of the tens digit can be 9 at most. So the maximum value of a is 4.

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

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

To satisfy the sum of the thousands, hundreds and the tens place digit, the value of the hundreds place should be equal to 1.

Hence the largest possible number is 4195