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

We represent our 4-digit number as abcd, where:

$a$ = thousands place digit

$b$ = hundreds place digit

$c$ = tens place digit

$d$ = units place digit

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From the given conditions, we can write:

Equation (1): $a + b + c = 14$

Equation (2): $b + c + d = 15$

Equation (3): $c = d + 4$

The problem states "tens place digit is 4 more than units place digit," which means $c = d + 4$.

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We subtract equation (1) from equation (2):

$(b + c + d) - (a + b + c) = 15 - 14$

This gives us: $d - a = 1$

Therefore: $a = d - 1$

This is a key insight! The thousands digit is always 1 less than the units digit.

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To get the highest possible 4-digit number, we need to maximize the leftmost digit first (thousands place), then hundreds, then tens, and finally units.

Since $a = d - 1$, to maximize $a$, we need to maximize $d$.

But we also have $c = d + 4$, and since digits can only be 0-9, we need:

$c \leq 9$ (maximum possible digit)

Therefore: $d + 4 \leq 9$

This means: $d \leq 5$

So the maximum value of $d$ is 5.

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With $d = 5$:

$d = 5$ (units place)

$c = d + 4 = 5 + 4 = 9$ (tens place)

$a = d - 1 = 5 - 1 = 4$ (thousands place)

Now using equation (1): $a + b + c = 14$

$4 + b + 9 = 14$

$b + 13 = 14$

$b = 1$ (hundreds place)

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Therefore, the highest possible 4-digit number is 4195.