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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
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Let, the $\text{4-digit}$ number be $abcd.$

  • $a+b+c = 14 \quad \longrightarrow (1)$
  • $b+c+d = 15 \quad \longrightarrow (2)$
  • $c = d+4 \quad \longrightarrow (3)$

Put the value of $c$ in the equation $(2).$

$\Rightarrow b+c+d = 15$

$\Rightarrow b+d+4+d = 15$

$\Rightarrow \underbrace{b}_{{\color{Red}{\text{Odd}}}} + \underbrace{2d}_{{\color{Magenta}{\text{Even}}}} = \underbrace{11}_{{\color{Red}{\text{Odd}}}} \quad \longrightarrow (4)$

Now, we can calculate the value of $a, b, c$ and $d.$

$\qquad \begin{array} {}  a & b & c & d \\ {\color{Blue}{4}} & {\color{Red}{1}} & {\color{Green}{9}} & {\color{Purple}{5}} \\ 3 & 3 & 8 & 4 \\ 2 & 5 & 7 & 3 \\ 1 & 7 & 6 & 2 \\ 0 & 9 & 5 & 1  \end{array}$

$\therefore$ The highest possible $\text{4-digit}$ number satisfying the above conditions is $4195.$

Correct Answer $: 4195$

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