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$A$ number is of two digits. The position of digits is interchanged and the new number is added to the original number. The resultant number will always be divisible by

$140 \%$ of $56 + 56 \%$ of $140 = ?$

$1601 \times 198 \div 49 - 1399 + 3878 = ?$

$\left[(7.99)^{2}-(13.001)^{2}+(4.01)^{3}\right]^{2}=?$

If $50$ is subtracted from two-third of a number,the result is equal to the sum of $40$ and one-fourth of that number. What is the number?

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