A number lies between 300 and 400 . If the number is added to the number formed by reversing the digits, the sum is 888 and if the unit’s digit and the ten’s digit change places, the new number exceeds the original number by 9 . Find the number.

Given the constraint that the sum of the number and its reversed form is 888 and focusing on the clue that the units digit in the sum scenario must add up to 8, we look directly at the options provided:

– (a) 339: The sum of 3 and 9 does not lead to an end digit of 8 in the sum.
– (b) 341: The sum of 1 and 4 does not lead to an end digit of 8 in the sum.
– (c) 378: The sum of 8 and 7 does not directly address the end digit sum condition.
– (d) 345: When 345 is added to its reverse 543, the unit digits 5 and 3 indeed add up to 8, meeting the immediate condition.

The specific insight about the unit’s digits adding up to 8 being satisfied only by option (d) “345” (since \(3+5=8\)) simplifies the approach significantly. However, to validate this option fully in the context of the entire problem:

– If the original number is 345 and its reverse is 543, their sum is indeed 888 (\(345 + 543 = 888\)), which satisfies one of the problem’s conditions.
– The second condition mentioned is that swapping the unit’s and ten’s digit of the original number results in a number that is 9 more than the original. Swapping the units and tens digit of 345 gives 354, which is indeed 9 more than 345 (\(354 – 345 = 9\)).

Therefore, considering both conditions and the insight provided about the sum leading to the last digits adding up to 8, the correct number is indeed option (d) **345**. This choice fulfills both specified conditions of the problem: the sum with its reversed form equals 888, and swapping the tens and units digits results in a number that is 9 more than the original.