I am a number less than 40,000. All my digits are multiples of 3. My first,third,and fifth digits are the same. both my thousands digits and my tens digit are six more than the others.
Only 3, 6 and 9 are multiples of 3. The only way any one can be 6 more than the others is 9-3.
39 393 LOl use ur brain kids
To find the number that matches the given criteria, we will break down the problem step by step.
1. The first, third, and fifth digits are the same: Since all the digits are multiples of 3, the only multiples of 3 less than 4 are 0 and 3. Since the thousands and tens digits are six more than the others, we can deduce that the thousands and tens digits must be 6 and the remaining digits are 0 and 3. Note that the order of the digits is not specified yet.
2. The thousands and tens digit are six more than the others: The only pairs of digits that satisfy this condition are (0, 6) or (3, 9).
Now let's consider the possible scenarios:
Scenario 1: Thousands digit is 0 and tens digit is 6
In this case, we have three digits left to fill the remaining three positions. Since the first, third, and fifth digits are the same, the only possibility is 03663.
Scenario 2: Thousands digit is 6 and tens digit is 0
In this case, we again have three digits left to fill the remaining three positions. Since the first, third, and fifth digits are the same, the only possibility is 66336.
Therefore, the number that satisfies all the given conditions is either 03663 or 66336.