We are tasked with finding the number that neither divides by 3 nor by 4 among three 2-digit numbers formed from the digits 1 to 6, with certain conditions. The sum of the three numbers is 93, exactly one number divides by 3, exactly one other divides by 4, and none of them divides by 5.

Step 1: Divisibility rules

• Divisibility by 3: A number is divisible by 3 if the sum of its digits is divisible by 3.
• Divisibility by 4: A number is divisible by 4 if the number formed by its last two digits is divisible by 4.
• Divisibility by 5: A number is divisible by 5 if it ends in 0 or 5. Since none of the numbers divides by 5, none of the numbers will end in 0 or 5.

Step 2: Possible 2-digit numbers

The digits we are working with are 1, 2, 3, 4, 5, and 6. These digits must be arranged into three 2-digit numbers. The sum of these digits is: $1 + 2 + 3 + 4 + 5 + 6 = 21$ This information will help us check our final answers.

Step 3: Dividing the numbers

Let the three 2-digit numbers be $N_1$, $N_2$, and $N_3$. According to the problem:

• One of these numbers divides by 3.
• Another divides by 4.
• The third does not divide by either 3 or 4.

Step 4: Strategy for forming numbers

We know that the total sum of $N_1 + N_2 + N_3 = 93$, and we will try to form numbers that satisfy the divisibility conditions.

Step 5: Trial and error approach

We will form various combinations of numbers using the digits 1 to 6, ensuring that the sum of the numbers equals 93. After checking different possibilities, one possible valid solution is:

• $N_1 = 36$: divisible by 4 (since 36 is divisible by 4).
• $N_2 = 21$: divisible by 3 (since the sum of the digits, 2 + 1 = 3, is divisible by 3).
• $N_3 = 36$.

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