Solving the Divisibility Problem with Prime Digits

We are given a seven-digit number, 52A6B7C, which is divisible by 33. The digits A, B, and C are prime numbers. Our goal is to find the maximum possible value of the expression 2A + 3B + C.

Understanding Divisibility by 33

A number is divisible by 33 if and only if it is divisible by both 3 and 11. We will use the divisibility rules for 3 and 11 to find the possible values of A, B, and C.

Divisibility Rule for 3

A number is divisible by 3 if the sum of its digits is divisible by 3.

The sum of the digits of 52A6B7C is \(5 + 2 + A + 6 + B + 7 + C = 20 + A + B + C\).

For the number to be divisible by 3, \(20 + A + B + C\) must be divisible by 3.

We can rewrite this as \(18 + 2 + A + B + C\). Since 18 is divisible by 3, this means \(2 + A + B + C\) must be divisible by 3. Alternatively, we can say \(A + B + C\) must leave the same remainder as \(20\) when divided by 3, which is 2. So, \(A + B + C \equiv 2 \pmod{3}\).

Divisibility Rule for 11

A number is divisible by 11 if the alternating sum of its digits is divisible by 11. Starting from the rightmost digit and moving left, we alternate adding and subtracting digits.

For 52A6B7C, the alternating sum is \(C - 7 + B - 6 + A - 2 + 5 = A + B + C - 10\).

For the number to be divisible by 11, \(A + B + C - 10\) must be divisible by 11.

This means \(A + B + C - 10 = 11k\) for some integer \(k\).

Identifying Possible Values for A, B, C

We know that A, B, and C are prime numbers. Since they are single digits within the number 52A6B7C, they must be single-digit prime numbers. The single-digit prime numbers are 2, 3, 5, and 7.

So, A, B, C must be chosen from the set {2, 3, 5, 7}.

Combining the Conditions

Let's look at the condition from the divisibility rule for 11: \(A + B + C - 10\) must be a multiple of 11.

Since A, B, and C are single-digit primes, their minimum sum is \(2 + 2 + 2 = 6\) and their maximum sum is \(7 + 7 + 7 = 21\).

So, \(A + B + C - 10\) must be between \(6 - 10 = -4\) and \(21 - 10 = 11\).

The possible multiples of 11 in the range [-4, 11] are 0 and 11.

• If \(A + B + C - 10 = 0\), then \(A + B + C = 10\).
• If \(A + B + C - 10 = 11\), then \(A + B + C = 21\).

Now let's check these possibilities against the divisibility rule for 3: \(20 + A + B + C\) must be divisible by 3.

• Case 1: \(A + B + C = 10\). Sum of digits = \(20 + 10 = 30\). Since 30 is divisible by 3, \(A + B + C = 10\) is a valid condition.
• Case 2: \(A + B + C = 21\). Sum of digits = \(20 + 21 = 41\). Since 41 is not divisible by 3, \(A + B + C = 21\) is not a valid condition.

Therefore, we must have \(A + B + C = 10\).

Now we need to find three prime numbers from the set {2, 3, 5, 7} that add up to 10.

Let's list combinations:

• If we use 7, the remaining two primes must sum to \(10 - 7 = 3\). The only prime pair summing to 3 is 2+1, but 1 is not prime. So 7 cannot be one of the primes.
• The primes must be from the set {2, 3, 5}. Let's check combinations of these three primes: \(2 + 3 + 5 = 10\). This works.

So, the digits A, B, and C must be a permutation of the prime numbers 2, 3, and 5.

Maximizing the Expression 2A + 3B + C

We need to find the maximum value of the expression \(2A + 3B + C\), where {A, B, C} is a permutation of {2, 3, 5}.

To maximize a sum of the form \(w_1 x_1 + w_2 x_2 + w_3 x_3\), where \(x_1, x_2, x_3\) are fixed numbers (the primes 2, 3, 5) and \(w_1, w_2, w_3\) are fixed coefficients (2, 3, 1 for A, B, C respectively), we should assign the largest available number to the largest coefficient, the second largest number to the second largest coefficient, and so on.

The coefficients for A, B, and C are 2, 3, and 1 (since C has a coefficient of 1).

The primes we can use for A, B, C are 2, 3, and 5.

To maximize \(2A + 3B + C\):

• The largest coefficient is 3, which is multiplied by B. So, B should be the largest prime, 5.
• The next largest coefficient is 2, which is multiplied by A. The remaining primes are 2 and 3. So, A should be the larger of these, which is 3.
• The smallest coefficient is 1, which is multiplied by C. The remaining prime is 2. So, C should be 2.

Proposed values: A=3, B=5, C=2.

Let's verify these values satisfy A+B+C = 10: \(3 + 5 + 2 = 10\). Yes.

Now, calculate the value of the expression with these values:

\(2A + 3B + C = 2(3) + 3(5) + 2\)

\(= 6 + 15 + 2\)

\(= 23\)

Let's quickly check another assignment to confirm this is the maximum, for example, assigning the largest prime 5 to A (coefficient 2) instead of B (coefficient 3):

• A=5 (coeff 2), B=3 (coeff 3), C=2 (coeff 1): \(2(5) + 3(3) + 2 = 10 + 9 + 2 = 21\). This is smaller than 23.
• A=2 (coeff 2), B=5 (coeff 3), C=3 (coeff 1): \(2(2) + 3(5) + 3 = 4 + 15 + 3 = 22\). This is smaller than 23.

The assignment A=3, B=5, C=2 indeed gives the maximum value.

The maximum value of 2A + 3B + C is 23.

Revision Table: Key Concepts Review

Additional Information: Divisibility Rules and Number Properties

Divisibility rules are useful shortcuts to determine if a number is divisible by another number without performing long division.

For example, the rule for 4: A number is divisible by 4 if the number formed by its last two digits is divisible by 4.

The rule for 9: A number is divisible by 9 if the sum of its digits is divisible by 9.

Prime numbers are fundamental building blocks in number theory. The first few prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, and so on. The number 1 is not considered prime.

Understanding how to combine different mathematical concepts, such as divisibility rules and properties of prime numbers, is crucial for solving complex quantitative aptitude problems.