Finding Unknown Digits in a Number Divisible by 72

The problem asks us to find the value of an expression involving the digits P and Q from a 9-digit number 83P93678Q, given that this number is divisible by 72.

A number is divisible by 72 if and only if it is divisible by both 8 and 9, since 72 = 8 × 9, and 8 and 9 are coprime numbers.

Step 1: Apply Divisibility Rule for 8

The divisibility rule for 8 states that a number is divisible by 8 if the number formed by its last three digits is divisible by 8. In the given number 83P93678Q, the last three digits form the number 78Q.

So, 78Q must be divisible by 8. We can write 78Q as \(780 + Q\).

Let's test values for Q from 0 to 9 to see which one makes \(780 + Q\) divisible by 8:

• If Q=0, 780 is divisible by 4 but not 8 (780 = 8 × 97 + 4).
• If Q=1, 781 leaves remainder when divided by 8.
• If Q=2, 782 leaves remainder when divided by 8.
• If Q=3, 783 leaves remainder when divided by 8.
• If Q=4, \(780 + 4 = 784\). \(784 \div 8 = 98\). So, 784 is divisible by 8.
• If Q=5, 785 leaves remainder when divided by 8.
• If Q=6, 786 leaves remainder when divided by 8.
• If Q=7, 787 leaves remainder when divided by 8.
• If Q=8, 788 leaves remainder when divided by 8.
• If Q=9, 789 leaves remainder when divided by 8.

Alternatively, we know that \(780 = 8 \times 97 + 4\). So, \(780 + Q = (8 \times 97 + 4) + Q\). For this to be divisible by 8, \(4 + Q\) must be divisible by 8. The possible values for Q (0-9) that make \(4+Q\) divisible by 8 are when \(4+Q\) is 8 (which gives Q=4) or 16 (which would give Q=12, not a single digit) etc. The only single digit value for Q is 4.

Thus, from the divisibility rule for 8, we find that Q must be 4.

Step 2: Apply Divisibility Rule for 9

The divisibility rule for 9 states that a number is divisible by 9 if the sum of its digits is divisible by 9.

The sum of the digits of the number 83P93678Q is:

\(8 + 3 + P + 9 + 3 + 6 + 7 + 8 + Q\)

Substitute the value Q=4 that we found:

\(8 + 3 + P + 9 + 3 + 6 + 7 + 8 + 4\)

Adding the known digits: \(8+3+9+3+6+7+8+4 = 48\).

So, the sum of digits is \(48 + P\).

For the number to be divisible by 9, the sum of digits \(48 + P\) must be divisible by 9. We need to find a single digit value for P (0-9) that satisfies this condition.

Let's check values for P from 0 to 9:

• If P=0, \(48 + 0 = 48\) (not divisible by 9).
• If P=1, \(48 + 1 = 49\) (not divisible by 9).
• If P=2, \(48 + 2 = 50\) (not divisible by 9).
• If P=3, \(48 + 3 = 51\) (not divisible by 9).
• If P=4, \(48 + 4 = 52\) (not divisible by 9).
• If P=5, \(48 + 5 = 53\) (not divisible by 9).
• If P=6, \(48 + 6 = 54\). \(54 \div 9 = 6\). So, 54 is divisible by 9.
• If P=7, \(48 + 7 = 55\) (not divisible by 9).
• If P=8, \(48 + 8 = 56\) (not divisible by 9).
• If P=9, \(48 + 9 = 57\) (not divisible by 9).

The only single digit value for P that makes \(48 + P\) divisible by 9 is P=6.

Thus, from the divisibility rule for 9, we find that P must be 6.

Step 3: Calculate the Expression Value

We have found that P=6 and Q=4. The question asks for the value of \(\sqrt {P^2+Q^2+12}\).

Substitute the values of P and Q into the expression:

\(\sqrt {6^2 + 4^2 + 12}\)

Calculate the squares:

\(\sqrt {36 + 16 + 12}\)

Add the numbers inside the square root:

\(\sqrt {52 + 12}\)

\(\sqrt {64}\)

Calculate the square root:

\(\sqrt {64} = 8\)

Therefore, the value of \(\sqrt {P^2+Q^2+12}\) is 8.

The final answer is 8.

Revision Table: Key Divisibility Rules

Additional Information: Number Properties and Calculations

This problem combines number theory concepts (divisibility rules) with basic algebraic calculation (evaluating an expression with square roots).

• Understanding prime factorization is key to composite number divisibility rules. Since 72 = \(2^3 \times 3^2\), divisibility by 72 implies divisibility by \(2^3=8\) and \(3^2=9\). Because 8 and 9 are coprime (their greatest common divisor is 1), we can apply their divisibility rules independently.
• The sum of digits rule for divisibility by 9 (or 3) is based on modular arithmetic. Any number N can be written as \(N = a_n 10^n + ... + a_1 10^1 + a_0 10^0\). Since \(10 \equiv 1 \pmod 9\), \(10^k \equiv 1^k \equiv 1 \pmod 9\) for any k >= 0. Thus, \(N \equiv a_n (1) + ... + a_1 (1) + a_0 (1) \pmod 9\), which means \(N \equiv (a_n + ... + a_1 + a_0) \pmod 9\). Therefore, a number is divisible by 9 if and only if the sum of its digits is divisible by 9.
• The divisibility rule for 8 is related to \(1000\). Since \(1000 = 8 \times 125\), any number can be written as \(1000k + last\_three\_digits\). For the number to be divisible by 8, the last three digits must be divisible by 8.
• Calculating square roots involves finding a number that, when multiplied by itself, equals the number under the radical sign. \(\sqrt{64} = 8\) because \(8 \times 8 = 64\).