If the number 413283P759387 is divisible by 13, then what is the value of P?

Question

Prepp Team · Accepted Answer

Finding the Value of P for Divisibility by 13

We are given the number 413283P759387 and need to determine the value of the digit P such that this entire number is divisible by 13.

A number is divisible by 13 if the weighted sum of its digits is divisible by 13. The weights are determined by the powers of 10 modulo 13, starting from the rightmost digit with a weight of \(10^0 \equiv 1\). The sequence of weights for powers of 10 modulo 13 is \(10^0 \equiv 1\), \(10^1 \equiv 10\), \(10^2 \equiv 9\), \(10^3 \equiv 12 \equiv -1\), \(10^4 \equiv 3\), \(10^5 \equiv 4\), \(10^6 \equiv 1\), and this sequence (1, 10, 9, 12, 3, 4) repeats every 6 terms.

Let's list the digits of the number from right to left (\(d_0\) to \(d_{12}\)) and their corresponding weights modulo 13.

Digit Position (from right)	Digit	Power of 10	Weight (\(10^{\text{position}} \pmod{13}\))
0	7	\(10^0\)	1
1	8	\(10^1\)	10
2	3	\(10^2\)	9
3	9	\(10^3\)	12
4	5	\(10^4\)	3
5	7	\(10^5\)	4
6	P	\(10^6\)	1
7	3	\(10^7\)	10
8	8	\(10^8\)	9
9	2	\(10^9\)	12
10	3	\(10^{10}\)	3
11	1	\(10^{11}\)	4
12	4	\(10^{12}\)	1

The number is divisible by 13 if the sum of each digit multiplied by its corresponding weight modulo 13 is divisible by 13. Let's calculate this sum:

Sum \(\equiv \left(7 \times 1\right) + \left(8 \times 10\right) + \left(3 \times 9\right) + \left(9 \times 12\right) + \left(5 \times 3\right) + \left(7 \times 4\right) + \left(P \times 1\right) + \left(3 \times 10\right) + \left(8 \times 9\right) + \left(2 \times 12\right) + \left(3 \times 3\right) + \left(1 \times 4\right) + \left(4 \times 1\right) \pmod{13}\)

Sum \(\equiv 7 + 80 + 27 + 108 + 15 + 28 + P + 30 + 72 + 24 + 9 + 4 + 4 \pmod{13}\)

Now, let's find the remainder of each term (except P) when divided by 13:

\(7 \equiv 7 \pmod{13}\)
\(80 = 6 \times 13 + 2 \equiv 2 \pmod{13}\)
\(27 = 2 \times 13 + 1 \equiv 1 \pmod{13}\)
\(108 = 8 \times 13 + 4 \equiv 4 \pmod{13}\)
\(15 = 1 \times 13 + 2 \equiv 2 \pmod{13}\)
\(28 = 2 \times 13 + 2 \equiv 2 \pmod{13}\)
\(P \times 1 = P \pmod{13}\)
\(30 = 2 \times 13 + 4 \equiv 4 \pmod{13}\)
\(72 = 5 \times 13 + 7 \equiv 7 \pmod{13}\)
\(24 = 1 \times 13 + 11 \equiv 11 \pmod{13}\)
\(9 \equiv 9 \pmod{13}\)
\(4 \equiv 4 \pmod{13}\)
\(4 \equiv 4 \pmod{13}\)

The sum of these remainders (excluding P) is:

\(7 + 2 + 1 + 4 + 2 + 2 + 4 + 7 + 11 + 9 + 4 + 4 \pmod{13}\)

Sum of remainders \(= 57\).

Now, find the remainder of 57 when divided by 13:

\(57 = 4 \times 13 + 5\)

So, \(57 \equiv 5 \pmod{13}\).

The total weighted sum modulo 13 is the sum of the remainders of the known terms plus the term with P:

Total Sum \(\equiv 5 + P \pmod{13}\)

For the original number to be divisible by 13, the total weighted sum must be divisible by 13, which means the sum must be congruent to 0 modulo 13:

\(5 + P \equiv 0 \pmod{13}\)

To solve for P, we subtract 5 from both sides of the congruence:

\(P \equiv -5 \pmod{13}\)

Since \(-5 \equiv -5 + 13 \equiv 8 \pmod{13}\), we get:

\(P \equiv 8 \pmod{13}\)

As P is a single digit (0-9), the only value of P that satisfies this congruence is 8.

Therefore, the value of P is 8.

Revision Table: Divisibility Check Steps

Step	Description
1	Identify the digits and the position of the unknown digit P.
2	Determine the weights for each digit position modulo the divisor (13), starting from \(10^0\) from the right.
3	Calculate the weighted sum of all digits.
4	Find the remainder of the weighted sum when divided by the divisor. This remainder must be 0 for divisibility.
5	Set up a congruence equation involving P and the calculated sum.
6	Solve the congruence equation for P.
7	Ensure the value of P is a valid digit (0-9).

Additional Information on Modular Arithmetic

Modular arithmetic is a system of arithmetic for integers, where numbers "wrap around" after reaching a certain value—the modulus. It is fundamental in number theory and crucial for understanding divisibility rules. The congruence relation \(a \equiv b \pmod{m}\) means that \(a\) and \(b\) have the same remainder when divided by \(m\), or equivalently, \(a-b\) is divisible by \(m\). This principle allows us to simplify calculations involving large numbers when checking for divisibility.