Solving Divisibility Problems: The 9m2365n48 Number

The question asks us to find the value of $(m^2 \times n^2)$ for a nine-digit number 9m2365n48 that is completely divisible by 88. We are given that m and n are natural numbers and we need to find the value for the smallest possible value of n.

A number is divisible by 88 if it is divisible by both 8 and 11, because $88 = 8 \times 11$ and 8 and 11 are co-prime numbers.

Understanding Divisibility Rules

• Divisibility by 8: A number is divisible by 8 if its last three digits are divisible by 8.
• Divisibility by 11: A number is divisible by 11 if the difference between the sum of the digits at odd places (from the right) and the sum of the digits at even places (from the right) is either 0 or a multiple of 11.

Step 1: Apply Divisibility Rule of 8

The given number is 9m2365n48. The last three digits are n48.

For the number to be divisible by 8, n48 must be divisible by 8.

We are given that n is a natural number, which means n can be 1, 2, 3, 4, 5, 6, 7, 8, 9 (since it is a single digit in the nine-digit number).

Let's check possible values of n starting from 1:

• If n = 1, the number is 148. $148 \div 8 = 18$ with a remainder. Not divisible by 8.
• If n = 2, the number is 248. $248 \div 8 = 31$. Divisible by 8.
• If n = 3, the number is 348. $348 \div 8 = 43$ with a remainder. Not divisible by 8.
• If n = 4, the number is 448. $448 \div 8 = 56$. Divisible by 8.
• If n = 5, the number is 548. $548 \div 8 = 68$ with a remainder. Not divisible by 8.
• If n = 6, the number is 648. $648 \div 8 = 81$. Divisible by 8.
• If n = 7, the number is 748. $748 \div 8 = 93$ with a remainder. Not divisible by 8.
• If n = 8, the number is 848. $848 \div 8 = 106$. Divisible by 8.
• If n = 9, the number is 948. $948 \div 8 = 118$ with a remainder. Not divisible by 8.

The natural number values of n for which n48 is divisible by 8 are 2, 4, 6, and 8.

The question asks for the smallest value of n. The smallest such value is 2.

So, we take n = 2.

Step 2: Apply Divisibility Rule of 11

The number is 9m2365n48. With n=2, the number is 9m2365248.

Let's find the alternating sum of the digits starting from the rightmost digit:

$\text{Sum} = 8 - 4 + 2 - 5 + 6 - 3 + 2 - m + 9$

Calculate the sum:

$\text{Sum} = (8 - 4) + (2 - 5) + (6 - 3) + (2 - m) + 9$

$\text{Sum} = 4 + (-3) + 3 + (2 - m) + 9$

$\text{Sum} = 4 - 3 + 3 + 2 - m + 9$

$\text{Sum} = 1 + 3 + 2 - m + 9$

$\text{Sum} = 4 + 2 - m + 9$

$\text{Sum} = 6 - m + 9$

$\text{Sum} = 15 - m$

For the number to be divisible by 11, the alternating sum $(15 - m)$ must be 0 or a multiple of 11.

We are given that m is a natural number and it is a single digit in the nine-digit number. So, m can be 1, 2, 3, 4, 5, 6, 7, 8, or 9.

Let's check the possible values for $15 - m$ based on the possible values of m:

• If m = 1, $15 - 1 = 14$. Not a multiple of 11.
• If m = 2, $15 - 2 = 13$. Not a multiple of 11.
• If m = 3, $15 - 3 = 12$. Not a multiple of 11.
• If m = 4, $15 - 4 = 11$. This is a multiple of 11. So m=4 is a possible value.
• If m = 5, $15 - 5 = 10$. Not a multiple of 11.
• If m = 6, $15 - 6 = 9$. Not a multiple of 11.
• If m = 7, $15 - 7 = 8$. Not a multiple of 11.
• If m = 8, $15 - 8 = 7$. Not a multiple of 11.
• If m = 9, $15 - 9 = 6$. Not a multiple of 11.

The only value of m that satisfies the condition (m is a natural single digit and $15-m$ is a multiple of 11) is 4.

So, we have m = 4 and n = 2 (the smallest natural number value for n).

Step 3: Calculate the required value

We need to find the value of $(m^2 \times n^2)$.

Substitute the values of m and n we found:

$m^2 = 4^2 = 16$

$n^2 = 2^2 = 4$

$(m^2 \times n^2) = 16 \times 4 = 64$

The value of $(m^2 \times n^2)$ for the smallest natural number value of n is 64.

Revision Table: Divisibility by 88

Additional Information on Divisibility Rules

Understanding divisibility rules helps solve number theory problems efficiently. Here's a quick recap of the rules used:

• Divisibility by 8: Check if the number formed by the last three digits is divisible by 8. This works because $1000 = 8 \times 125$. Any number can be written as $1000a + b$, where b is the number formed by the last three digits. If 1000a is always divisible by 8, the divisibility depends only on b.
• Divisibility by 11: The rule is based on the property that powers of 10 alternate in their remainder when divided by 11 ($10 \equiv -1 \pmod{11}$, $100 \equiv 1 \pmod{11}$, $1000 \equiv -1 \pmod{11}$, and so on). The alternating sum reflects this property.

To check divisibility by a composite number like 88, find its prime factorization ($88 = 2^3 \times 11$). Since 8 and 11 are relatively prime, a number is divisible by 88 if and only if it is divisible by both 8 and 11.