Finding the Greatest Seven-Digit Number with Specific Remainders

The problem asks for the greatest seven-digit number that leaves specific remainders when divided by a set of numbers. The divisors are 10, 15, 20, 24, and 30, and the corresponding remainders are 6, 11, 16, 20, and 26.

Let the required number be $N$. We are given the following conditions based on the problem statement:

• $N \equiv 6 \pmod{10}$
• $N \equiv 11 \pmod{15}$
• $N \equiv 16 \pmod{20}$
• $N \equiv 20 \pmod{24}$
• $N \equiv 26 \pmod{30}$

Let's observe the difference between each divisor and its corresponding remainder:

• $10 - 6 = 4$
• $15 - 11 = 4$
• $20 - 16 = 4$
• $24 - 20 = 4$
• $30 - 26 = 4$

The difference is constant, which is 4. This implies that if we add 4 to the number $N$, the result $(N+4)$ will be perfectly divisible by each of the divisors: 10, 15, 20, 24, and 30.

So, $(N+4)$ must be a common multiple of 10, 15, 20, 24, and 30. To find the number $N$, we first need to find the least common multiple (LCM) of these divisors.

Calculating the LCM of the Divisors

We find the prime factorization of each divisor:

• $10 = 2 \times 5$
• $15 = 3 \times 5$
• $20 = 2^2 \times 5$
• $24 = 2^3 \times 3$
• $30 = 2 \times 3 \times 5$

The LCM is the product of the highest powers of all prime factors present in the factorizations. The prime factors are 2, 3, and 5.

• Highest power of 2: $2^3$ (from 24)
• Highest power of 3: $3^1$ (from 15, 24, 30)
• Highest power of 5: $5^1$ (from 10, 15, 20, 30)

LCM$(10, 15, 20, 24, 30) = 2^3 \times 3^1 \times 5^1 = 8 \times 3 \times 5 = 120$.

Thus, $(N+4)$ must be a multiple of 120. We can write this as:

$N + 4 = 120k$

where $k$ is a positive integer. Rearranging for $N$, we get:

$N = 120k - 4$

Finding the Greatest Seven-Digit Number

The greatest seven-digit number is 9,999,999. We need to find the largest value of $k$ such that $N$ is a seven-digit number, and specifically, the greatest possible seven-digit number of the form $120k - 4$.

We want to find the largest $k$ such that $120k - 4 \le 9999999$.

$120k \le 9999999 + 4$

$120k \le 10000003$

$k \le \frac{10000003}{120}$

Let's perform the division:

$k \le 83333.358\dots$

Since $k$ must be an integer, the largest possible integer value for $k$ is 83333.

Calculating the Required Number

Now we substitute the largest integer value of $k$ (which is 83333) back into the equation for $N$:

$N = 120k - 4$

$N = 120 \times 83333 - 4$

$N = 9999960 - 4$

$N = 9999956$

This number, 9999956, is a seven-digit number. Because we used the largest possible integer value for $k$ that keeps $N$ within the range $\le 9999999$, this is the greatest such number that satisfies the form $120k-4$.

Verification of Remainders

Let's quickly check if 9999956 leaves the correct remainders:

• $9999956 \div 10$: $9999956 = 10 \times 999995 + 6$. Remainder is 6. (Correct)
• $9999956 \div 15$: $9999956 = 15 \times 666663 + 11$. Remainder is 11. (Correct)
• $9999956 \div 20$: $9999956 = 20 \times 499997 + 16$. Remainder is 16. (Correct)
• $9999956 \div 24$: $9999956 = 24 \times 416664 + 20$. Remainder is 20. (Correct)
• $9999956 \div 30$: $9999956 = 30 \times 333331 + 26$. Remainder is 26. (Correct)

All remainder conditions are met. Thus, 9999956 is the greatest seven-digit number with these properties.

Revision Table: Key Concepts for Remainder Problems

Additional Information: Number Theory and Remainders

Problems involving finding a number that satisfies multiple remainder conditions are classic number theory problems. The pattern observed in this problem, where the difference between the divisor and the remainder is constant, is a common variation.

If the remainders were arbitrary (i.e., the difference $d-r$ was not constant), the problem would typically be solved using the Chinese Remainder Theorem (CRT). The CRT provides a method to find a solution (if one exists) to a system of linear congruences.

In this specific case, because the difference $(d-r)$ is constant (4), the problem simplifies significantly. Instead of dealing with a system of congruences directly, we transform it into a single congruence related to the LCM:

$N \equiv 6 \pmod{10} \implies N+4 \equiv 10 \pmod{10} \implies N+4 \equiv 0 \pmod{10}$

$N \equiv 11 \pmod{15} \implies N+4 \equiv 15 \pmod{15} \implies N+4 \equiv 0 \pmod{15}$

and so on for all divisors. This shows that $(N+4)$ is a multiple of every divisor, and hence a multiple of their LCM.

Finding the greatest seven-digit number requires identifying the largest multiple of the LCM (minus the constant difference) that fits within the seven-digit range. This is done by finding the largest multiple of the LCM just below or equal to the upper limit (9,999,999), or slightly above it depending on the constant difference, and then adjusting by subtracting the difference.