Which Number When Divided by 7 Gives a Remainder 0, and Divided by Any Other Number Between 1-9 Gives a Remainder 1?

Mathematically, we are looking for a number N that satisfies the following conditions:

Condition 1: Remainder of 0 When Divided by 7

N 7k for some integer k.

Condition 2: Remainder of 1 When Divided by 2, 3, 4, 5, 6, 7, 8, and 9

The number N should give a remainder of 1 when divided by 2, 3, 4, 5, 6, 8, and 9. This can be expressed as:

N equiv 1 pmod{2} N equiv 1 pmod{3} N equiv 1 pmod{4} N equiv 1 pmod{5} N equiv 1 pmod{6} N equiv 1 pmod{7} N equiv 1 pmod{8} N equiv 1 pmod{9}

Converting Conditions to a Simplified Form

By subtracting 1 from both sides, we get:

N - 1 equiv 0 pmod{2} N - 1 equiv 0 pmod{3} N - 1 equiv 0 pmod{4} N - 1 equiv 0 pmod{5} N - 1 equiv 0 pmod{6} N - 1 equiv 0 pmod{7} N - 1 equiv 0 pmod{8} N - 1 equiv 0 pmod{9}

This means that N - 1 must be a multiple of the least common multiple (LCM) of the numbers from 2 to 9.

Calculating the LCM

The prime factorization of the numbers from 2 to 9 is:

2 2^1 3 3^1 4 2^2 5 5^1 6 2^1 cdot 3^1 7 7^1 8 2^3 9 3^2

Therefore, the LCM is:

LCM(2, 3, 4, 5, 6, 7, 8, 9) 2^3 cdot 3^2 cdot 5^1 cdot 7^1 8 cdot 9 cdot 5 cdot 7 2520

Formulating the Number N

Thus, we can express N - 1 as a multiple of 2520:

N - 1 2520m for some integer m. Consequently, N 2520m 1.

Ensuring Divisibility by 7

Given that N must be divisible by 7, we rewrite the congruence:

2520m 1 equiv 0 pmod{7} Rightarrow 1 equiv 0 pmod{7}

This is a contradiction, indicating that m must be chosen to make N divisible by 7. Therefore, 2520m 1 must be divisible by 7.

Evaluating Small Values of m

Checking small values of m to find the smallest N:

For m 1: N 2520 cdot 1 1 2521, not divisible by 7. For m 2: N 2520 cdot 2 1 5041, not divisible by 7. For m 3: N 2520 cdot 3 1 7561, not divisible by 7. For m 4: N 2520 cdot 4 1 10081, divisible by 7.

The smallest N that satisfies both conditions is 10081.