Find the least five digit number which leaves a remainder 9 in each case when divided by 12, 40, and 75.



In order to find the least number that satisfies the given condition, we need to apply the Chinese Remainder Theorem.



The Chinese Remainder Theorem states that if we have a system of congruences:

x ≡ a₁(mod m₁)

x ≡ a₂(mod m₂)

...

x ≡ aₙ(mod mₙ)

where m₁, m₂, ..., mₙ are pairwise co-prime, then the system has a unique solution modulo M, where M = m₁m₂...mₙ.



We need to find the least five-digit number that leaves a remainder of 9 when divided by 12, 40, and 75. Let x be the required number. Then we can write:

x ≡ 9(mod 12)

x ≡ 9(mod 40)

x ≡ 9(mod 75)

We can see that 12, 40, and 75 are pairwise co-prime, so we can apply the Chinese Remainder Theorem to find the solution.

M = 12 * 40 * 75 = 36000

Now, we need to find the values of a₁, a₂, and a₃ such that the above congruences hold.



We need to find the smallest multiple of 12 that leaves a remainder of 9. Let's start with 9 and keep adding 12 until we get a multiple of 12.

9, 21, 33, 45, 57, ...

57 is the smallest multiple of 12 that leaves a remainder of 9. So, a₁ = 57.



We need to find the smallest multiple of 40 that leaves a remainder of 9. Let's start with 9 and keep adding 40 until we get a multiple of 40.

9, 49, 89, 129, ...

129 is the smallest multiple of 40 that leaves a remainder of 9. So, a₂ = 129.



We need to find the smallest multiple of 75 that leaves a remainder of 9. Let's start with 9 and keep adding 75 until we get a multiple of 75.

9, 84, 159, ...

159 is the smallest multiple of 75 that leaves a remainder of 9. So, a₃ = 159.



Now, we can substitute the values of a₁, a₂, and a₃ in the Chinese Remainder Theorem and get the unique solution modulo M.

x ≡ 57 * 250 * 159 + 129 * 225 * 159 + 9 * 112 * 40(mod 36000)

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