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Find the smallest positive integer $N$ such that
N &\equiv 4 \pmod{6}, \\
N &\equiv 4 \pmod{10}, \\
N &\equiv 4 \pmod{15}, \\
N &\equiv 4 \pmod{24}.

 Jul 7, 2024
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We are starting with this:

 

\(N \equiv 4 \pmod{6}, \\ N \equiv 4 \pmod{10}, \\ N \equiv 4 \pmod{15}, \\ N \equiv 4 \pmod{24}.\)

 

We can rewrite this as:

 

\(N = 6a +4\\N=10b+4\\N=15c+4\\N=24d+4\)

 

\(N -4 = 6a\\N-4=10b\\N-4=15c\\N-4=24d\)

 

Therefore, N - 4 is a factor of 6, 10, 15, and 24. To find the minimum value of N-4, we find the least common multiple of 6, 10, 15, 24. We can do this by listing out the factors:

 

\(6 = 2^13^1\)

\(10=2^15^1\)

\(15=3^15^1\)

\(24=2^33^1\)

 

\(N-4= \text{lcm}(6, 10, 15, 24) = 2^33^15^1=120\)

 

Therefore, the least possible value of N is 124.

 Jul 7, 2024

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