+1186 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}.
+135 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.
