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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
 #1
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We are starting with this:

 

N4(mod6),N4(mod10),N4(mod15),N4(mod24).

 

We can rewrite this as:

 

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

 

N4=6aN4=10bN4=15cN4=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=2131

10=2151

15=3151

24=2331

 

N4=lcm(6,10,15,24)=233151=120

 

Therefore, the least possible value of N is 124.

 Jul 7, 2024

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