Find the least number that leaves remainder 2 when divided by 3, 4 and 5.
\(62\)
Step-by-step explanation (beginner friendly):
“Remainder 2 when divided by 3, 4, and 5” means:
\(n \equiv 2 \pmod{3}\)
\(n \equiv 2 \pmod{4}\)
\(n \equiv 2 \pmod{5}\)
This tells us \(n - 2\) is divisible by 3, 4, and 5.
So \(n - 2\) is a multiple of their LCM.
Find \(\operatorname{LCM}(3,4,5)\):
\(3 = 3\)
\(4 = 2 \times 2\)
\(5 = 5\)
Take highest powers: \(2^2,\; 3,\; 5\)
\(\operatorname{LCM} = 2^2 \times 3 \times 5 = 4 \times 3 \times 5 = 60\)
Least positive multiple is \(60\). So:
\(n - 2 = 60\)
\(n = 60 + 2 = 62\)
Quick check:
\(62 \div 3 = 20\) remainder \(2\)
\(62 \div 4 = 15\) remainder \(2\)
\(62 \div 5 = 12\) remainder \(2\)
Answer: \(62\)