Answer :
To determine the least number that, when divided by 3, 4, and 5, leaves a remainder of 2 in each case, follow these steps:
1. Understanding the Problem:
- The number we seek, say [tex]\( N \)[/tex], should satisfy the following conditions:
[tex]\[ N \equiv 2 \pmod{3} \][/tex]
[tex]\[ N \equiv 2 \pmod{4} \][/tex]
[tex]\[ N \equiv 2 \pmod{5} \][/tex]
2. Express with Modulo Equations:
- These congruences imply that [tex]\( N - 2 \)[/tex] is divisible by 3, 4, and 5. Essentially, we can write:
[tex]\[ N - 2 \equiv 0 \pmod{3} \][/tex]
[tex]\[ N - 2 \equiv 0 \pmod{4} \][/tex]
[tex]\[ N - 2 \equiv 0 \pmod{5} \][/tex]
3. Least Common Multiple (LCM):
- To find [tex]\( N - 2 \)[/tex], we need a number that is a common multiple of 3, 4, and 5. The least common multiple of these numbers can be computed.
- The prime factorization of 3 is [tex]\( 3 \)[/tex].
- The prime factorization of 4 is [tex]\( 2^2 \)[/tex].
- The prime factorization of 5 is [tex]\( 5 \)[/tex].
- Therefore, the LCM can be calculated by taking the highest power of each prime involved:
[tex]\[ \text{LCM}(3, 4, 5) = 3^1 \times 2^2 \times 5^1 = 3 \times 4 \times 5 = 60 \][/tex]
4. Finding the Number:
- Hence, [tex]\( N - 2 \)[/tex] must be a multiple of 60. The smallest multiple of 60 is [tex]\( 60 \times 1 = 60 \)[/tex].
5. Calculate the Desired Number:
- Adding back the remainder 2 to this [tex]\( N - 2 \)[/tex]:
[tex]\[ N = 60 + 2 = 62 \][/tex]
6. Verification:
- Let's verify if [tex]\( 62 \)[/tex] satisfies the original conditions:
[tex]\[ 62 \div 3 = 20 \text{ remainder } 2 \][/tex]
[tex]\[ 62 \div 4 = 15 \text{ remainder } 2 \][/tex]
[tex]\[ 62 \div 5 = 12 \text{ remainder } 2 \][/tex]
Thus, the least number which, when divided by 3, 4, and 5, leaves a remainder of 2 in each case, is [tex]\( \boxed{62} \)[/tex].
1. Understanding the Problem:
- The number we seek, say [tex]\( N \)[/tex], should satisfy the following conditions:
[tex]\[ N \equiv 2 \pmod{3} \][/tex]
[tex]\[ N \equiv 2 \pmod{4} \][/tex]
[tex]\[ N \equiv 2 \pmod{5} \][/tex]
2. Express with Modulo Equations:
- These congruences imply that [tex]\( N - 2 \)[/tex] is divisible by 3, 4, and 5. Essentially, we can write:
[tex]\[ N - 2 \equiv 0 \pmod{3} \][/tex]
[tex]\[ N - 2 \equiv 0 \pmod{4} \][/tex]
[tex]\[ N - 2 \equiv 0 \pmod{5} \][/tex]
3. Least Common Multiple (LCM):
- To find [tex]\( N - 2 \)[/tex], we need a number that is a common multiple of 3, 4, and 5. The least common multiple of these numbers can be computed.
- The prime factorization of 3 is [tex]\( 3 \)[/tex].
- The prime factorization of 4 is [tex]\( 2^2 \)[/tex].
- The prime factorization of 5 is [tex]\( 5 \)[/tex].
- Therefore, the LCM can be calculated by taking the highest power of each prime involved:
[tex]\[ \text{LCM}(3, 4, 5) = 3^1 \times 2^2 \times 5^1 = 3 \times 4 \times 5 = 60 \][/tex]
4. Finding the Number:
- Hence, [tex]\( N - 2 \)[/tex] must be a multiple of 60. The smallest multiple of 60 is [tex]\( 60 \times 1 = 60 \)[/tex].
5. Calculate the Desired Number:
- Adding back the remainder 2 to this [tex]\( N - 2 \)[/tex]:
[tex]\[ N = 60 + 2 = 62 \][/tex]
6. Verification:
- Let's verify if [tex]\( 62 \)[/tex] satisfies the original conditions:
[tex]\[ 62 \div 3 = 20 \text{ remainder } 2 \][/tex]
[tex]\[ 62 \div 4 = 15 \text{ remainder } 2 \][/tex]
[tex]\[ 62 \div 5 = 12 \text{ remainder } 2 \][/tex]
Thus, the least number which, when divided by 3, 4, and 5, leaves a remainder of 2 in each case, is [tex]\( \boxed{62} \)[/tex].