A certain whole number less than 100 leaves the remainders \( 1,2,3 \), and 4 when divided respectively by \( 2,3,4 \), and 5 . What is the whole number? The whole number is

To find the whole number less than 100 that leaves remainders 1 when divided by 2, 2 when divided by 3, 3 when divided by 4, and 4 when divided by 5, we can set up a system of congruences: 1. \( x \equiv 1 \mod 2 \) 2. \( x \equiv 2 \mod 3 \) 3. \( x \equiv 3 \mod 4 \) 4. \( x \equiv 4 \mod 5 \) We can express each congruence in terms of the modulus and rearrange to find a pattern: - The first congruence indicates that \( x \) is odd. - The second states \( x \) is one more than a multiple of 3: \( x = 3k + 2 \). - The third states \( x \) is one more than a multiple of 4: \( x = 4m + 3 \). - The fourth states \( x \) is one more than a multiple of 5: \( x = 5n + 4 \). We notice that the last congruence implies \( x + 1\) is a multiple of 5, \( x + 2 \) is a multiple of 3, \( x + 3 \) is a multiple of 4, and \( x\) is therefore structured like \( x + 1\): 1. \( x \equiv -1 \mod 2 \) (equivalent to \( x \equiv 1 \mod 2 \)) 2. \( x \equiv -1 \mod 3 \) (equivalent to \( x \equiv 2 \mod 3 \)) 3. \( x \equiv -1 \mod 4 \) (equivalent to \( x \equiv 3 \mod 4 \)) 4. \( x \equiv -1 \mod 5 \) (equivalent to \( x \equiv 4 \mod 5 \)) Thus, \( x + 1 \) must be a common multiple (the least common multiple of the moduli 2, 3, 4, and 5). Calculating the LCM: - The prime factors of 2, 3, 4 (which is \( 2^2 \)), and 5 are: - \( 2^2, 3^1, 5^1 \) - The LCM is \( 2^2 \times 3^1 \times 5^1 = 4 \times 3 \times 5 = 60 \). Now \( x + 1\) can be expressed as \( 60k \) for \( k \in \mathbb{Z} \). Thus, \( x = 60k - 1 \). To keep \( x \) less than 100, let’s check values of \( k \): - For \( k = 1: x = 60 \cdot 1 - 1 = 59 \) (Valid) - For \( k = 2: x = 60 \cdot 2 - 1 = 119 \) (Not valid) Therefore, the only solution under 100 is: The whole number is 59.