The least number which when divided by 5, 6, 7 and 9 leaves a remainder 2, but when divided by 4 leaves no remainder is

1. Break Down the Problem

We need to find the least number $x$ that meets the following conditions:

• When divided by 5, 6, 7, and 9, it leaves a remainder of 2.
• When divided by 4, it leaves no remainder.

2. Relevant Concepts

1. Remainder Condition: For a number $x$ to leave a remainder of 2 when divided by $n$, it can be expressed as: $x \equiv 2 \pmod{n}$ where $n$ can be 5, 6, 7, and 9.

2. Divisibility Condition: For a number $x$ to leave no remainder when divided by 4, we express it as: $x \equiv 0 \pmod{4}$

3. Analysis and Detail

1. Set up the congruences from the remainder condition:

   • $x \equiv 2 \pmod{5}$
   • $x \equiv 2 \pmod{6}$
   • $x \equiv 2 \pmod{7}$
   • $x \equiv 2 \pmod{9}$

   This implies that: $x - 2 \equiv 0 \pmod{5}, \; x - 2 \equiv 0 \pmod{6}, \; x - 2 \equiv 0 \pmod{7}, \; x - 2 \equiv 0 \pmod{9}$ Therefore, $x - 2$ must be a common multiple of 5, 6, 7, and 9.

2. Find the least common multiple (LCM) of these numbers:

   • The prime factorization:
     • $5 = 5^1$
     • $6 = 2^1 \times 3^1$
     • $7 = 7^1$
     • $9 = 3^2$

   The LCM is calculated by taking the highest power of each prime: $\text{LCM}(5, 6, 7, 9) = 2^1 \times 3^2 \times 5^1 \times 7^1 = 630$

   Thus: $x - 2 = 630k \quad \text{for some integer } k$

   Therefore: $x = 630k + 2$

3. Now apply the second condition: $630k + 2 \equiv 0 \pmod{4}$ Since $630 \equiv 2 \pmod{4}$: $2k + 2 \equiv 0 \pmod{4} \implies 2k \equiv -2 \pmod{4} \implies k \equiv 1 \pmod{2}$ Therefore $k$ must be odd.

4. Verify and Summarize

The smallest odd integer for $k$ is 1: $x = 630 \times 1 + 2 = 632$

• Check if 632 meets the conditions:
  • $632 \div 5 = 126$ remainder $2$
  • $632 \div 6 = 105$ remainder $2$
  • $632 \div 7 = 90$ remainder $2$
  • $632 \div 9 = 70$ remainder $2$
  • $632 \div 4 = 158$ remainder $0$

Final Answer

The least number is $\boxed{632}$.