Finding the Smallest Number with a Remainder

To find the smallest number that leaves a specific remainder when divided by several numbers, we first find the Least Common Multiple (LCM) of the divisors and then add the remainder.

• Step 1: Calculate the LCM of the divisors (8, 9, 12).

  Find the prime factorization of each divisor:

  • $8 = 2^3$
  • $9 = 3^2$
  • $12 = 2^2 \times 3$

  The LCM is the product of the highest powers of all prime factors involved:

  LCM$(8, 9, 12) = 2^3 \times 3^2 = 8 \times 9 = 72$.

• Step 2: Add the common remainder.

  The question states that the remainder is 5 in each case. Add this remainder to the LCM:

  Smallest Number = LCM + Remainder

  Smallest Number = $72 + 5 = 77$.

• Step 3: Verify the result.

  Check if 77 leaves a remainder of 5 when divided by 8, 9, and 12:

  • $77 \div 8 = 9$ remainder $5$ ($77 = 8 \times 9 + 5$)
  • $77 \div 9 = 8$ remainder $5$ ($77 = 9 \times 8 + 5$)
  • $77 \div 12 = 6$ remainder $5$ ($77 = 12 \times 6 + 5$)

  Since 77 is the smallest number obtained by adding the remainder to the LCM, it is the smallest number satisfying the condition.

The smallest number is 77.