Problem 12 - Entrance Test

The number 123 in base b can be written in base 10 as: 1 * b^2 + 2 * b^1 + 3 * b^0 = b^2 + 2b + 3. We are given that this is equal to 38 in base 10: b^2 + 2b + 3 = 38. Rearrange the equation into a standard quadratic form: b^2 + 2b + 3 - 38 = 0 b^2 + 2b - 35 = 0. We can solve this quadratic equation by factoring or using the quadratic formula. Factoring: We look for two numbers that multiply to -35 and add to 2. These numbers are 7 and -5. So, (b + 7)(b - 5) = 0. This gives two possible solutions for b: b = -7 or b = 5. Since a base 'b' must be a positive integer, b = -7 is not a valid base. Also, the digits used in base b must be less than b. In 123_b, the largest digit is 3, so b must be greater than 3. Therefore, the only valid solution is b = 5. Let's verify: In base 5, 123_5 = 1*5^2 + 2*5^1 + 3*5^0 = 1*25 + 2*5 + 3*1 = 25 + 10 + 3 = 38.