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    Error in solution in the explanation problem

    Day 8: Mental Math Warm-Up
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      sensibleocelot M0★ M1★ M2★ M3★ M4 M5
      last edited by

      I was doing the math problem in the Week 2 Day 8 explanation (Workout 1A: Algebra Basics) when I saw an error in the problem and the solution.

      This is the problem: "The number 369,246,369 equals the product xy, where x is a 6-digit number and y is a 3-digit number. What is the sum of the digits of 4x?"

      This is the solution for the problem: "The number 369,246,369 is composed of 123 × 3 shifted to the far left, 123 × 2 shifted to the middle, and 123 × 3 on the right side. Thus 369246369 = 3,002,003 × 123. How do we know that there isn't a different way to factor 369,246,369 as the product of a 6-digit and 3-digit number? Well, we can multiply 123 with any digit up to 7 and still get a 3-digit number. All we need to do is check whether 3,002,003 has any factors that are less than or equal to 7. A quick check shows that 3,002,003 isn't divisible by 2, 3, 4, 5, 6, or 7. Thus 369,246,369 = 3,002,003 × 123 is the only viable factorization, and the 6-digit number is 3,002,003. Multiplying this by 4 gives 12,008,012, which has a digit sum of 1 + 2 + 8 + 1 + 2 = 14. The answer is choice v.)"

      The error in the solution is that 3,002,003 is not a 6-digit number. It's a 7-digit number! In both the problem and the solution, it states that 369,246,369 could be factored into a product of a 6-digit number and a 3-digit number, but that isn't possible since 3,002,003 not only can't be divided by 2, 3, 4, 5, 6, or 7, but 3002003 is a 7-digit prime (the prime factorization of 369,246,369 is 3 * 41 * 3002003).

      quacker88Q 1 Reply Last reply Reply Quote 0
      • quacker88Q
        quacker88 MOD @sensibleocelot
        last edited by

        @sensibleocelot Great catch! Yeah, \(3,002,003\) is definitely not a \(6\)-digit number, I'll go ahead and report that 🙂

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