Math Problem Statement

Context: When 419 is added to the three-digit number 3p4, we get 7q3. If 7q3 is divisible by 3, what is the greatest possible value of p? p and q are digits.

Answer question

Solution

To solve this problem, let's break it down step by step using the given information and the rules for divisibility by 3.

Given Information:

  1. We have a three-digit number 3p43p4, where pp is a digit.
  2. When 419 is added to 3p43p4, the result is another three-digit number 7q37q3, where qq is a digit.
  3. The number 7q37q3 is divisible by 3.

Step-by-Step Solution:

  1. Set up the equation:

    The problem states that when 419 is added to 3p43p4, we get 7q37q3. Therefore, we can write the equation as: 3p4+419=7q33p4 + 419 = 7q3

  2. Convert the numbers into expressions:

    Represent the numbers in a form suitable for calculations:

    • 3p43p4 can be expressed as 300+10p+4300 + 10p + 4.
    • 7q37q3 can be expressed as 700+10q+3700 + 10q + 3.

    The equation becomes: (300+10p+4)+419=700+10q+3(300 + 10p + 4) + 419 = 700 + 10q + 3

  3. Simplify the equation:

    Simplify both sides: 723+10p=703+10q723 + 10p = 703 + 10q

    Subtract 703 from both sides: 20+10p=10q20 + 10p = 10q

    Divide the entire equation by 10: 2+p=q2 + p = q

  4. Condition for divisibility by 3:

    For 7q37q3 to be divisible by 3, the sum of its digits must be divisible by 3. Calculate the sum of the digits in 7q37q3: 7+q+3=10+q7 + q + 3 = 10 + q

    For divisibility by 3: 10+q0(mod3)10 + q \equiv 0 \pmod{3}

    Since q=p+2q = p + 2, substitute qq in the divisibility condition: 10+(p+2)=12+p10 + (p + 2) = 12 + p

    For 12+p12 + p to be divisible by 3: 12+p0(mod3)12 + p \equiv 0 \pmod{3}

  5. Find the possible values of pp:

    The number 12 is already divisible by 3. Therefore, pp must also be divisible by 3 for 12+p12 + p to be divisible by 3. The possible values for pp (as pp is a digit) are 0, 3, 6, and 9.

  6. Determine the greatest possible value of pp:

    The greatest possible value of pp that is divisible by 3 is 9.

  7. Verify the solution with p=9p = 9:

    • If p=9p = 9, then q=p+2=9+2=11q = p + 2 = 9 + 2 = 11. But qq must be a digit, so qq cannot be 11.
    • If p=6p = 6, then q=6+2=8q = 6 + 2 = 8.

    With p=6p = 6 and q=8q = 8, calculate 7q37q3 as: 7q3=7837q3 = 783

    Check divisibility by 3: 7+8+3=187 + 8 + 3 = 18

    18 is divisible by 3, so 7q3=7837q3 = 783 is divisible by 3.

Conclusion:

The greatest possible value of pp is 6.


Would you like more details or have any questions? Here are some related questions to further your understanding:

  1. What would be the value of qq if p=4p = 4?
  2. How does changing the digits of the number 3p43p4 affect the divisibility of 7q37q3?
  3. What is the smallest possible value of pp that makes 7q37q3 divisible by 3?
  4. If 419419 was subtracted from 7q37q3 instead of added to 3p43p4, how would the solution change?
  5. How many different valid pairs of (p,q)(p, q) exist such that 7q37q3 is divisible by 3?

Tip: When checking for divisibility by a number, understanding the properties of that number can simplify the calculations.

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Math Problem Analysis

Mathematical Concepts

Equations
Divisibility
Number Theory

Formulas

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Theorems

Divisibility rule for 3

Suitable Grade Level

Grades 7-9