Math Problem Statement

Zizi had some cards. She could place them into sets of 4 cards or sets of 6 cards without leaving any remainder. When she placed them into sets of 7 cards instead, she needed one more card to make the last set of 7. What was the smallest possible number of cards she had?

Solution

To solve the problem, we need to find the smallest number of cards NN that meets the given conditions. Specifically:

  1. NN can be evenly divided by 4 and 6.
  2. When NN is divided by 7, the remainder is 1.

First, let's find the least common multiple (LCM) of 4 and 6, as NN must be a multiple of this LCM to satisfy the first condition.

The prime factorizations are:

  • 4=224 = 2^2
  • 6=2×36 = 2 \times 3

The LCM is obtained by taking the highest power of each prime factor: LCM(4,6)=22×3=12\text{LCM}(4, 6) = 2^2 \times 3 = 12

So, NN must be a multiple of 12: N=12kN = 12k for some integer kk.

Next, we use the second condition: N1(mod7)N \equiv 1 \pmod{7}

Substituting N=12kN = 12k into the congruence: 12k1(mod7)12k \equiv 1 \pmod{7}

Simplifying 12(mod7)12 \pmod{7}: 125(mod7)12 \equiv 5 \pmod{7}

So the equation becomes: 5k1(mod7)5k \equiv 1 \pmod{7}

We need to find the multiplicative inverse of 5 modulo 7, which is a number xx such that: 5x1(mod7)5x \equiv 1 \pmod{7}

We can test values for xx to find the inverse: 5×1=55(mod7)5 \times 1 = 5 \equiv 5 \pmod{7} 5×2=103(mod7)5 \times 2 = 10 \equiv 3 \pmod{7} 5×3=151(mod7)5 \times 3 = 15 \equiv 1 \pmod{7}

Therefore, the multiplicative inverse of 5 modulo 7 is 3. Hence: k3(mod7)k \equiv 3 \pmod{7} k=7m+3k = 7m + 3 for some integer mm.

Now, substituting k=7m+3k = 7m + 3 back into N=12kN = 12k: N=12(7m+3)N = 12(7m + 3) N=84m+36N = 84m + 36

The smallest NN occurs when m=0m = 0: N=84(0)+36N = 84(0) + 36 N=36N = 36

Thus, the smallest possible number of cards Zizi had is 36\boxed{36}.

Would you like details or have any questions?

Here are some related questions you might find interesting:

  1. How do you find the LCM of two numbers?
  2. What is the significance of congruences in number theory?
  3. Can you explain the Chinese Remainder Theorem?
  4. How do you determine the multiplicative inverse modulo nn?
  5. What are some practical applications of modular arithmetic?

Tip: When solving modular arithmetic problems, finding the least common multiple (LCM) and using properties of congruences are often very helpful strategies.

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Math Problem Analysis

Mathematical Concepts

Number Theory
Modular Arithmetic

Formulas

Least Common Multiple (LCM)
Modular Congruences

Theorems

Chinese Remainder Theorem
Multiplicative Inverse Modulo

Suitable Grade Level

Grades 9-12