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 determine the smallest possible number of cards Zizi had, let's denote the number of cards as NN.

We know the following:

  1. NN is divisible by 4.
  2. NN is divisible by 6.
  3. NN leaves a remainder of 1 when divided by 7.

From the first two conditions, we can conclude that NN must be a common multiple of 4 and 6. The least common multiple (LCM) of 4 and 6 is:

LCM(4,6)=12\text{LCM}(4, 6) = 12

Thus, NN must be a multiple of 12. We can express NN as:

N=12kN = 12k

for some integer kk.

Now, considering the third condition:

N1(mod7)N \equiv 1 \pmod{7}

Substituting N=12kN = 12k into the congruence gives:

12k1(mod7)12k \equiv 1 \pmod{7}

We need to find the smallest positive integer kk such that this congruence holds. To solve this, we can simplify 12mod712 \mod 7:

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

Thus, the congruence becomes:

5k1(mod7)5k \equiv 1 \pmod{7}

We need to find the multiplicative inverse of 5 modulo 7, which is an integer kk such that:

5k1(mod7)5k \equiv 1 \pmod{7}

By testing values, we find that:

5×3=151(mod7)5 \times 3 = 15 \equiv 1 \pmod{7}

Thus, k=3k = 3 is the multiplicative inverse. Therefore, the smallest kk is 3. Substituting k=3k = 3 back into N=12kN = 12k:

N=12×3=36N = 12 \times 3 = 36

To verify:

  • 36 is divisible by 4: 36÷4=936 \div 4 = 9
  • 36 is divisible by 6: 36÷6=636 \div 6 = 6
  • 36 leaves a remainder of 1 when divided by 7: 36÷7=536 \div 7 = 5 remainder 1

Therefore, the smallest possible number of cards Zizi had is:

36\boxed{36}

Would you like any more details or have any questions?

Here are some questions for you to consider:

  1. What is the least common multiple (LCM) of two numbers?
  2. How do you find the multiplicative inverse in modular arithmetic?
  3. What is the process to verify if a number satisfies given conditions in a problem?
  4. How can congruences be used to solve real-world problems?
  5. What are some other methods to solve modular arithmetic problems?

Tip: When dealing with divisibility and remainders, modular arithmetic is a powerful tool that simplifies complex problems.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Modular Arithmetic
Divisibility

Formulas

LCM (Least Common Multiple)
Multiplicative Inverse in Modular Arithmetic

Theorems

-

Suitable Grade Level

Grades 9-12