Math Problem Statement

How many integers between 1 and 100 leave a remainder of 3 when divided by 5 and a remainder of 5 when divided by 7?

Solution

We are tasked with finding how many integers between 1 and 100 leave:

  • a remainder of 3 when divided by 5, and
  • a remainder of 5 when divided by 7.

Let’s break this down step by step.

Step 1: Represent the conditions as congruences

The conditions can be written as two congruences:

  1. x3(mod5)x \equiv 3 \pmod{5} (i.e., xx leaves a remainder of 3 when divided by 5)
  2. x5(mod7)x \equiv 5 \pmod{7} (i.e., xx leaves a remainder of 5 when divided by 7)

Step 2: Solve the system of congruences using the Chinese Remainder Theorem

The system of congruences is: x3(mod5)x \equiv 3 \pmod{5} x5(mod7)x \equiv 5 \pmod{7}

We will solve this system step by step.

Step 3: Use the first congruence x3(mod5)x \equiv 3 \pmod{5}

Let x=5k+3x = 5k + 3 for some integer kk. We now substitute this into the second congruence.

Step 4: Substitute into the second congruence

Substitute x=5k+3x = 5k + 3 into x5(mod7)x \equiv 5 \pmod{7}: 5k+35(mod7)5k + 3 \equiv 5 \pmod{7} Simplifying: 5k53(mod7)5k \equiv 5 - 3 \pmod{7} 5k2(mod7)5k \equiv 2 \pmod{7}

Step 5: Solve 5k2(mod7)5k \equiv 2 \pmod{7}

Now, we need to solve 5k2(mod7)5k \equiv 2 \pmod{7}. To do this, we can multiply both sides by the modular inverse of 5 modulo 7.

The inverse of 5 modulo 7 can be found using trial and error or extended Euclidean algorithm. The inverse of 5 modulo 7 is 3, because: 5×3=151(mod7)5 \times 3 = 15 \equiv 1 \pmod{7}

Now multiply both sides of the equation 5k2(mod7)5k \equiv 2 \pmod{7} by 3: 3×5k3×2(mod7)3 \times 5k \equiv 3 \times 2 \pmod{7} k6(mod7)k \equiv 6 \pmod{7}

Thus, k=7m+6k = 7m + 6 for some integer mm.

Step 6: Substitute back into x=5k+3x = 5k + 3

Substitute k=7m+6k = 7m + 6 into x=5k+3x = 5k + 3: x=5(7m+6)+3x = 5(7m + 6) + 3 x=35m+30+3x = 35m + 30 + 3 x=35m+33x = 35m + 33

Step 7: Find values of xx between 1 and 100

We need to find the integers x=35m+33x = 35m + 33 that are between 1 and 100. Solving for mm: 135m+331001 \leq 35m + 33 \leq 100 Subtract 33 from all parts: 3235m67-32 \leq 35m \leq 67 Divide by 35: 3235m6735-\frac{32}{35} \leq m \leq \frac{67}{35} m{0,1}m \in \{0, 1\}

Step 8: Calculate the corresponding xx values

For m=0m = 0: x=35(0)+33=33x = 35(0) + 33 = 33 For m=1m = 1: x=35(1)+33=68x = 35(1) + 33 = 68

Step 9: Conclusion

The integers between 1 and 100 that satisfy both conditions are 3333 and 6868. Therefore, there are 2 integers that satisfy the given conditions.


Do you need more details or have any questions? Here are some related problems you might find interesting:

  1. How many integers between 1 and 200 leave a remainder of 4 when divided by 6 and 3 when divided by 8?
  2. What are the smallest three integers that satisfy x2(mod4)x \equiv 2 \pmod{4} and x1(mod3)x \equiv 1 \pmod{3}?
  3. How many integers between 1 and 500 satisfy x5(mod7)x \equiv 5 \pmod{7} and x2(mod9)x \equiv 2 \pmod{9}?
  4. Solve the system x1(mod2)x \equiv 1 \pmod{2} and x3(mod5)x \equiv 3 \pmod{5}.
  5. How many integers between 1 and 1000 satisfy x1(mod11)x \equiv 1 \pmod{11} and x5(mod13)x \equiv 5 \pmod{13}?

Tip: The Chinese Remainder Theorem is a powerful tool for solving systems of simultaneous congruences.

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Math Problem Analysis

Mathematical Concepts

Modular Arithmetic
Chinese Remainder Theorem

Formulas

x ≡ 3 (mod 5)
x ≡ 5 (mod 7)

Theorems

Chinese Remainder Theorem

Suitable Grade Level

Grades 10-12