If a number leaves a remainder of 3 when divided by 5 and a remainder of 2 when

If a number leaves a remainder of 3 when divided by 5 and a remainder of 2 when divided by 7, what is the smallest such number?

If a number leaves a remainder of 3 when divided by 5 and a remainder of 2 when divided by 7, what is the smallest such number?

Step 1: Understand the problem. We need to find a number that gives a remainder of 3 when divided by 5 and a remainder of 2 when divided by 7.
Step 2: Write down the first condition. If a number 'x' leaves a remainder of 3 when divided by 5, we can express this as: x = 5k + 3, where k is any whole number (0, 1, 2, ...).
Step 3: Write down the second condition. If the same number 'x' leaves a remainder of 2 when divided by 7, we can express this as: x = 7m + 2, where m is any whole number (0, 1, 2, ...).
Step 4: Set the two expressions for 'x' equal to each other: 5k + 3 = 7m + 2.
Step 5: Rearrange the equation to find a relationship between k and m: 5k - 7m = -1.
Step 6: Solve for k in terms of m. We can try different values of m to find a corresponding k that is a whole number.
Step 7: Start with m = 0: 5k - 7(0) = -1 → 5k = -1 (not a whole number).
Step 8: Try m = 1: 5k - 7(1) = -1 → 5k = 6 → k = 6/5 (not a whole number).
Step 9: Try m = 2: 5k - 7(2) = -1 → 5k = 13 → k = 13/5 (not a whole number).
Step 10: Try m = 3: 5k - 7(3) = -1 → 5k = 20 → k = 4 (this is a whole number).
Step 11: Substitute k = 4 back into the equation for x: x = 5(4) + 3 = 20 + 3 = 23.
Step 12: Verify the solution: 23 divided by 5 gives a remainder of 3, and 23 divided by 7 gives a remainder of 2.
Step 13: Conclude that the smallest number satisfying both conditions is 23.

Modular Arithmetic – The problem involves finding a number that satisfies specific modular conditions, which is a fundamental concept in number theory.
Chinese Remainder Theorem – This problem can be approached using the principles of the Chinese Remainder Theorem, which deals with solving systems of congruences.