A number leaves a remainder of 1 when divided by 5 and a remainder of 2 when divided by 7. What is the smallest such number?

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Q: A number leaves a remainder of 1 when divided by 5 and a remainder of 2 when divided by 7. What is the smallest such number?

Solution: The smallest number satisfying both conditions is 22.

Steps: 14

Step 1: Understand the problem. We need to find a number that gives a remainder of 1 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 1 when divided by 5, we can express this as: x = 5k + 1, 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 + 1 = 7m + 2.
Step 5: Rearrange the equation to find a relationship between k and m: 5k - 7m = 1.
Step 6: Now, we can test different values of k to find a corresponding m that satisfies the equation.
Step 7: Start with k = 0: 5(0) - 7m = 1 → -7m = 1 (not valid).
Step 8: Try k = 1: 5(1) - 7m = 1 → 5 - 7m = 1 → 7m = 4 (not valid).
Step 9: Try k = 2: 5(2) - 7m = 1 → 10 - 7m = 1 → 7m = 9 → m = 1.2857 (not valid).
Step 10: Try k = 3: 5(3) - 7m = 1 → 15 - 7m = 1 → 7m = 14 → m = 2 (valid).
Step 11: Now substitute k = 3 back into the equation for x: x = 5(3) + 1 = 15 + 1 = 16.
Step 12: Check if x = 16 satisfies the second condition: 16 divided by 7 gives a remainder of 2 (valid).
Step 13: Continue testing higher values of k until you find the smallest valid x. After testing, you find that k = 4 gives x = 21, and k = 5 gives x = 22.
Step 14: The smallest number that satisfies both conditions is 22.