If a number leaves a remainder of 3 when divided by 7 and a remainder of 5 when

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Question: If a number leaves a remainder of 3 when divided by 7 and a remainder of 5 when divided by 9, what is the smallest such number? (2023)

Options:

Correct Answer: 26

Exam Year: 2023

Solution:

The smallest number that satisfies both conditions is 26 (26 % 7 = 5 and 26 % 9 = 8).

If a number leaves a remainder of 3 when divided by 7 and a remainder of 5 when divided by 9, what is the smallest such number? (2023)

If a number leaves a remainder of 3 when divided by 7 and a remainder of 5 when divided by 9, what is the smallest such number? (2023)

Step 1: Understand the problem. We need to find a number that gives a remainder of 3 when divided by 7 and a remainder of 5 when divided by 9.
Step 2: Write down the first condition: If we call the number 'x', then x % 7 = 3. This means when we divide x by 7, the remainder is 3.
Step 3: Write down the second condition: x % 9 = 5. This means when we divide x by 9, the remainder is 5.
Step 4: Start testing numbers that satisfy the first condition (x % 7 = 3). The numbers that satisfy this condition are 3, 10, 17, 24, 31, ... (these are found by adding 7 repeatedly to 3).
Step 5: Check each of these numbers to see if they also satisfy the second condition (x % 9 = 5).
Step 6: Test 3: 3 % 9 = 3 (not a match). Test 10: 10 % 9 = 1 (not a match). Test 17: 17 % 9 = 8 (not a match). Test 24: 24 % 9 = 6 (not a match). Test 31: 31 % 9 = 4 (not a match). Test 38: 38 % 9 = 2 (not a match). Test 45: 45 % 9 = 0 (not a match). Test 52: 52 % 9 = 7 (not a match). Test 59: 59 % 9 = 5 (match!).
Step 7: The smallest number that satisfies both conditions is 59.

Modular Arithmetic – The question tests the understanding of remainders when dividing numbers, specifically using modular arithmetic to find a number that satisfies multiple conditions.
System of Congruences – The problem involves solving a system of congruences, which is a common topic in number theory.