Number systems questions are very common in competitive exams like the CSAT and state PSCs. Often, examiners will ask you to find remainders for massive numbers. For example, they might ask if a 10-digit number is divisible by 7, 11, or 13.
Using long division on these large numbers takes too much time. Therefore, you need to use quick mathematical shortcuts. By memorizing these divisibility rules, you can solve complex problems in seconds.
The Standard Rules (A Quick Review)
First, let us review the basic rules you already know.
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Divisibility by 3 and 9: The sum of all digits must be divisible by 3 or 9.
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Divisibility by 4: The last two digits must be divisible by 4.
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Divisibility by 8: The last three digits must be divisible by 8.
Advanced Rules for CSAT Exams
Next, let us look at the advanced rules that frequently appear in the CSAT.
1. The Alternating Sum Rule for 11 A number is completely divisible by 11 if it meets a specific condition. You must find the difference between the sum of its odd-position digits and the sum of its even-position digits. If that difference is 0 or a multiple of 11, the number is divisible.
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Example: Is the number 856416 divisible by 11?
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First, find the sum of digits at odd places: 8 + 6 + 1 = 15.
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Second, find the sum of digits at even places: 5 + 4 + 6 = 15.
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Finally, find the difference: 15 – 15 = 0.
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Conclusion: Yes, 856416 is perfectly divisible by 11.
2. The Triplet Grouping Rule for 7, 11, and 13 For very large numbers, there is an amazing shortcut to check for 7, 11, and 13 all at once. First, group the digits into sets of three starting from the right. After that, subtract the smaller group from the larger group.
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Example: Check if 124,783 is divisible by 7.
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Group 1 (Right side): 783.
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Group 2 (Left side): 124.
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Next, subtract them: 783 – 124 = 659.
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Finally, test the small number for 7: 659 ÷ 7 = 94 with a remainder of 1.
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Conclusion: Because 659 is not perfectly divisible by 7, the original number is also not divisible by 7.
Master Divisibility Checklist
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Divisibility by 3: The sum of digits is a multiple of 3. (Example: 123 -> 1 + 2 + 3 = 6).
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Divisibility by 6: The number must be even AND divisible by 3.
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Divisibility by 8: The last 3 digits are divisible by 8. (Example: 41256 -> 256 ÷ 8 = 32).
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Divisibility by 9: The sum of digits is a multiple of 9.
In summary, practicing these structural rules will turn tough arithmetic problems into easy logical puzzles.