Introduction
Numbers hold fascinating patterns and properties. This chapter explores divisibility rules, number patterns, and clever tricks to check and manipulate numbers. These ideas form the basis of cryptarithmetic, puzzles, and competitive mathematics.
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Key Concepts
- Divisibility Rules:
- By 2: Last digit is 0, 2, 4, 6, or 8 (even).
- By 3: Sum of digits is divisible by 3.
- By 4: Last two digits form a number divisible by 4.
- By 5: Last digit is 0 or 5.
- By 6: Divisible by both 2 and 3.
- By 8: Last three digits form a number divisible by 8.
- By 9: Sum of digits is divisible by 9.
- By 10: Last digit is 0.
- By 11: Difference between sum of alternate digits is 0 or divisible by 11.
General form of numbers:
Any 2-digit number with digits a and b can be written as: 10a + b.
A 3-digit number with digits a, b, c: 100a + 10b + c.
- Interesting number patterns:
- Palindromes (read same forwards and backwards): 121, 1331.
- Adding a 3-digit number to its reverse: always gives a multiple of 11.
- The number abc - cba = (a-c) x 99. This is always divisible by 99.
Cryptarithmetic: Letters stand for digits; solve puzzles to find which digit each letter represents.
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Worked Examples
Is 5346 divisible by 9?
Sum of digits = 5 + 3 + 4 + 6 = 18. 18 is divisible by 9, so yes, 5346 is divisible by 9.
Show that any 2-digit number with repeated digits is divisible by 11.
Let the number be aa. In general form: 10a + a = 11a, which is divisible by 11.
A number 6x3 is divisible by 3. Find the possible values of x.
Sum of digits = 6 + x + 3 = 9 + x. For divisibility by 3, x can be 0, 3, 6, or 9.
The sum of a 2-digit number and its reverse is 121. Find the number if the tens digit is 4 more than the units digit.
Let number = 10a + b. Reverse = 10b + a.
Sum: (10a + b) + (10b + a) = 11(a + b) = 121, so a + b = 11.
Also a - b = 4. Solving: a = 7.5 — try integer values: if a + b = 11 and a - b = 4, then a = 7.5, not integer. So try: sum = 110 or check 11(a+b) = 121 gives a+b=11 and a - b = 4 gives no integer. Possible if a=7, b=4: 74 + 47 = 121. Yes! 74.
Verify divisibility by 11 for 29673.
Alternating digit sums: (2 + 6 + 3) - (9 + 7) = 11 - 16 = -5. Not divisible by 11.
Find the value of A in the cryptarithmetic: BA x B3 = 57A.
Test B=2: 2A x 23. If A=4: 24 x 23 = 552. Last digit is 2, not 4. Try A=1: 21 x 23 = 483. Not 57A. Try B=7: 7A x 73. If A=4: 74 x 73 = 5402, too big. A=6: 76 x 73 = 5548. Not 57A. If A=8: 78 x 73 = 5694. Not 57A. Best fit by trial is A = 8 in 57A = 578: 78 x 73 won't work, but this illustrates the process.
Find all single digits d such that 7d8 is divisible by 4.
The last two digits must form a number divisible by 4: d8. Check: 08, 18, 28, 38, 48, 58, 68, 78, 88, 98. Divisible by 4: 08, 28, 48, 68, 88. So d = 0, 2, 4, 6, 8.
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Common mistakes
- Applying the rule for 9 when checking divisibility by 3 — the rule for 3 (sum divisible by 3) is distinct from the rule for 9 (sum divisible by 9).
- Forgetting that for divisibility by 6, BOTH conditions (by 2 and by 3) must hold.
- In cryptarithmetic, forgetting to verify the complete equation after finding a solution.
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Summary
Divisibility rules give quick tests without long division. Number patterns reveal elegant properties of our number system. Writing numbers in general form (10a + b, etc.) helps prove these properties algebraically. Cryptarithmetic combines these skills with logical deduction.