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Lesson 7 of 10

Algebra Play · Lesson 7 of 10

Decoding Divisibility Tricks

“Use place-value algebra to prove divisibility patterns involving reversed, cycled, and repeated digits.”

Learning Objectives

• Write two-digit and three-digit numbers in expanded place-value form. • Prove that the difference of a two-digit number and its reverse is divisible by 9. • Prove that the sum of a two-digit number and its reverse is divisible by 11. • Explain divisibility patterns created by cycling three digits. • Use 1001 = 7×11×13 to explain why a repeated three-digit block has several guaranteed factors.

Many divisibility tricks come from place value. A two-digit number with tens digit a and ones digit b is not a+b; it is 10a+b. Once we write the number in expanded form, reversing or repeating digits becomes an algebraic operation rather than a mystery.

Reversing a Two-Digit Number

Written numberExpanded form
ab10a + b
ba10b + a

If b>a, then the reversed number ba is larger. The difference is (10b+a)-(10a+b)=9b-9a=9(b-a). Because the result has a factor of 9, it is divisible by 9.

Difference of a two-digit number and its reverseLaTeX
If a>b, the positive difference is 9(a-b). Either way, the magnitude is a multiple of 9.
Why 74−47 is divisible by 9

Problem
Explain the divisibility without relying only on calculation.

  1. 1.47 = 10(4)+7 and 74 = 10(7)+4.
  2. 2.74−47 = [10(7)+4]-[10(4)+7].
  3. 3.This simplifies to 9(7−4)=27.
  4. 4.The factor 9 is built into the place-value difference, so the same reasoning works for any two digits.

The quotient after dividing the positive difference by 9 is simply the difference between the two digits. For 47 and 74, the digit difference is 7−4=3, which matches 27÷9=3.

Adding a Number and Its Reverse

Now add instead of subtracting. (10a+b)+(10b+a)=11a+11b=11(a+b). Therefore the sum is always divisible by 11.

A sum that must be divisible by 11

Problem
Why is 28+82 guaranteed to be divisible by 11?

  1. 1.28 = 10(2)+8 and 82=10(8)+2.
  2. 2.Their sum is 11(2+8)=11×10=110.
  3. 3.The factor 11 appears for any pair of digits a and b.
Sum with the reverseLaTeX

Cycling Three Digits

Let abc represent 100a+10b+c. Cycling the digits gives bca=100b+10c+a and cab=100c+10a+b. Add all three expressions.

Sum of cyclic three-digit numbersLaTeX
This proves divisibility by both 3 and 37.
Cycling 2,4,7

Problem
Check the result using 247, 472, and 724.

  1. 1.247+472+724=1443.
  2. 2.The digit sum is 2+4+7=13.
  3. 3.111×13=1443.
  4. 4.1443÷37=39 and 1443÷3=481, confirming both guaranteed factors.

Repeating a Three-Digit Block

If N is the three-digit number abc, then abcabc means 1000N+N=1001N. Since 1001=7×11×13, the repeated six-digit number is divisible by 7, then 11, then 13, regardless of the original three-digit block.

Repeated block identityLaTeX
The repeated-block trick

Problem
Start with 352 and form 352352. What happens after dividing by 7, then 11, then 13?

  1. 1.352352 = 1001×352.
  2. 2.Because 1001 = 7×11×13, dividing by those three factors cancels the multiplier 1001.
  3. 3.The final result is 352.
  4. 4.The trick returns the original three-digit number.

Why Algebra Is Better Than Many Tests

Checking 12, 47, and 83 may reveal the reverse-number pattern, but it cannot explain why the result must persist. The place-value expressions expose the common factor directly. Divisibility statements are especially suited to algebra because a visible factor proves the claim.

Common mistake

Do not treat abc as the product a×b×c. In place-value notation, abc means 100a+10b+c. The digits are positions, not factors.

Try this

Choose any three-digit number with non-zero hundreds digit. Form its two cyclic rearrangements and add all three. Predict the quotient after division by 37 before calculating.

Reading the Quotient, Not Just the Divisibility

The reverse-difference trick contains more information than 'divisible by 9'. After dividing the positive difference by 9, the quotient is |a−b|, the difference between the digits. So someone who knows the quotient learns how far apart the two digits are, even without knowing the original number.

Similarly, the cyclic three-digit sum reveals the digit sum: dividing abc+bca+cab by 111 gives a+b+c. Factorisation can therefore explain both divisibility and what the resulting quotient means.

Quiz

Quick check

What is the expanded form of a two-digit number ab?

Quick check

The positive difference between ab and ba is always divisible by which number?

Quick check

The sum of ab and ba is always divisible by which number?

Quick check

Why is abc+bca+cab divisible by 37?

Quick check

Why does dividing abcabc successively by 7,11,13 return abc?

Practice Problems

Practice Problems
  1. Use algebra to show that the positive difference between 83 and 38 is divisible by 9, then identify the quotient in terms of the digits.
  2. Prove that 31+13 is divisible by 11 without first adding the two numbers.
  3. For digits 3,5,8, add 358,583,835 and verify divisibility by 37 and 3.
  4. Explain why abcabc/7/11/13 returns abc for any three-digit abc.
  5. A student claims that the difference of a number and its reverse is always divisible by 11. Give a counterexample and state the correct guaranteed divisor.
  6. Create a new digit-reversal claim, test it with examples, and then decide whether you can justify it algebraically.

83−38=9(8−3)=45. The quotient by 9 is the digit difference 5.

Key Takeaways

Key Takeaways

• Place-value expansion turns digit tricks into ordinary algebra. • A two-digit number and its reverse differ by 9 times the digit difference. • A two-digit number plus its reverse equals 11 times the digit sum. • The sum abc+bca+cab is 111(a+b+c), so it is always divisible by 3 and 37. • Repeating a three-digit block gives abcabc=1001×abc, and 1001=7×11×13.

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The Largest Product

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Algebraic Story Puzzles