Patterns & Reasoning

Depth · one concept from first principles · step 2 of 4

Over 11 means the block is a multiple of 9

The repeating decimal 0.xy‾0.\overline{xy} equals z/11 in lowest terms, with x a one-digit odd number. Count the possible natural numbers z.
Number systemOperations
Your answer
How to solve
Strategy Convert to Algebra — Write both sides over the same denominator 99 to get an equation in whole numbers. It says the two-digit number xy is a multiple of 9, which a digit-sum list handles quickly.
1STEP 1

Put both sides over 99

A two-digit repeating block sits over 99, and z/11 = 9z/99.

(10x + y)/99 = 9z/99 → 10x + y = 9z
2STEP 2

Use the digit-sum test

A number is a multiple of 9 exactly when its digit sum is. With x ≤ 9 and y ≤ 9, the digit sum x + y is 9 or 18.

x + y = 9 or x + y = 18
3STEP 3

List the odd x

x + y = 9 gives (1, 8), (3, 6), (5, 4), (7, 2), (9, 0); x + y = 18 gives (9, 9), but 0.99‾0.\overline{99} = 1, so it is out.

(x, y) = (1, 8), (3, 6), (5, 4), (7, 2), (9, 0)
4STEP 4

Read off z

Divide each block by 9: 18, 36, 54, 72, 90 give z = 2, 4, 6, 8, 10. None is a multiple of 11.

z = 2, 4, 6, 8, 10
Answer
5
Spot-check: 0.36‾0.\overline{36} = 36/99 = 4/11 and 0.90‾0.\overline{90} = 90/99 = 10/11, both in lowest terms with denominator 11.
Takeaway

Over 11 means over 99 -- so the repeating block has to be a multiple of 9.

  • Put both sides over 99
  • Use the digit-sum test
  • List the odd x
  • Read off z
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