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

Reversed repeating blocks add to the digit sum

$a$ and $b$ are even numbers less than $10$ with $a > b > 0$ . The repeating decimals $0.\overline{ab}$ and $0.\overline{ba}$ (the two-digit blocks $ab$ and $ba$ repeat) add up to $0.\overline{6}$ . Write the difference of the two repeating decimals as a repeating decimal.

Number systemExpressions

Your answer

How to solve

Strategy Convert to Algebra — Write each repeating decimal as a fraction over 99 using place value. Adding the reversed blocks collapses to a statement about a + b alone; then the evenness condition pins down the digits.

1STEP 1

Write both decimals as fractions

A two-digit repeating block over 99: the block ab is the number 10a + b.

0.\overline{ab}

= (10a + b)/99,

0.\overline{ba}

= (10b + a)/99

2STEP 2

Add them and solve for a + b

The sum is (11(a + b))/99 = (a + b)/9, and $0.\overline{6}$ = 6/9.

((10a + b) + (10b + a))/99 = (11(a + b))/99 = (a + b)/9 = 6/9 → a + b = 6

3STEP 3

Find the digits

List even pairs with a ＞ b ＞ 0 and a + b = 6: only a = 4, b = 2 works (6 + 0 is ruled out because b ＞ 0).

a = 4, b = 2

4STEP 4

Take the difference

The larger decimal is $0.\overline{42}$ and the smaller is $0.\overline{24}$ .

42/99 - 24/99 = 18/99 =

0.\overline{18}

Answer

0.\overline{18}

Check the sum: 42/99 + 24/99 = 66/99 = 2/3 =

0.\overline{6}

. The difference 0.1818… is also 0.4242… - 0.2424… digit by digit.

Takeaway

Swap the digits and add -- place value leaves only the digit sum behind.

Write both decimals as fractions
Add them and solve for a + b
Find the digits
Take the difference

Where next?

Level upsuggested

Over 11 means the block is a multiple of 9

Another one like thissuggested

12 practice problems

Step back and solidifysuggested

Equal remainders: divisor divides every difference Smallest fraction: LCM of denominators over GCF

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▶ Practice — 12 problems