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Hello:

When a repeating decimal is changed to a fraction, the following calculation can be used.

0.33333...

x = 0.333 or 1 * x = 0.33333

10 * x = 10 * 0.3333...

10 * x = 3.3333

subtract 1 * x = 0.3333 from both side of the calculation.

The result is 9 * x = 3. Now determine the value of x by dividing both sides by 9. The result is 3/9 or 1/3.

How is this type of calculation used with 2.1343434... and 0.5270270?

I thank you for your reply.

For 2.1343434, it repeats every two digits, so multiply by 100 and get,

if x = 2.1343434... , that 100x = 213.43434... .

Thus, 100x - x = 213.43434.. - 2.1343434... .

This gives us 99x = 213.43 - 2.13 = 211.30 = 211.3.

This means that x = 211.3/99 = 2113/990.

Since 990 = 2*3*3*5*11 and 2113 is prime, that is my final answer.

The way to see 2113 is prime is to divide it by prime numbers until the result is less than the number which is being used to divide it. Since the first few prime numbers are known to be

2, 3, 5, 7, 11, 13, 17, 19, 23, 29,31, 37, 41, 43 and 47 {and 47*47 = 2209, which is greater than 2113), it can be seen (using Excel), that

2113/2 = 1056.5,

2113/3 = 704.3333333,

2113/5 = 422.6,

2113/7 = 301.8571429,

2113/11 = 192.0909091,

2113/13 = 162.5384615,

2113/17 = 124.2941176,

2113/19 = 111.2105263,

2113/23 = 91.86956522,

2113/29 = 72.86206897,

2113/31 = 68.16129032,

2113/37 = 57.10810811,

2113/41 = 51.53658537,

2113/43 = 49.13953488, and

2113/47 = 44.95744681.

Form this, it can then be seen since none of the equations have an integer solution,

the number 2113 is prime.

For x = 0.5270270... , the number repeat every 3 digits, so multiply x by 10³ = 1000.

This gives 1000x - x = 527.027027027... - 0.527027027... .

Doing this subtraction gives us 999x = 526.5.

Multiplying both sides by 2/999 (to get rid of the 999 and the .5) gives

2x = 1053/999. This gives x = 1053/1998.

It is known that 1998 is divisible by two and 3, and it turns out that the

prime factors of 1998 are 2*3*3*3*37.

It is also known 1053 is divisible by 3 (since the sum of 1, 0, 5, and 3 is 9).

Doing this many times gives 1053 = 3*3*3*3*13.

Cancelling 3 3's gives 1053/1998 = 3*13/(2*37) = 39/74.

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