What fraction of whole numbers is equal to 17100 and that has a denominator that is a power of 10?

Answer 1

17100 = 17100/1

= 17100/1 × 10/10 = 171000/10

= 17100/1 × 10²/10² = 1710000/100

= 17100/1 × 10³/10³ = 17100000/1000

...

= 17100/1 × 10ⁿ/10ⁿ

There is no upper limit to this sequence: n = 0, 1, 2, 3, ....

These numbers can be put in a one-to-one relationship with the counting numbers 1, 2, 3, ...

The counting numbers can also be put in a one-to-one relationship with the whole numbers

Therefore the sequence of fractions can be put in a one-to-one relationship with the whole numbers.

Therefore the fraction of whole numbers which are in the sequence is 1.

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Let's try another approach:

As 10 to some power is positive, there is no way we can change a negative number to make it positive by dividing by 10 to some power, therefore the solution numbers are all positive.

As negative whole numbers make up half of the whole numbers, the fraction of whole numbers which are equivalent to 17100 when divided by a power of 10 is at most ½.

The fraction of the ½ of the whole numbers which are positive can now be considered:

After the nth whole number has been found which matches the criteria that is it equivalent to 17100 when it is divided by a power of 10, in total there are n numbers matching out of a total of 17100 × 10ⁿ⁻³ numbers (the value of the nth number); thus:

The first number which meets the criteria is 171/10⁻² → the fraction is 1/171

The second number to meet the criteria is 1710/10⁻¹ → the fraction is 2/1710

The third number to meet the criteria is 17100/10⁰ → the fraction is 3/17100

The fourth number to match the criteria is 171000/10 → the fraction is 4/171000

The fifth number to match the criteria is 1710000/10²→ the fraction is 5/1710000

→ For the nth match, the fraction is: n/(17100 × 10ⁿ⁻³) = (1/17100) × n/10ⁿ⁻³

This gives us a sequence of fractions:

1/17100 × 1/10⁻², 1/17100 × 2/10⁻¹, 1/17100 × 3/10⁰, 1/17100 × 4/10¹, 1/17100 × 5/10², ...

= 100000/17100000, 20000/17100000, 3000/17100000, 400/17100000, 35/17100000, 6/17100000, ....

Consider terms n and n+1:

U{n} = (1/17100) × n/10ⁿ⁻³ = (1/17100) × 10 × n/10ⁿ⁻²

U{n+1} = (1/17100) × (n+1)/10ⁿ⁻²

U{n} - U{n+1} = (1/17100) × 10 × n/10ⁿ⁻² - (1/17100) × (n+1)/10ⁿ⁻²

= (10n - (n+1))/(17100 × 10ⁿ⁻²)

= (9n - 1)/(17100 × 10ⁿ⁻²)

As n ≥ 1,

9n - 1 ≥ 9×1 - 1 = 8

→ term n - term n+1 ≥ 8 > 0

→ term n is larger than term n+1

for all n ≥ 1

Each of these terms is less than 1, and each term of the sequence is smaller than the previous one and so as n increases the value of the each term (the fraction of whole numbers which meet the criteria) tends towards 0.

→ The fraction of all whole numbers which equate to 17100 when divided by a power of 10 is as near enough to zero as make no odds.

ie the fraction is so small it is effectively none of the whole numbers.

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This is a problem of dealing with the infinite.

Answer 2

The required fraction is 1/0000584795321637426900, with the underlined string repeating. A repeating decimal does not have a fixed denominator because every time you add another repeating digit, the denominator goes up by another power of 10.

However, the above number can be approximated by 1/00005847953216374269, which is 5,847,953,216,374,269/100,000,000,000,000,000,000 or 5,847,953,216,374,269/1020.

Answer 3

Any of the following:17100 / 1

171000 / 10

1710000 / 100

17100000 / 1000

etc.