Question 53 of 58advanced🔍 AnalyzeLong Answer5 marks

A rational number has a terminating decimal expansion whose last non-zero digit occurs in the 4th decimal place. Show that such a number can be written in the form p104\frac{p}{10^4}, where pp is an integer not divisible by 10. Is it necessary that the denominator of this rational number, when written in the lowest form, is divisible by 242^4 or 545^4? Give reasons.

Correct Answer

A rational number with a terminating decimal expansion whose last non-zero digit is in the 4th decimal place can be expressed with a denominator that is a power of 10. Since the last non-zero digit occurs in the 4th decimal place, any digits after this must be zero. Therefore, the number can be written as an integer divided by 10410^4. Let this integer be pp. Thus, the number is of the form:

p104\frac{p}{10^4}

Furthermore, if pp were divisible by 10, the last digit of pp would be 0. This would mean that the last non-zero digit of the decimal expansion actually occurs in the 3rd decimal place or earlier, which contradicts the given condition. Hence, pp must be an integer that is not divisible by 10.

Regarding the second part, it is necessary that the denominator of this rational number, when written in its lowest form, is divisible by 242^4 or 545^4. The denominator in its lowest form is obtained by cancelling the common factors of pp and 10410^4 (which is 24×542^4 \times 5^4). Since pp is not divisible by 10, it cannot be divisible by both 2 and 5, but it may be divisible by one of them.

For example, consider the rational number:

116=0.0625\frac{1}{16} = 0.0625

The last non-zero digit is in the 4th decimal place. Its denominator in lowest form is 16=2416 = 2^4, which is divisible by 242^4 but not 545^4. Similarly,

1625=0.0016\frac{1}{625} = 0.0016

has a lowest form denominator of 625=54625 = 5^4, which is divisible by 545^4 but not 242^4. Thus, the lowest form denominator must be divisible by either 242^4 or 545^4 (or both), but it is not necessary for it to be divisible by any specific one of them.

Exercise: END-OF-CHAPTER EXERCISES | Q: 10 | (Page 25)
For More Understanding

Explanation

The question tests the understanding of terminating decimal expansions and their relationship with the prime factorization of the denominator. A terminating decimal can always be written as an integer divided by a power of 10. If the last non-zero digit is in the 4th decimal place, the denominator is 10410^4, and the numerator cannot be a multiple of 10 (otherwise, the decimal would terminate earlier).

For the second part, the lowest form denominator is obtained by removing common factors from 24×542^4 \times 5^4. Since the numerator is not a multiple of 10, it doesn't contain both 2 and 5 as factors, but it might contain one of them. Therefore, the lowest form denominator could be 242^4 (if 5 is cancelled), 545^4 (if 2 is cancelled), or 24×542^4 \times 5^4 (if neither is cancelled).

It is necessary that the denominator is divisible by 242^4 or 545^4 to ensure the decimal terminates at the 4th place, but it is not necessary for it to be divisible by both or any specific one.

Solution Steps

  1. Step 1: A terminating decimal with the last non-zero digit in the 4th decimal place can be written as an integer divided by 10410^4, i.e.,

    p104\frac{p}{10^4}

  2. Step 2: If pp were divisible by 10, the last digit of pp would be 0, meaning the decimal would terminate at the 3rd place or earlier. Hence, pp is not divisible by 10.

  3. Step 3: The denominator in lowest form is found by cancelling common factors of pp and 104=24×5410^4 = 2^4 \times 5^4.

  4. Step 4: Since pp is not divisible by 10, it cannot contain both 2 and 5 as factors. Therefore, when cancelling common factors from 104=24×5410^4 = 2^4 \times 5^4, the lowest form denominator must retain at least 242^4 or 545^4 (or both). For instance,

    116=0.0625\frac{1}{16} = 0.0625

    has a lowest form denominator of 242^4 (divisible by 242^4 but not 545^4), and

    1625=0.0016\frac{1}{625} = 0.0016

    has a lowest form denominator of 545^4 (divisible by 545^4 but not 242^4). Thus, it IS necessary for the denominator to be divisible by 242^4 or 545^4, though not necessarily by any specific one of them.

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