If \(\dfrac{31}{2^a5^b}\) terminates exactly after 10 decimal places and \(b>a\), what is the value of \(b\)?
Answer and explanation
Correct answer: 10
A rational number in lowest terms has a terminating decimal iff its denominator's prime factors are only 2 and/or 5. Here 31 is coprime to 2 and 5, so the denominator remains \(2^a5^b\). The number of decimal places for termination equals \(\max(a,b)\). Given that this equals 10 and that \(b>a\), the larger exponent must be \(b\), so \(b=10\). The distractor \(a+10\) is incorrect because the termination length is not a sum involving \(a\) but the maximum of the exponents. Exam tip: always reduce the fraction first and use "termination length = max(exponents of 2 and 5 in denominator" ).
Frequently asked questions
What is the correct answer to this question?
10
Why is this the correct answer?
A rational number in lowest terms has a terminating decimal iff its denominator's prime factors are only 2 and/or 5. Here 31 is coprime to 2 and 5, so the denominator remains \(2^a5^b\). The number of decimal places for termination equals \(\max(a,b)\). Given that this equals 10 and that \(b>a\), the larger exponent must be \(b\), so \(b=10\). The distractor \(a+10\) is incorrect because the termination length is not a sum involving \(a\) but the maximum of the exponents. Exam tip: always reduce the fraction first and use "termination length = max(exponents of 2 and 5 in denominator" ).
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Real Numbers. Topic: Decimal expansion of rational numbers.
Student feedback
Was this question useful?
👍 0 Helpful 👎 0 Not helpful
Yes 0% No 0%
0 responsesStudent Reviews
No published reviews yet.
