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, where p 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 2⁴ or 5⁴? Give reasons.

Step-by-step solution

Idea: A number with 4 decimal places is a whole number of ten-thousandths. In p2⁴ × 5⁴, cancelling can remove 2s only if p is even and 5s only if 5 divides p; it cannot do both, because 10 does not divide p.

1. Let the number be x = N.d₁d₂d₃d₄, with d₄ ≠ 0 (the last non-zero digit is in the 4th place). Then 10⁴x = Nd₁d₂d₃d₄ is an integer; call it p. So x = p10⁴.1 mark
2. The units digit of p is d₄ ≠ 0, so p does not end in 0: p is not divisible by 10. (Example: 3.1415 = 3141510⁴, and 31415 ends in 5.)1 mark
3. Now p10⁴ = p2⁴ × 5⁴. Since 10 ∤ p, p is not divisible by both 2 and 5. If 2 ∤ p, no factor 2 cancels and 2⁴ stays in the denominator. If 5 ∤ p, no factor 5 cancels and 5⁴ stays.1 mark
4. So yes, the lowest-form denominator is always divisible by 2⁴ or by 5⁴ (at least one of them). Not necessarily both: 0.0005 = 510⁴ = 12000 = 12⁴ × 5³ keeps 2⁴ only, and 0.0016 = 1610⁴ = 1625 = 15⁴ keeps 5⁴ only.1 mark

Check: 0.0005: 2000 = 2⁴ × 125 ✓ divisible by 16. 0.0016: 625 = 5⁴ ✓. 0.0001 = 110⁴: divisible by both ✓.

Answer to write in the exam

Common mistakes that cost marks

• Saying the denominator must be divisible by both 2⁴ and 5⁴. Examples like 0.0016 = 1625 show it need not.
• Saying p is not divisible by 2 (or by 5). p can be even (16) or a multiple of 5 (5); it just cannot be a multiple of 10.
• Forgetting the whole-number part: 12.0034 = 12003410⁴, not 3410⁴.

How this can come in the exam

Try one yourself

Write 2.0048 as p10⁴ and in lowest form. Which of 2⁴, 5⁴ divides the denominator?

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