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Answer: 70$\frac13$
Maximum possible distance: 3$\times$75 = 225
Maximum possible milometer reading: 29792 + 225 = 30017
Largest palindromic number before 30017 will have first digit 3 $\therefore$ last digit 3
second and third digits 0 so fourth digit 0
Final milometer reading: 30003
Distance travelled: 30003 $-$ 29792 = 211 (or 14 less than maximum)
Speed: 211$\div$3 = 70$\frac{1}{3}$ (or 14$\div$3 = 4$\frac23$ less than speed limit)
Palindromic Milometer
Age 11 to 14
ShortChallenge Level 
Answer: 70$\frac13$
Maximum possible distance: 3$\times$75 = 225
Maximum possible milometer reading: 29792 + 225 = 30017
Largest palindromic number before 30017 will have first digit 3 $\therefore$ last digit 3
second and third digits 0 so fourth digit 0
Final milometer reading: 30003
Distance travelled: 30003 $-$ 29792 = 211 (or 14 less than maximum)
Speed: 211$\div$3 = 70$\frac{1}{3}$ (or 14$\div$3 = 4$\frac23$ less than speed limit)
You can find more short problems, arranged by curriculum topic, in our short problems collection.
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