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Big Powers

Three people chose this as a favourite problem. It is the sort of problem that needs thinking time - but once the connection is made it gives access to many similar ideas.

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Diggits

Can you find what the last two digits of the number $4^{1999}$ are?

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Tenth Power

Do these powers look odd...?

Not a Zero

Age 11 to 14 Short Challenge Level:

Answer: $6$


The zeros on the end of the number are caused by factors of $10$. 

$2^{57} \times 3^4 \times 5^{53} = 2^4 \times 3^4 \times 2^{53} \times 5^{53} = 2^4 \times 3^4 \times 10^{53} = 6^4\times10^{53}$

Last non-zero digit comes from $6^4$

$6^1=6$
$6^2=36$
$6^3=\ ...6$
$6^4=\ ...6$

This means the last non-zero digit will be a $6$.


This problem is taken from the UKMT Mathematical Challenges.
You can find more short problems, arranged by curriculum topic, in our short problems collection.