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We received lots of good solutions to this problem. Thanks to everyone who submitted a solution! Unfortunately there were so many we can't mention you all by name. A special well done to the pupils of Beaconsfield High School for all their great solutions - we're glad you enjoyed the problem so much!

Here is a really nice solution submitted by Oliver from Loreto College:

Assuming n is an integer, there are no values of n so that $2^n$ is a multiple of 10 because $2^n$ doesn't contain any necessary factors of 5.

The unit digits of $2^n$ for n=1,2,3... are 2, 4, 8, 6 then repeat. For $3^n$ they go 3, 9, 7, 1 then repeat. If n is odd, the units that are being added are either 2 + 3 or 7 + 8, which both end in 5. So $2^n + 3^n$ where n is odd always ends in 5. This is a stronger conclusion than saying 'it's a multiple of 5' as a multiple of 5 can also end in 0.

If n is a multiple of 4 the units being added are 6 + 1 so in this case it will always end in 7.

$1^n + 2^n + 3^n$ is even for all values of n. This is obvious because $1^n$ and $3^n$ are always odd and $2^n$ is always even, and odd + odd + even =even.

$1^n + 2^n + 3^n + 4^n$ is a multiple of 10 for when 4 does not divide n:

The unit digits of the powers of 1, 2, 3 and 4 are:

Summing down the columns gives 10, 20, 20 and 14. This shows that when n is not divisible by 4, the last digit of $1^n + 2^n + 3^n + 4^n$ is 0 so it's a multiple of 10. (When n is divisible by 4, the last digit is 4).

$1^n + 2^n + 3^n + 4^n + 5^n$ ends in 5 for n not divisible by 4. This is obvious if we consider the previous result because $5^n$ ends in 5 for all n, so adding this to the multiples of 10 will give a final digit of 5.

Ryan from Renaissance College Hong Kong had another good explanation for why $2^n$ cannot be a multiple of 10:

For which values of n will $2^n$ be a multiple of 10?

As can be seen from the following table, the unit digits of the powers of

two are in a repetitious pattern of 2, 4, 8, 6, 2, 4, 8, 6…

All multiples of 10 have a unit digit of 0. However, as seen from the

pattern, none of the powers of 2 have their unit digit ending in a 0.

Therefore, no power of 2 is a multiple of 10.

Alexander from University College School noticed the following about the extension questions:

If the power n in $4^n + 5^n + 6^n$ is an odd number, then the last digit of this sum is 5; if the power n is even then the last digit of the sum is 7.

$3^n + 8^n$: For consecutive values of n (n = 1, 2, 3, …) the last digit of the sum goes in a pattern of 1, 3, 9, 7.

$2^n + 4^n + 6^n$: For consecutive values of n (n = 1, 2, 3, …) the last digit of the sum goes in a pattern of 2, 6, 8, 8.

$3^n +5^n +7^n$: For consecutive values of n (n = 1, 2, 3, …) the last digit of the sum goes in a pattern of 5, 3, 5, 7.

$3^n-2^n$: For consecutive values of n (n = 1, 2, 3, …) the last digit of the sum goes in a pattern of 1, 5, 9, 5.

Can you explain why this might be true?

Ryan also had some interesting ideas about how he could extend his results:

The unit digit of $1^n$ is 1, 1, 1, 1…

The unit digit of $2^n$ is 2, 4, 8, 6, 2, 4, 8, 6,...

The unit digit of $3^n$ is 3, 9, 7, 1, 4, 9, 7, 1,...

The unit digit of $4^n$ is 4, 6, 4, 6…

The unit digit of $5^n$ is 5, 5, 5, 5…

The unit digit of $6^n$ is 6, 6, 6, 6…

The unit digit of $7^n$ is 7, 9, 3, 1, 7, 9, 3, 1,...

The unit digit of $8^n$ is 8, 4, 2, 6, 8, 4, 2, 6,...

The unit digit of $9^n$ is 9, 1, 9, 1…

The unit digit of $10^n$ is 0, 0, 0, 0…

Thus the unit digits of the powers of any number repeat in a cycle four digits long, as all numbers end with one of the above 10 digits.Let the pattern of the unit digits of the powers of x be a, b, c, d... and let The pattern of the unit digits of the powers of another number y be e, f, g, h…

As a result, the sum/difference of the unit digits of the powers of x and y are in a repeating pattern of: a + e, b + f, c + g, and d + h, or a - e, b - f, c - g and d – h.

As a result, no matter what combination of powers, adding or subtracting, the unit digits are always in a pattern of 4 repeating numbers.

Well done everyone!

Here is a really nice solution submitted by Oliver from Loreto College:

Assuming n is an integer, there are no values of n so that $2^n$ is a multiple of 10 because $2^n$ doesn't contain any necessary factors of 5.

The unit digits of $2^n$ for n=1,2,3... are 2, 4, 8, 6 then repeat. For $3^n$ they go 3, 9, 7, 1 then repeat. If n is odd, the units that are being added are either 2 + 3 or 7 + 8, which both end in 5. So $2^n + 3^n$ where n is odd always ends in 5. This is a stronger conclusion than saying 'it's a multiple of 5' as a multiple of 5 can also end in 0.

If n is a multiple of 4 the units being added are 6 + 1 so in this case it will always end in 7.

$1^n + 2^n + 3^n$ is even for all values of n. This is obvious because $1^n$ and $3^n$ are always odd and $2^n$ is always even, and odd + odd + even =even.

$1^n + 2^n + 3^n + 4^n$ is a multiple of 10 for when 4 does not divide n:

The unit digits of the powers of 1, 2, 3 and 4 are:

x^1 | x^2 | x^3 | x^4 |

1 | 1 | 1 | 1 |

2 | 4 | 8 | 6 |

3 | 9 | 7 | 1 |

4 | 6 | 4 | 6 |

Summing down the columns gives 10, 20, 20 and 14. This shows that when n is not divisible by 4, the last digit of $1^n + 2^n + 3^n + 4^n$ is 0 so it's a multiple of 10. (When n is divisible by 4, the last digit is 4).

$1^n + 2^n + 3^n + 4^n + 5^n$ ends in 5 for n not divisible by 4. This is obvious if we consider the previous result because $5^n$ ends in 5 for all n, so adding this to the multiples of 10 will give a final digit of 5.

Ryan from Renaissance College Hong Kong had another good explanation for why $2^n$ cannot be a multiple of 10:

Power of 2 | Answer | Units Digit |

1 | 2 | 2 |

2 | 4 | 4 |

3 | 8 | 8 |

4 | 16 | 6 |

5 | 32 | 2 |

6 | 64 | 4 |

7 | 128 | 8 |

8 | 256 | 6 |

For which values of n will $2^n$ be a multiple of 10?

As can be seen from the following table, the unit digits of the powers of

two are in a repetitious pattern of 2, 4, 8, 6, 2, 4, 8, 6…

All multiples of 10 have a unit digit of 0. However, as seen from the

pattern, none of the powers of 2 have their unit digit ending in a 0.

Therefore, no power of 2 is a multiple of 10.

Alexander from University College School noticed the following about the extension questions:

If the power n in $4^n + 5^n + 6^n$ is an odd number, then the last digit of this sum is 5; if the power n is even then the last digit of the sum is 7.

$3^n + 8^n$: For consecutive values of n (n = 1, 2, 3, …) the last digit of the sum goes in a pattern of 1, 3, 9, 7.

$2^n + 4^n + 6^n$: For consecutive values of n (n = 1, 2, 3, …) the last digit of the sum goes in a pattern of 2, 6, 8, 8.

$3^n +5^n +7^n$: For consecutive values of n (n = 1, 2, 3, …) the last digit of the sum goes in a pattern of 5, 3, 5, 7.

$3^n-2^n$: For consecutive values of n (n = 1, 2, 3, …) the last digit of the sum goes in a pattern of 1, 5, 9, 5.

Can you explain why this might be true?

Ryan also had some interesting ideas about how he could extend his results:

The unit digit of $1^n$ is 1, 1, 1, 1…

The unit digit of $2^n$ is 2, 4, 8, 6, 2, 4, 8, 6,...

The unit digit of $3^n$ is 3, 9, 7, 1, 4, 9, 7, 1,...

The unit digit of $4^n$ is 4, 6, 4, 6…

The unit digit of $5^n$ is 5, 5, 5, 5…

The unit digit of $6^n$ is 6, 6, 6, 6…

The unit digit of $7^n$ is 7, 9, 3, 1, 7, 9, 3, 1,...

The unit digit of $8^n$ is 8, 4, 2, 6, 8, 4, 2, 6,...

The unit digit of $9^n$ is 9, 1, 9, 1…

The unit digit of $10^n$ is 0, 0, 0, 0…

Thus the unit digits of the powers of any number repeat in a cycle four digits long, as all numbers end with one of the above 10 digits.Let the pattern of the unit digits of the powers of x be a, b, c, d... and let The pattern of the unit digits of the powers of another number y be e, f, g, h…

As a result, the sum/difference of the unit digits of the powers of x and y are in a repeating pattern of: a + e, b + f, c + g, and d + h, or a - e, b - f, c - g and d – h.

As a result, no matter what combination of powers, adding or subtracting, the unit digits are always in a pattern of 4 repeating numbers.

Well done everyone!