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# Power Countdown

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Age 14 to 16

Challenge Level

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- Getting Started
- Student Solutions
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**Other ways of making 8**

Samuel from Verulam School in the UK: $\left(4^\frac12\right)^3$

Johnny from HKIS in Hong Kong: $\left(32^\frac15 \times 16^\frac14 \times \frac12\right)^3$

Tom from St Georges Church of England Academy: $\left(2^3\times16^\frac14\right)\div32^\frac15$

**Ways of making 125 from 2, 4, 5, 25, 27, 81**

Yuktha from Wallington High School for Girls in England: $25 \times 5 = 125$

Haneen from Ponteland High School in the UK and Yuktha both got this answer:

$5^{\left(81^\frac14\right)}=125$ because $81^\frac14 = 3$ and $5^3 = 125$

**Numbers you can make using 2, 5, 16, 243, 343, 512**

Niall from JAPS, Also Jai and Charlotte from Tytherington High School Macclesfield and Haren from Jebel Ali Primary School all got 1024 in the same way. Haren wrote:

16 to the power of 5 is 1,048,576

1,048,576 to the power of $\frac12$ is 1024

Equivalently, $16^\frac12=4$ and $4^5=1024$.

Jai and Charlotte made 64: $16^\frac12 = 4,$ $243^\frac15=3,$ $4^3 = 64$

Daniel, Alex, Nick and George from Tytherington High School also made 64:

$512^\frac13 =8$

$8^2$

Tom from Devonport Boys made 49:

$243 ^ \frac15 = 3$

$343 ^ \frac13 = 7$

$7 ^ 2= 49$

Daniel, Alex, Nick and George said it is impossible to make 89:

89 cannot get exactly as it's prime 81 is as close as we got.

Tom went into more detail.:

89 is impossible to achieve exactly:

$2=2^1$

$5=5^1$

$16=2^4$

$243=3^5$

$343=7^3$

$512=2^9$

- all the numbers we are using can be expressed as a single prime to a power.
- raising these numbers to any (pos. integer) power will only increase the value to which the prime has been raised.
- roots of primes are irrational so rooting these numbers beyond the powers which are already there will only give irrational results: no use to us.
- 89 is prime but we don't have any numbers which we can use that have 89 as a factor and because we cannot multiply or divide there is no way of changing these factors to achieve a multiple of 89.
- I believe a similar argument can be applied for 216 as 216=6$^3$ and we don't have [enough] usable numbers with 6 (or specifically 2 and 3) as a factor.

Find the polynomial p(x) with integer coefficients such that one solution of the equation p(x)=0 is $1+\sqrt 2+\sqrt 3$.

The squares of any 8 consecutive numbers can be arranged into two sets of four numbers with the same sum. True of false?