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What fractions can you find between the square roots of 65 and 67? ### Root to Poly

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

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

# Power Countdown

##### Age 14 to 16Challenge Level

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.