NUMBER

$$2^3\times2^4 = 2^{3+4} = 2^7 = 128$$

$$5^{6-2} = 5^4 = 625$$

$$3^{2\times3} = 3^6 = 729$$

$$2^n$$

$$2^8 = 256$$

$2^{13}=8192$ and $2^{14}=16384$, so it takes until hour 14 to exceed 10000 — that's $14-8=6$ more hours.

$2,4,8,6,2,4,8,6$ (for powers 1 through 8).

The last digits repeat in a cycle of 4: $2,4,8,6$, then repeat.

$23 \div 4 = 5$ remainder $3$, so $2^{23}$ has the same last digit as $2^3$, which is $8$.

$2^3\times3^2 = 8\times9=72$. $6^5=7776$. These are not equal, so the student's simplification is incorrect.

The index law $a^m\times a^n=a^{m+n}$ only applies when the BASE is the same on both terms. Here the bases (2 and 3) are different, so the exponents cannot simply be combined — each power must be evaluated separately before multiplying.

Amount $=500\times2^t$. After 4 periods: $500\times2^4=500\times16=8000$.

$500\times2^t \ge 50000 \Rightarrow 2^t\ge100$. Since $2^6=64$ and $2^7=128$, $t=7$ periods are needed, i.e. $7\times3=21$ years.

$P(h)=3\times2^h$. After 10 hours: $3\times2^{10}=3\times1024=3072$.

Original: $3\times2^h$. Rival: $1\times3^h=3^h$. Testing values: $h=3$: original$=24$, rival$=27$ (rival exceeds!). $h=2$: original$=12$, rival$=9$ (rival still behind). So the rival colony first exceeds the original after 3 whole hours.

$7^1=7$(7), $7^2=49$(9), $7^3=343$(3), $7^4=2401$(1), $7^5=16807$(7), $7^6=\ldots$(9), $7^7=\ldots$(3), $7^8=\ldots$(1). Units digits: $7,9,3,1,7,9,3,1$.

The cycle length is 4 ($7,9,3,1$ repeating). $35\div4=8$ remainder $3$, so $7^{35}$ has the same units digit as $7^3$, which is $\boxed{3}$.

A number is a multiple of 10 only if its units digit is 0. Since $7^{35}$ has units digit 3 (never 0, as the full 4-digit cycle $7,9,3,1$ never includes 0), $7^{35}$ can never be a multiple of 10.

$2^{10}$ means $2$ multiplied by itself $10$ times: $\underbrace{2\times2\times2\times\cdots\times2}_{10\text{ factors}} = 1024$, which is very different from $2\times10=20$.

The two values differ by a factor of over 50 ($1024$ vs $20$) — in a real context like calculating compound growth, data storage (e.g. computer memory, where $2^{10}$ 'kilobytes' is a standard unit), or population growth, this confusion could lead to a drastically wrong final answer, potentially by orders of magnitude.

Generation 5 exponent $=8+4=12$. Capacity $=2^{12}=4096$ MB.

Need $2^{8+n}\ge1000000$. Testing: $2^{20}=1048576\ge1000000$ ? (generation where exponent$=20$, i.e. Generation $20-8+1=13$). This 'doubling' pattern (similar to Moore's Law in computing) has historically held reasonably well over shorter timeframes, but is not realistic to extrapolate indefinitely — physical and economic limits eventually slow or halt this kind of exponential growth in real technology.

$P(h)=3\times2^h$. After 10 hours: $3\times2^{10}=3\times1024=3072$.

Original: $3\times2^h$. Rival: $1\times3^h=3^h$. Testing values: $h=3$: original$=24$, rival$=27$ (rival exceeds!). $h=2$: original$=12$, rival$=9$ (rival still behind). So the rival colony first exceeds the original after 3 whole hours.

$7^1=7$(7), $7^2=49$(9), $7^3=343$(3), $7^4=2401$(1), $7^5=16807$(7), $7^6=\ldots$(9), $7^7=\ldots$(3), $7^8=\ldots$(1). Units digits: $7,9,3,1,7,9,3,1$.

The cycle length is 4 ($7,9,3,1$ repeating). $35\div4=8$ remainder $3$, so $7^{35}$ has the same units digit as $7^3$, which is $\boxed{3}$.

A number is a multiple of 10 only if its units digit is 0. Since $7^{35}$ has units digit 3 (never 0, as the full 4-digit cycle $7,9,3,1$ never includes 0), $7^{35}$ can never be a multiple of 10.

$2^{10}$ means $2$ multiplied by itself $10$ times: $\underbrace{2\times2\times2\times\cdots\times2}_{10\text{ factors}} = 1024$, which is very different from $2\times10=20$.

The two values differ by a factor of over 50 ($1024$ vs $20$) — in a real context like calculating compound growth, data storage (e.g. computer memory, where $2^{10}$ 'kilobytes' is a standard unit), or population growth, this confusion could lead to a drastically wrong final answer, potentially by orders of magnitude.

Generation 5 exponent $=8+4=12$. Capacity $=2^{12}=4096$ MB.

Need $2^{8+n}\ge1000000$. Testing: $2^{20}=1048576\ge1000000$ ? (generation where exponent$=20$, i.e. Generation $20-8+1=13$). This 'doubling' pattern (similar to Moore's Law in computing) has historically held reasonably well over shorter timeframes, but is not realistic to extrapolate indefinitely — physical and economic limits eventually slow or halt this kind of exponential growth in real technology.