Complicated but easy problem solving?
The next palindrome is $16061$, not $16961$.
$7$ of $64$ cubes were removed, leaving $\frac{57.1539}{ 57} = 27$, so each small cube is $3×3×3\,\text{cm}$. The big cube was divided into $4×4×4$ cubes, so it was $12×12×12\text{ cm}$. The surface is six sides times $144\text{ cm}^2$, or $864\text{ cm}^2$.
(Q1)
this number between 0..50 and 100..150 is the same, so count the numbers with 3's in 0..50 twice and 51..99 once.
The first one is
- where 3 occurs as 1st digit (10 total)
- where 3 is a 2nd digit (3, 13, etc, 43) - 5 total
- we counted 33 twice, but it has $3$ twice also, so there is no overcount.
- subtotal $10+5 = 15$
The second one is just where the 3 occurs as a second digit with $\{5,6,7,8,9\}$ as a first one, so 5 total (53, 63, etc).
The final total then is $2 \cdot 15 + 5 = 35$.
(Q2) Hint the next one is 16061 not 16961
(Q3) Hint The surface area of the cube with side $a$ is $S=6a^2$ and volume is $V=a^3$. You have to figure out how removing these cubes impacts the surface area, compute $S$, the surface area of the original cube and use above info to infer $a$ and $V$.
(Q2) The next palindromic number is in fact 16061.
(Q3) In reality we want to find the volume of the $64$ smaller cubes. It says that the volume of $57$ smaller cubes is $1539\mathrm{cm}^{3}$, so we can find the volume of one small cube then multiply it by $64$ to find the volume $V$ of the original cube. Now we know the volume we can find the surface area. Recall that the volume of a cube with sides $a$ is $a^{3}$. So $a = (V)^{\frac{1}{3}}$. Then recall the surface area of a cube is $S = 6a^{2}$.