Tilted Square Numbers!

We note that the tilted square of dots is equal to two triangle less a diagonal of dots. The number of dots on the diagonal $=2n-1$ .

The number of dots on the $n^{th}$ term is $D(n) = 2\left(\dfrac{n(1+2n-1)}{2}\right) - (2n-1) = 2n^2 - 2n+1$

Therefore,

$\begin{aligned} D(10^{10^3}) & = D(10^{1000}) \\ & = 2\times 10^{2000}-2\times 10^{1000} + 1 \\ & = 2\times 10^{2000} - 10^{1001} + 10^{1001} - 2\times 10^{1000} +1 \\ & = (2\times 10^{2000} - 10^{1001}) + 8\times 10^{1000} +1 \\ & = (2\times 10^{999}-1)10^{1001} + 8 \times 10^{1000} + 1 \\ & = \text{1 followed by 999} \times \text{9 followed by 8} \\ & \quad \quad \text{followed by 999} \times \text{0 followed by 1}\end{aligned}$

Therefore, the sum of digits of $D(10^{1000})$ is:

$N = 1 + 999\times 9 + 8 + 0 + 1 = \boxed{9001}$

No. of orange dots= ${ \left( n-1 \right) }^{ 2 }$ No. of green dots= ${ n }^{ 2 }$

Total no. of dots= ${ \left( n-1 \right) }^{ 2 }+{ n }^{ 2 }$

Putting $n={ 10 }^{ { 10 }^{ 3 } }$

We can observe a pattern while finding the sum of digits of numbers in the form ${ \left( { 10 }^{ a } \right) }^{ 2 }+{ \left( { 10 }^{ a }-1 \right) }^{ 2 }$

${ \left( { 10 }^{ 1 } \right) }^{ 2 }+{ \left( { 10 }^{ 1 }-1 \right) }^{ 2 }=181$ , sum of digits=10.

${ \left( { 10 }^{ 2 } \right) }^{ 2 }+{ \left( { 10 }^{ 2 }-1 \right) }^{ 2 }=19801$ , sum of digits=19

${ \left( { 10 }^{ 3 } \right) }^{ 2 }+{ \left( { 10 }^{ 3 }-1 \right) }^{ 2 }=1998001$ , sum of digits=28

${ \left( { 10 }^{ 4 } \right) }^{ 2 }+{ \left( { 10 }^{ 4 }-1 \right) }^{ 2 }=199980001$ , sum of digits=37

So this forms an AP with initial term $1+9$ and common difference of $9$ .

So we can observe that for numbers in the form ${ \left( { 10 }^{ a } \right) }^{ 2 }+{ \left( { 10 }^{ a }-1 \right) }^{ 2 }$ , sum of digits is $1+9a$

So for $n={{10}^{10}}^{3}$ , the sum of digits is $1+9\times {10}^{3}$ which is equal to $\boxed{9001}$

The difference between consecutive terms are in AP. The general term can be written as $T_n = 1 + 4(1+2+...+(n-1)) = 1 + 2n(n-1)$
For $n=10^r, T_n = 1+2\times 10^r \times (10^r -1)$ .

$10^r - 1$ has $r$ consecutive $9$ 's in it. Multiplying it by two will give us a number of the form $\overline { 1999...9998 }$ with $r-1$ consecutive $9$ 's in it. Multiplying it by $10^r$ will add $r$ zeroes to the number and the sum of digits will be $9r$ Finally, we add the $1$ and the sum of digits will be $9r+1 = 9\times 1000 + 1 = \boxed{9001}$ .

First we should say that ${10^{10}}^{3} = 10^{1000}$ .

So the change in the number of dots increases by $4$ each term, so dividing that by $2$ will gives us $a = 2$ . Next, since we know the rate of increase of change is $4$ , when $x=0$ , $y=1$ . So we have $y=2x^{2} + bx + 1$ . Replacing the $x$ and $y$ values... we find that $b = -2$ . We now have our quadratic equation $y=2x^{2} - 2x + 1$ .

When we have $x=10^{1000}$ , the sum of digits of $2x^{2}$ is $2 + 0\times2000$ . Now when we subtract $2x$ , we know that there are $1000$ digits, in $2\times10^{1000}$ , and the first (from the right) $999$ are zeroes. However, the first non-zero digit is the thousandth digit, and it is $8$ (since we are subtracting $2$ from $10$ ). Since that would leave a $1$ , all the other digits (excepting the first from the left) would be $9$ ( $9 + 1 = 10$ ). The first digit must be less than $2$ , but greater than $0$ , so it is $1$ . So now we have the sum of digits is $1 + 9\times999 + 8$ . Finally, we have $+1$ , so our sum of digits is now

$1 + 9\times999 + 9 = 1 + 9\times1000 = \boxed{9001}$ .