Find any two consecutive $100$-digit numbers whose sum of digits is a perfect square

Question

Find any two consecutive $100$-digit numbers whose sum of digits is a perfect square.

this is all I did and I am stuck

lets say $abc \dots m$ is a $100$ digit number whose sum of its digits is a perfect square

$$a+b+c+.....+m = k^2$$

lets say $abc \dots m+1$ is a $100$ digit number whose sum of its digits is a perfect square

$$a+b+c+.....+m+1 = q^2$$

$$k^2 = q^2 - 1$$

and I am just stuck

Answer 1

Let the lower number end in $n$ $9$s. Then, the digit sum of that number is $9n-1$ more than the digit sum of the number above, so $9n-1$ is a difference of two squares.

If $n=2$ then $9\times2-1=17=9^2-8^2$.

A potential lower number could be: $$\underbrace{11\ldots11}_{63\,1\mathrm{s}}\underbrace{00\ldots00}_{35\,0\mathrm{s}}99$$ so the higher number is $$\underbrace{11\ldots11}_{63\,1\mathrm{s}}\underbrace{00\ldots00}_{34\,0\mathrm{s}}100$$

Answer 2

Let $Q(n)$ (from German Quersumme) denote the sum of the (base-ten) digits in the integer $n$.

It is a well-known fact that $Q(n)$ is congruent modulo 9 to $n$.

Also, if you look down the diagonal of the base-9 multiplication table, you'll see that all the perfect squares end in 0, 1, 4, 7. Put another way, if $n$ is a perfect square, then:

$$Q(n) \equiv n \in \{0, 1, 4, 7\} \pmod{9}$$

For both $Q(n)$ and $Q(n+1)$ to be perfect squares, the only possible combination is $Q(n) \equiv 0 \pmod{9}$, so that $Q(n+1) \equiv 1 \pmod{9}$.

Now, let's choose two perfect squares, one of which is a multiple of 9, and one that's one more than a multiple of 9. I'll pick $Q(n) = 81$ and $Q(n + 1) = 64$.

Note that in order for $Q$ to decrease when $n$ is incremented, we need a rollover from 9 to 0 in the unit's place. So, we have:

• $n$ = (99 digits adding to 72)$9$
• $n + 1$ = (99 digits adding to 64)$0$

Since $72 > 63$, we'll need another place-value column affected by rollover.

• $n$ = (98 digits adding to 63)$99$
• $n + 1$ = (98 digits adding to 64)$00$

OK, so now we just need two consecutive digits in the hundred's place. I'll pick $8$ and $9$.

• $n$ = (97 digits adding to 55)$899$
• $n + 1$ = (97 digits adding to 55)$900$

And the “97 digits adding to 55” must be the same string of digits for the two numbers. I'll choose to construct it as 1 + 90 0's + 6 9's. giving the pair of numbers:

• 1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000999999899
• 1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000999999900

So there you go: Two consecutive 100-digit integers whose digit sums (81 and 64) are both perfect squares.

Answer 3

Use $n=4040404...$

The sum of the digits would $4\times 50 = 200$

Since $15$ squared is $225$, you'll need to add $25$

Divide $n$ by two $=2020202...$ and make two copies, then change $12$ of the $0$s to $1$s (same digits for each copy) and add an extra $1$ to one of the numbers.