Is it bigger than a googol?

Number Theory Level 4

$\large 192021222\ldots 919293$

The number above is formed by concantenating the positive integers from 19 to 93 inclusive.

If $a$ is the largest integer that such that $3^a$ divides the number described above, find the value of $a$ .

The answer is 1.

1 solution

Trung Đặng Đoàn Đức
Jul 3, 2015

WARNING : This solution is a bit weird , viewer discretion is advised.

First, notice that $19202122...919293 = \sum _{ n=19 }^{ 93 }{ n\ast { 100 }^{ 93-n } }$ .

Second, also notice that ${ 100 }^{ x }\equiv 1(mod\quad 9),\forall x\in N$ . Thus, $n\ast { 100 }^{ 93-n }\equiv n\ast 1\equiv n\quad (mod\quad 9)\quad (19\le n\le 93)$ .

Combining both two, $19202122...919293 \equiv \sum _{ n=19 }^{ 93 }{ n\ast { 100 }^{ 93-n } } \equiv \sum _{ n=19 }^{ 93 }{ n } \equiv 4200\equiv 6\quad (mod\quad 9)$ .

So, $19202122...919293$ is not divisible by 9, and using the same method, $19202122...919293$ is divisible by 3. Hence, the answer is $a=1$

Actually, we have $3^a=3$ i.e., $a=1$ . I believe you have a typo in your last sentence.

EDIT: I see that the typo has been fixed. Thanks.

By the way, there's actually a much simpler way using the divisibility rules of $3$ and $9$ . Denote the given value by $x$ and its digit sum as $\sigma(x)$ . We have,

$\begin{aligned}\sigma(x)&=(1+9)+(2+0)+\ldots +(9+2)+(9+3)\\&=(1+9)+\left[\sum_{k=2}^8\sum_{j=0}^9 (k+j)\right]+(9+0)+(9+1)+(9+2)+(9+3)\\&=52+\sum_{k=2}^8\sum_{j=0}^9 (k+j)=52+\sum_{k=2}^8(10k+45)=717\end{aligned}$

The last few equalities follows by elementary summation identities and sum of first $n$ positive integers formula.

All you have to do now is to simply show that $3\mid 717$ but $9\not\mid 717$ to conclude your answer by virtue of divisibility rules of $3$ and $9$ . This is trivial and is left as an exercise to the interested reader.

Prasun Biswas - 5 years, 11 months ago

Thank you for your solution. It's actually pretty nice and also neater than my solution :)

Trung Đặng Đoàn Đức - 5 years, 11 months ago

Your solution is pretty neat too. Upvoted! :)

Prasun Biswas - 5 years, 11 months ago

Is it bigger than a googol?

The answer is 1.

1 solution

0 pending reports