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Question

All the numbers from $19$ to $93$ are written consecutively to form the number $N =19202122...........919293$. Find the largest power of $3$ that divides $N$.

The following hint has been provided.

The square of any integer is either divisible by $4$ or leaves a remainder $1$ divided by $4$. Thus, an integer which leaves a remainder $2$ or $3$ when divided by $4$ can never be a square. If a prime p divides a square number the $p^2$ also divides that number.

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closed as off-topic by Namaste, Jyrki Lahtonen, Xander Henderson, José Carlos Santos, Adrian Keister Jul 29 '18 at 0:46

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    $\begingroup$ What is your attempt? $\endgroup$ – Mythomorphic Jul 28 '18 at 12:43
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    $\begingroup$ 1 st power of 3 only $\endgroup$ – maths student Jul 28 '18 at 13:00
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With a bit more rigour and without a calculator, we have:

$$N =\sum_{k=0}^{93-19 } (93-k)\cdot 10^{2k}$$

Now if we inspect this $\bmod 9$ we can use the fact that $10^j \equiv 1 \mod 9$ to find:

$$N \equiv\sum_{k=0}^{74} (93-k) \mod 9$$ $$N \equiv 75\cdot 93 - \sum_{k=0}^{74} k \mod 9$$ $$N \equiv 3\cdot 3 - \frac{1}{2}\cdot 74\cdot 75 \mod 9$$ $$N \equiv 9 - 37\cdot 3 \mod 9$$ $$N \equiv 9 - 1\cdot 3 \mod 9$$ $$N \equiv 6 \mod 9$$

Thus we know that $N$ is not divisible by $9$. But since $(N \bmod 9) \equiv N \mod 3$ we can re-use the above calculation to immediately find that $N \equiv 0 \mod 3$ and thus is divisible by $3$. That is the largest power of $3$ that divides $N$.

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First of all calculate sum of digits that sum of natural numbers from 19 to 93 which is 4200. It is divisible by 3 but not 9. For divisibility of higher power it has to be divisible by 9 but it is not divisible by 9. So higher power should be 1.

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  • $\begingroup$ Brilliant! Wonder how it could be done if the sum were divisible by $9$ $\endgroup$ – rsadhvika Jul 28 '18 at 13:26
  • $\begingroup$ @rsadhvika hint might help. $\endgroup$ – maths student Jul 28 '18 at 13:41

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