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I have to solve a couple of problems involving very large numbers, ie. those that are too big to use in a calculator. I clearly need a general approach for this sort of problem. Any pointers would be appreciated, ie. no need to solve the problems completely, just give me some direction. (If I get desperate I'll ask for actual solutions :)

The two problems are:

1) Is $2222^{5555} + 5555^{2222}$ prime?

and

2) Can you demonstrate that the digital representation of $7^{35}$ contains at least one digit that occurs four times?

Based on a hint from Chas, I have an idea for (2).

For a number to contain one digit four times, it must have at least 31 digits. That is, a 30-digit number may not meet the criterion as it could have each of the ten digits 3 times. But as soon as we have 31-digits, a fourth occurrence of one of the digits must be present.

Therefore we meet the criterion if:

\begin{align} 7^{35} &> 10^{30} \\ log(7^{35}) &> log(10^{30}) \\ 35 *log(7) &> 30 * log(10) \\ 35 * 0.845 &> 30 \\ 29.58 &> 30 \end{align}

The inequality is not true. Therefore I cannot prove the criterion can be met. Am I on the right track?

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For Question 1, you can easily prove it is not prime by considering that both 5555 and 2222 are divisible by 1111

Continuing your proof on question 2:

You have shown correctly that $7^{35}$ has exactly 30 digits and that it is not enough to say there is at least one digit that repeats 4 times. By pigeon hole, the only time when this happens is when all the 10 digits repeat 3 times. So in this case, when we take the sum of the digits, it is a multiple of 3, and hence the number is divisible by 3.This is not possible has $7^{35}$ is not a multiple of 3. Thereby, we have proved that this case is not possible

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Hint for the first question:

What is $2222^{5555} \mod 3$? What is $5555^{2222} \mod 3$?

Hint for the second question:

How many digits long is $7^{35}$?

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  • $\begingroup$ Chas, thanks for the tips. You've given me an idea for the second Q which I've added to my question. Does it look right? Still no idea about the first one. $\endgroup$ – dave Oct 10 '16 at 10:52

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