Get each of the numbers in the form 10ⁿ.

This is easy for (A), (C), (D), (F), where the powers of 10 are respectively 8 × 10⁸, 100, 10¹⁰⁰, 10¹²³.
So C < A < D < F.

For the others, make use of 2¹⁰ ≈ 10³.

(G) is 2⁶⁵⁵³⁶ ≈ 10^{(0.3 × 65536)} ≈ 10²⁰⁰⁰⁰ and so lies between (C) and (A).

(B) is 272 × 2²⁵⁶ ≈ 272 × 10⁷⁸ ≈ 10⁸⁰ and so is less than (C).

(H) is 2 to the power of about 10²⁰⁰⁰⁰. A rather surprising fact is that because this huge number is 10 to the power of about 0.3 × 10²⁰⁰⁰⁰, we may as well say this is ≈ 10²⁰⁰⁰⁰. It is much greater than (F).

What is (E)? The largest number you can make out of 4 4's comes from putting them in a tower of powers, i.e. 4 to the power of (4 to the power of 4⁴). This is 4 to the power of 4²⁵⁶, which is 4 to the power of 2⁵¹² ≈ 4 to the power of 10¹⁵⁶. Again, this is ≈ 10 to the power of 10¹⁵⁶. So (E) is bigger than (F), but not as big as (H).

The final order is B, C, G, A, D, F, E, H.

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Notice first that 6 and 28 are 110 and 11100 in binary, so fall into this pattern, with 3 (11 in binary) and 7 (111 in binary) as Mersenne primes. Use the number 28 as a model for the general case. See the pattern in the binary. Because 111 is prime, the factors either have 111 in them or not. The factors without a 111 are (1, 10, 100), and these add up to 111. The factors with a 111 are: (111, 1110) and now 111+111+1110=1110+1110=11100 as requred for perfection.

It is known that any even perfect number must be of this form. It is not known whether there exist any odd perfect numbers.

The second question is also easily seen from the binary pattern. The question in Chapter 1 asked for 7 × 9 in binary, giving 111111 representing 63. We can look at this in reverse: because 111111 has six 1's, and six is not a prime, it can be split up as 11-11-11 showing that 11 is a factor, or as 111-111 showing that 111 is a factor. That is, 111111 = 11 × 10101 = 111 × 1001. The same argument works for a string of n 1's whenever n is not prime.

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The point is that 65535 = 257 × 255 = 257 × 17 × 15 = 257 × 17 × 5 × 3. So the product 65535 × 65536 × 65537 involves each of the Fermat primes just once, plus 16 2's.

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The first question just needs a count. Remember that the sequence is a cycle, so the last 0 counts as being followed by 0001... taken from the beginning.

The second question can also be done just by trying them out. But it is more useful to see a pattern in the results. We have already found that modulo 17, the powers of 10 are:
10¹=10, 10²=15, 10³=14, 10⁴=4, 10⁵=6, 10⁶=9, 10⁷=5, 10⁸=16,
10⁹=7, 10¹⁰=2, 10¹¹=3, 10¹²=13, 10¹³=11, 10¹⁴=8, 10¹⁵=12, 10¹⁶=1.

So looking at the powers of 15, modulo 17, is the same as looking at the powers of 10², and clearly these will only give the even powers of 10: 10⁴, 10⁶, 10⁸... 10¹⁶.

The last question is just another way of expressing this observation. The numbers 4, 9, 16, 2, 13, 8 are all squares, modulo 17, and so lie on the diagonal of the base-17 multiplication table.