1) For 2.7, remember the proof method is a proof by contradiction. So we're assuming we CAN write 1 as the sum of the reciprocals of k odd integers. Hence, all the n1, n2, n3, ..., nk are odd. Then think about what happens when we multiple odd integers together for the rest.

Note: If it helps, try to first do the problem for a sample value of k, such as k=4 (it must be even). The same type of argument then can work for larger even values of k.

2) First off, some of these questions can be pretty challenging, so don't get discouraged! It's good that you're thinking about the problems and trying them. Especially problems that are proofs can take some time to get used to. Here's some quick hints for the problems

• 2.24: Remember that if something is a multiple of 7 we can add or subtract it from a number and it doesn't change whether it is a multiple of 7. For example, 1473 is divisible by 7 if and only if 3 is divisible by 7 because we know 1470 is a multiple of 7. Hence, can we write 10a + b = (some multiple of 7) + (3a + b) for the divisibility rule.
• 2.22: Similar to 2.24, can you write the $\overline{abcd}$ = (some multiple of 11) + (d - c + b - a). Hint: What is the nearest multiple of 11 to 1000? to 100? to 10? The multiple could be above or below.
• 2.27: If n is odd and n=a^2 + b^2, what can you say about the parity of a and b?
• 2.28: This one is a little tricky to prove, so it's more important to make sure you understand the statements. That being said, for part (a), remember that repeated addition can be thought of as multiplication, and in a similar way, repeated subtraction can be thought of as division. For part (b), the proof will be similar to Example 2.8. Let d = gcd(b,a) and e = gcd(a, r). Try to prove that d <= e and e <= d, so in fact d = e.
• 2.29: I think you had another discussion about this already.
• 2.30: Note 44 = 4 * 11, so the number must be divisible by 4 and 11. Recall we do have divisibility rules for 4 and 11 we can try to take advantage of.

Hope this helps!