Exercise 3: Proving 7 | 5n and Applying the Sum Property — Possible Solution
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PART 1: PROVE THAT IF 7 | n, THEN 7 | 5n
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This is a direct application of this chapter's own "multiple"
property: if a | b, then a | (bc) for any integer c.

Here, a = 7, b = n, and c = 5. Since 7 | n is given, the property
says directly that 7 | (n * 5), which is the same thing as
7 | 5n. No further work is needed - this is exactly what the
property states, applied with c = 5.

PART 2: SHOW 7 | (21 + 5n) FOR n = 21 (SO 5n = 105) WITHOUT DIVIDING 126 DIRECTLY
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Given: 7 | 21 (since 21 = 7*3) and 7 | 5n where 5n = 105 (since
105 = 7*15).

This chapter's own "sum" property states: if a | b and a | c, then
a | (b + c).

Here, a = 7, b = 21, c = 105 (which is 5n for n=21). Since 7 | 21
and 7 | 105 are both already established, the sum property
guarantees directly that 7 | (21 + 105) - that is, 7 | 126 -
without ever needing to actually compute 126 / 7 or check its
remainder.

(As a separate confirmation, not part of the proof itself:
126 / 7 = 18 exactly, confirming the property's conclusion was
correct - but the proof itself never needed this check to be
valid.)

WHY THIS WORKS AS AN ANSWER
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Both parts apply this chapter's own named properties directly and
explicitly (naming which property, and which values play the role
of a, b, and c in each), rather than simply computing the
divisions and declaring the results true - the entire point of the
exercise is to show these properties let a conclusion be reached
WITHOUT direct computation, which is exactly what both parts
demonstrate.