Answer: False. The number [tex]3^{40}+7^{40}[/tex] is not a multiple of 10
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Explanation:
Let's look at a fairly small table of powers of 3 and 7. Check out the diagram below.
Notice that the units digits for each power involve the items from this set: {1,3,7,9}.
Furthermore, note that in each blue row we have the units digits add to 3+7 = 10 or 7+3 = 10. In order to be a multiple of 10, we need to have 0 as the units digit. Eg: 90 is a multiple of 0 for this reason.
So [tex]3^1 + 7^1 = 3+7 = 10[/tex] is one multiple of 10
And so is [tex]3^3+7^3 = 27+343 = 370[/tex]
and so on. As the table shows, we have [tex]3^n+7^n[/tex] as some multiple of 10 as long as n is an odd positive integer. This contradicts the fact that 40 is an even integer for [tex]3^{40}+7^{40}[/tex]. So there's no way that this expression (whatever the massive number happens to be) is a multiple of 10. The units digit for this sum is not zero.
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You can use modular arithmetic as an alternative pathway to show that the given number isn't a multiple of 10
[tex]3^{40} \equiv 3^{2*20} \ (\text{ mod }10)\\\\3^{40} \equiv (3^2)^{20} \ (\text{ mod }10)\\\\3^{40} \equiv (9)^{20} \ (\text{ mod }10)\\\\3^{40} \equiv (-1)^{20} \ (\text{ mod }10)\\\\3^{40} \equiv 1 \ (\text{ mod }10)\\\\[/tex]
This shows that the units digit of [tex]3^{40}[/tex] is 1
Also,
[tex]7^{40} \equiv 7^{2*20} \ (\text{ mod }10)\\\\7^{40} \equiv (7^2)^{20} \ (\text{ mod }10)\\\\7^{40} \equiv (49)^{20} \ (\text{ mod }10)\\\\7^{40} \equiv (-1)^{20} \ (\text{ mod }10)\\\\7^{40} \equiv 1 \ (\text{ mod }10)\\\\[/tex]
which looks very similar to the steps of [tex]3^{40}[/tex]. This produces the same units digit. The two units digits add to 1+1 = 2, which is not zero and this is sufficient proof to show that [tex]3^{40}+7^{40}[/tex] is not a multiple of 10.
Going back to the table below, we see that the units digit of 1 show ups for both 3^n and 7^n whenever n is a multiple of 4. This applies to 40 because 40 = 4*10.
