3⁴⁰+ 7⁴⁰is a multiple of 10

Mathematics

2 answers:

Galina-37 [17]2 years ago

7 0

Answer: False. The number $3^{40}+7^{40}$ is <u>not</u> 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 $3^1 + 7^1 = 3+7 = 10$ is one multiple of 10

And so is $3^3+7^3 = 27+343 = 370$

and so on. As the table shows, we have $3^n+7^n$ 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 $3^{40}+7^{40}$ . 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

$3^{40} \equiv 3^{2*20} \ (\text{ mod }10)\\\\3^{40} \equiv (3^2)^{20} \ (\text{ mod }10)\\\\3^{40} \equiv (9)^{20} \ (\text{ mod }10)\\\\3^{40} \equiv (-1)^{20} \ (\text{ mod }10)\\\\3^{40} \equiv 1 \ (\text{ mod }10)\\\\$

This shows that the units digit of $3^{40}$ is 1

Also,

$7^{40} \equiv 7^{2*20} \ (\text{ mod }10)\\\\7^{40} \equiv (7^2)^{20} \ (\text{ mod }10)\\\\7^{40} \equiv (49)^{20} \ (\text{ mod }10)\\\\7^{40} \equiv (-1)^{20} \ (\text{ mod }10)\\\\7^{40} \equiv 1 \ (\text{ mod }10)\\\\$

which looks very similar to the steps of $3^{40}$ . 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 $3^{40}+7^{40}$ is <u>not</u> 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.