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3 years ago

3⁴⁰+ 7⁴⁰is a multiple of 10

Mathematics

2 answers:

Galina-37 [17]3 years ago

7 0

Answer: False. The number $3^{40}+7^{40}$ 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 $3^1 + 7^1 = 3+7 = 10$ 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 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.

Margarita [4]3 years ago

6 0

ANSWER :

It will be false , it is not the multiple of 10

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A fair coin is to be tossed 20 times. Find the probability that 10 of the tosses will fall heads and 10 will fall tails, (a) usi

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Using the distributions, it is found that there is a:

a) 0.1762 = 17.62% probability that 10 of the tosses will fall heads and 10 will fall tails.

b) 0% probability that 10 of the tosses will fall heads and 10 will fall tails.

c) 0.1742 = 17.42% probability that 10 of the tosses will fall heads and 10 will fall tails.

Item a:

Binomial probability distribution

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.
n is the number of trials.
p is the probability of a success on a single trial.

In this problem:

20 tosses, hence $n = 20$ .
Fair coin, hence $p = 0.5$ .

The probability is P(X = 10), thus:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 10) = C_{20,10}.(0.5)^{10}.(0.5)^{10} = 0.1762$

0.1762 = 17.62% probability that 10 of the tosses will fall heads and 10 will fall tails.

Item b:

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

It measures how many standard deviations the measure is from the mean.
After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
The binomial distribution is the probability of x successes on n trials, with p probability of a success on each trial. It can be approximated to the normal distribution with $\mu = np, \sigma = \sqrt{np(1-p)}$ .

The probability of an exact value is 0, hence 0% probability that 10 of the tosses will fall heads and 10 will fall tails.

Item c:

For the approximation, the mean and the standard deviation are:

$\mu = np = 20(0.5) = 10$

$\sigma = \sqrt{np(1 - p)} = \sqrt{20(0.5)(0.5)} = \sqrt{5}$

Using continuity correction, this probability is $P(10 - 0.5 \leq X \leq 10 + 0.5) = P(9.5 \leq X \leq 10.5)$ , which is the p-value of Z when X = 10.5 subtracted by the p-value of Z when X = 9.5.