Find the smallest power of 10 that will exceed M. Let M = 118,526.65902.

Mathematics

1 answer:

zavuch27 [327]3 years ago

5 0

Answer:

The smallest power of 10 that will exceed $M$ is $10^{6}$ .

Step-by-step explanation:

We can use the following approach to determine the smallest power of 10 that will exceed M. We can transform that number into scientific notation, which is of the form:

$x.y \times 10^{n}$ , $\forall \,x,y\in \mathbb{N}$

Where:

$x$ - Integer part, formed by a digit, which is of the highest order.

$y$ - Decimal part, formed by a digit onwards.

$n$ - Power grade.

The smallest power of 10 that will exceed M is $10^{n+1}$

If $M = 118,526.65902$ , then, the power grade is number of spaces that dot must be moved leftwards. In this case, dot must be moved 5 spaces on the left. The integer part is 1 and the decimal part is 1852665902. Then, the value of $M$ in scientific notation is:

$M = 1.1852665902\times 10^{5}$

Then, the smallest power of 10 that will exceed $M$ is $10^{6}$ .

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Feliz [49]

Recall Euler's theorem: if $\gcd(a,n) = 1$ , then

$a^{\phi(n)} \equiv 1 \pmod n$

where $\phi$ is Euler's totient function.

We have $\gcd(9,32) = 1$ - in fact, $\gcd(9,32^k)=1$ for any $k\in\Bbb N$ since $9=3^2$ and $32=2^5$ share no common divisors - as well as $\phi(9) = 6$ .

Now,

$37^{32} = (1 + 36)^{32} \\\\ ~~~~~~~~ = 1 + 36c_1 + 36^2c_2 + 36^3c_3+\cdots+36^{32}c_{32} \\\\ ~~~~~~~~ = 1 + 6 \left(6c_1 + 6^3c_2 + 6^5c_3 + \cdots + 6^{63}c_{32}\right) \\\\ \implies 32^{37^{32}} = 32^{1 + 6(\cdots)} = 32\cdot\left(32^{(\cdots)}\right)^6$