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- Properties of The Number 137 - Mathematics Stack Exchange
There are some other Properties for $137$ in Wikipedia I found another Properties such that $137=2^7+2^3+1$ or the only way to write the number $137$ as a summation of two square numbers is $137=4^2+11^2$ thanks for your advice and suggestions Edit: After reading comments and answers, I want to suggest a definition for such numbers like $137$
- Find an integer $r$ with $0 ≤ r ≤ 10$ such that $7^ {137 }≡ r (\text . . .
Your working is fine You need to end by noting that $-5 \equiv 6 \pmod {11}$ since they asked for a residue between $0$ and $10$ An alternative approach would be: $7^ {137} \equiv (-4)^ {137} \equiv -2^ {274} \equiv - (2^ {5})^ {54} \cdot 2^4 \equiv - (32)^ {54} \cdot 16 \equiv - (-1)^ {54} \cdot 16 \equiv - 16 \equiv -5 \equiv 6 \pmod {11}$ I prefer this approach to reduce the base in
- How do you calculate the modulo of a high-raised number?
I need some help with this problem: $$439^{233} \\mod 713$$ I can't calculate $439^{223}$ since it's a very big number, there must be a way to do this Thanks
- calculus - Exponential Decay Question: The Half-Life of Cesium-137 is . . .
Exponential Decay Question: The Half-Life of Cesium-137 is 30 Years Suppose We Have a 100-mg Sample Ask Question Asked 10 years, 3 months ago Modified 10 years, 3 months ago
- Generate all possible combinations of 3 digits without repetition
It's possible to generate all possible combinations of 3 digits by counting up from 000 to 999, but this produces some combinations of digits that contain duplicates of the same digit (for example,
- Upper and lower bounds - Mathematics Stack Exchange
By halving 5 (the number you are rounding to) = 2 5 Then to find the upper bound you add it to the number you are rounding so 135 + 2 5 = 137 5 ( this is a multiple of 5)
- geometry - What are the holosnubs of the regular polyhedra . . .
Snubbing usually is considered as vertex alternation, i e alternate vertices are maintained, the other ones get replaced by the sectioning facets underneath (which in case of vertex alternations then are nothing but the former's vertex figures) Whenever the pre-image contained an odd numbered polygon then surrounding that polygon by alternation rule, you'll come back in wrong parity Thence
- Expectation of an absolute value - Mathematics Stack Exchange
For a standard normal distribution (mean zero, standard deviation 1), the difference between two samples is also normal with mean zero, standard deviation $\sqrt2$ To get the distribution of the absolute value, you fold the negative half onto the positive half and then double it to renormalize You can then compute the expected value in the usual way, which I assume you know, and it leads to
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