Evaluating $$6^2\div 2(3)+4$$  I understand how people are getting $10$ but I am getting $58$ because I am not distributing the $2$ to the $3$ inside the parentheses. Is that correct?
8308
There is a huge debate on the internet on the value of $48\div2(9+3)$. I believe the answer $2$ as I believe it is part of the bracket operation in BEDMAS.  yields the same answer. I also believe that if $48\div2\times(9+3)$ was asked it would be $288$ which Mathway agrees with as well. However,  says it is $288$ either way. A friend of mine (who is better at math) told me that there is no such thing as 'implicit multiplication', only shorthand so that is in fact done after the division (going left to right, not necessarily because division occurs before multiplication. But he didn't explicitly give a reason) What is the answer and why?
5854
Why is it that a number such as $108$, that is divisible by $2$ and $3$, is also divisible by $6$?   Is this true that all numbers divisible by two integers are divisible by their product?
116650
Does $3 - 4 + 1 = 0$ or $3 - 4 + 1 = -2$? Makes sense that $(3 - 4) + 1 = 0$ and $3 - (4 + 1) = -2$, but what if there are no parenthesis? Also, if I have $4$ apples and I add $1$ more apple, then I have $5$ apples, but if I eat $3$ apples, then I have $2$ left. Any ideas?
108015
The answer is 2. But I want to learn how to simplify this expression without the use of calculator.
82626
Suppose,I have a number 1256. I want (1256 % 11)! That is 2 . But, here, if i try in this way,  ( (125 % 11 ) * 10 + 6 ) % 11 = 2. That is exactly the same answer 2 as above.  I am confused , how This two process gives the similar answer ??? Even if , I try this method for any number n,m  to find (n mod m) . How this works ?? Can anybody explain ?
123191
I was working my way through some number theoretic proofs and being a newbie am stuck on this problem :     If $(x, 4) = 2$ and $(y, 4) =2$, then $(x + y, 4) = 4$,     where $(a,b)$ denotes the greatest common divisor of $a$ and $b$.   My Solution  (Incorrect)   $x-4 = 2t$ $y-4 = 2p$ $x + y - 8 = 2(t+p) \Rightarrow x + y - 4 = 2f + 4 = 2(f+2)\Rightarrow 2$ divides $(x+y , 4)$   My Question:  My solution is definitely inadequate. Can someone help me out?
31892
Assume $p$ is prime and $p\ge 3$.  Through experimentation, I can see that it's probably true. Using Wilson's theorem and Fermat's little theorem, it's equivalent to saying $2^2 4^2 6^2 \cdots (p-1)^2 = (-1)^{\frac{p+1}{2}}$ mod $p$, but I can't figure out any more than that.
68048
We have to prove that  "the product of two numbers ending in 6 also ends with 6" mathematically.  I have no clue how to start. I don't want you to prove it for me! but some hints would be very helpful as I'm totally stuck on how to start this prove.
119123
Find the value of   $$1\sin 2^{\circ} +2\sin 4^{\circ} + 3\sin 6^{\circ}+\cdots+ 90\sin180^{\circ}$$   My attempt  I converted the $\sin$ functions which have arguments greater than $90^\circ$ to $\cos$ but I have gone no where with it!  I also tried using double angle formula for the angles which are even.
87190
Does math have an incorrect notation / syntax? I don't mean writing misaligned notation (google), but when you take something like a number to powers to powers to powers, $${{2^2}^2}^3$$ (I was told this is incorrect notation by a teacher). Is it really incorrect, or does it just need to be simplified with parentheses? Do people write maths like this?  a radical expression with the root being a radical expression? $$\sqrt[\sqrt{2^3}]{2}$$
105411
I have checked the solution ($x=26$).  Solving modulo $5$ gives  $$1978^{20}\equiv 1978^{2\cdot 10}\equiv 1\pmod{5}$$  Solving modulo $25$ also gives $$1978^{20}\equiv 1\pmod{5}$$  How to evaluate the remainder $x$?
60872
I've tried solving it, but I'm not familiar with these kind of progressions and I couldn't get to a result. Also, I've found that the result is just $N/2$, but I don't get why would that be the answer, as it doesn't work for $N=1$ or $N=2$. Appreciate your help :)
49659
Can anyone help me out here? Can't seem to find the right rules of divisibility to show this:  If $a \mid m$ and $(a + 1) \mid m$, then $a(a + 1) \mid m$.
69468
I solved this problem by manually adding parentheses and counting them, and got correct answer of 32. Is there a simple to find the answer? Thanks.     The value of the expression $1÷2÷3÷5÷7÷11÷13$ can be altered by including parentheses. If we are allowed to place as many parentheses as we want, how many distinct values can be obtained for this expression?
46697
So I'm first asked to compute, mod 19, the powers of 2,   $$2^{2},2^{3},2^{6},2^{9}$$  which I compute as   $$4,8,7,18$$  respectively.    I'm then asked to prove that 2 generates $(\mathbb{Z}/19\mathbb{Z})^{*}$ based on the above.  I'm not seeing how you can only look at these powers to know that 2 generates the group.  Of course I could compute the rest of the powers of 2 and show that all $1\leq a\leq 18$ are congruent to $2^{k}$ for some integer $k$, but I get the impression that this is not what I'm supposed to be seeing here.  What I am noticing is that I've basically computed $2^{2}$ and $2^{3}$ and then a few combinations of them.  Trivially I can get $1$ as $2^{0}$ and 2 likewise, and I'm allowed to take negative integer powers so I must get $2^{-1}$.  If I somehow knew that $2^{-1}$ were not $4, 8, 7,$ or $18$ then that would be nice, but I don't see how I can be assured of that without explicit calculation--and I get the feeling this is supposed to be an exercise in not explicitly calculating these.  Maybe this is supposed to be "half calculation", like showing that because I can get $4\cdot 8=32$ and $4\cdot 7=28$ I must therefore be able to obtain ... I don't know what.  Any hints?  Or am I over-thinking this and I should just calculate every power of 2?
7294
I've always heard this reasoning, and it makes obvious sense, but how do you actually show it for some arbitrary product? Would it be something like this?  $$(a(b(cd)))e=((ab)(cd))e=(((ab)c)d)e=abcde?$$  Do you just say that the grouping of the parentheses now corresponds to just multiplying straight through? Thanks.
35604
I need to solve $3x^2 + 6x +1 \equiv 0 \pmod {19}$  I saw the same problem here -     but didn't understand how he got to the conclusion   $x(x+2) \equiv 6 \pmod {19}$  and anyway im trying to solve it how we learned in class -   multiply both sides and the modulo by $4a$ then solve the two equations  $y^2 \equiv b^2 -4ac \pmod {4an}$  $2ax + b \equiv y \pmod {4an}$  so I tried multiplying the hole thing by $4a$ , that is $4 \times 3$  and got to $(2 \times 3x + 6)^2 \equiv 36 -4 \times 3 \pmod {19 \times 4 \times 3}$  now I am stuck . any help will be appreciated