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For the math geniuses...

For the math geniuses, if your into this sort of thing, help me do these problems:
http://bwsvi.com/math.jpg

This is a sample sheet from a previous contest, im taking it soon so i was trying to do these...if you could do you think you can post the steps and the answer on how you solved it? Thanks 🙂
 
Originally posted by: mastaki11a
#2 is 10
n/7=5
(n/7)7=5(7)
n=35

so
2n/7=x
2(35)/7=x
70/7=x
10=x

the question is what is the REMAINDER...
take a number n, say 12... divide it by 7, you get a remainder of 5
now 2n%7 = 24%7 = 3;

try with n = 19... 19%7 = 5
2n%7 = 38%7 = 3... etc


*edit* replaced the erroneous division operators with the correct remainder ones
 
Originally posted by: mastaki11a
#2 is 10
n/7=5
(n/7)7=5(7)
n=35

so
2n/7=x
2(35)/7=x
70/7=x
10=x

Um, no, but you're close... you can't have a remainder greater than what you're dividing by. let's assume that n = 12. Then 12/7 = 1 remainder 5. 2n = 24, so then 24/7 = 3 remainder 3.

so #2 is 3, not 10.


Another way that seems to work is that if you have a number and get a remainder of 5 when you divide by 7, twice that number should give you twice the remainder. So that would give you a remainder of 10. Since 10 is greater than 7, the number being divided by, the answer would be 10mod7, or 3.

[edit] Damn, I gotta type faster[/edit]

JW
 
#2 should be 3, since n/7=x+5/7, 2n/7=2x+5*2/7=2x+1+3/7
#4 -1, multiply out, a(b+1)+b(a+1)=ab, cancel out ab, ab+a+b=0, a+b=-ab, 1/a+1/b=a+b/b*a =-1
 
#1: a=1 b=2
#2: 3
#3: Is this some sort of trick question? 12 apprentices ...
#4: -1
#5: (118,80)
#6: 9989898989


Sorry for not explaining, but I did these on the run, and consequently, I can't confirm these results.

 
#3 seems a little too easy:

A= Time of Amature, P= Time of Professional, X= Time of Apprentice
P=.5A
X=2P=2(.5A)=A
Thus: X=A

So it takes 12 for both Apprentices and Amatures.
 
For #6 I got 99999785960 but I'm not sure it's right.

There are 9 single digit numbers, 1----9 and 51 two digit two numbers for a total of a 111 digit number. That means we need an 11 digit number to find. Preserving the 9 from 9, 19 29 39 and 49 that leaves 50,51,52,53,54,55,56,57,58,59,60 to find the biggest 6didget number and the biggest one I can see is 7,8, 5, 9, 6, 0.
 
Originally posted by: Moonbeam
For #6 I got 99999785960 but I'm not sure it's right.

There are 9 single digit numbers, 1----9 and 51 two digit two numbers for a total of a 111 digit number. That means we need an 11 digit number to find. Preserving the 9 from 9, 19 29 39 and 49 that leaves 50,51,52,53,54,55,56,57,58,59,60 to find the biggest 6didget number and the biggest one I can see is 7,8, 5, 9, 6, 0.

hmm that looks very rite to me, i can follow the logic...hmm...anyone confirm?
 
chiwawa626

#4 - multiply both sides by (a+1)(b+1)

u get [ (b+1)/b ] + [ (a+1)/a ] = 1

separate - [b/b] + [1/b] + [a/a] +[1/a] = 1

so you get - [1/a] + [1/b] = 1 - 1 - 1 = -1
 
Originally posted by: Krakerjak
chiwawa626

#4 - multiply both sides by (a+1)(b+1)

u get [ (b+1)/b ] + [ (a+1)/a ] = 1

separate - [b/b] + [1/b] + [a/a] +[1/a] = 1

so you get - [1/a] + [1/b] = 1 - 1 - 1 = -1


Yay i get it! thanks dude...

Anyone explain #5?

 
#5.

Look at the triangles formed between the squares and the axes. The smallest side of the triangle is of length 19 (you can tell by the x coordinate of the given point). From the y coordinate, you can see that 99 = length of smallest leg + length of other leg. From this you can determine that the length of the other leg is 80.

The x coordinate of the point in question is a distance of the small leg + the large leg away from the origin in the x direction or 80 + 19 = 99.

The y coordinate of the point is a distance of a large leg away from the origin in the y direction or 80. Thus:

(99, 80) is the correct answer.
 
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