Question

Let n(A) = m and n(B)=n . if A has 15360 more subset than B, then (5m+3n) : (5m-3n) is as?

Answer 1

@dpaInc @FoolForMath @apoorvk @eigenschmeigen @experimentX @mathmagician

Answer 2

Looks like i forgot the basics ..
is {1, 2} same as {2, 1}??

Answer 3

@experimentX yea think so

Answer 4

subsets*

Answer 5

n(A) = m , n(B) = n

number of subsets of set S where n(S) = s is 2^s i think

Answer 6

yes the NO of substet = 2^n @eigenschmeigen u r correct go on

Answer 7

oh .. kinds looks like
\[ \sum_{i=1}^{n} \binom{n}{i} - \sum_{i=1}^{m} \binom{m}{i} = 15360\]

Answer 8

2^m = 15360 + 2^n

Answer 9

n, m are integers

Answer 10

ok

Answer 11

m = 14 , n = 10 works

Answer 12

will there be other solutions?

Answer 13

@eigenschmeigen we need to find ratio

Answer 14

yes but if we solve then the ratio is trivial :D

Answer 15

5:2

Answer 16

but how u solved this plz show 2^m = 15360 + 2^n

Answer 17

are there any other solutions for n and m though is what i'm interested in

Answer 18

i kinda just saw that 15360 + 1024 = 16384

Answer 19

i'm trying to algebraically solve it right now

Answer 20

no u r correct....

Answer 21

but there may be other solutions as far as we know

Answer 22

no matter but how u solved this plz show 2^m = 15360 + 2^n

Answer 23

any way plz show

Answer 24

that's what i'm saying, i didnt really i just looked at it and figured it out. now i have a solution though

Answer 25

ok then show it let us check that!

Answer 26

notice that prime factorisation of 15360:
\[15360 = 2^{10} \times 3 \times 5\]


\[2^m = 15360 + 2^n\]\[2^m - 2^n = 15360 \]\[2^n(2^{m-n} -1) = 15360 \]\[2^n(2^{m-n} -1) = 2^{10} \times 3 \times 5 \]
\[2^n(2^{m-n} -1) = 15360 = 2^{10} \times 15\]\[2^n(2^{m-n} -1) = 2^{10} (16 -1)\]\[2^n(2^{m-n} -1) = 2^{10} (2^{4} -1)\]

so one solution is given by
n = 10 , m-n = 4

m = 14

Answer 27

i'm not sure whether i have proved this is unique, hopefully someone can confirm

Answer 28

yes u r correct this was the way my teacher i just forfot to take a note at that time thxxx

Answer 29

to show uniqueness it is necessary for
\[2^k \times 15 \neq (2^q - 1) \text{ , } 0<k \le 10\]

Answer 30

which is true as
\[2^k(2^4 -1) = 2^{k+4} - 2^k\]