Data Sufficiency

10/2^n find the remainder?
1. n is even number
2. n is multiples of 4

how do we deal with this kind of remainder questions, when something is like 10/2^n or maybe 12/4^n....

st(1) turns out to make two remainders when n=1 and n=2, the remainder is 2 And when n>3 the remainder is 10 Not Sufficient;
st(2) n is multiples of 4 means n>=4 and n is divided by 4 which makes viable only one possibility for remainder = 10 Sufficient

IOM B

neilcao wrote:10/2^n find the remainder?
1. n is even number
2. n is multiples of 4

how do we deal with this kind of remainder questions, when something is like 10/2^n or maybe 12/4^n....

just follow the decision stem above

I got the same thing. When N is a multiple of 4, then 2^N will always be greater than 10. This division will always yield the dividend as the remainder. Hence, 10.

here i got E
(1) n-even integer
if n=0, 10=2^0*10+0 remaider 0 ( 0 is even)
if n=2, 10=(2^2)*2+2 remaider 2
if n=4, 10=2^4*0+10 remaider 10
not suff
(2) n is multiple of 4
if n=0 remaider 0 (0 is a multiple of 4)
if n=4 then 10=2^4*0+10 remaider 10
not suff
together we have two versions 0 and 10 so both insuff

ahh, you're right, 0 is a multiple of all numbers.

@clock, technically you are absolutely right!!! ... and welcome to the GMAT math conventions. Please read below a fragment from the BTG thread, the content copied/pasted belongs to Ron Purewal and hear how an informed test taker could reason one's solution (though, it doesn't help to excel in exam )

"... under the traditional mathematical definition, yes, 0 and negative multiples are counted as 'multiples'.

HOWEVER,

to date, as far as i know, there has never been an official problem requiring the use of 0 or negative numbers as 'multiples' of positive integers. in fact, every single problem dealing with factors, multiples, primes, divisibility, and the like has been restricted, by fiat, to positive integers."

Source https://www.beatthegmat.com/negative-mul ... 12560.html

Therefore, my friend I chose B-opsss, you could also mention -ve multiples in your analysis of st(2)

clock60 wrote:here i got E
(1) n-even integer
if n=0, 10=2^0*10+0 remaider 0 ( 0 is even)
if n=2, 10=(2^2)*2+2 remaider 2
if n=4, 10=2^4*0+10 remaider 10
not suff
(2) n is multiple of 4
if n=0 remaider 0 (0 is a multiple of 4)
if n=4 then 10=2^4*0+10 remaider 10
not suff
together we have two versions 0 and 10 so both insuff

above post looks very impressive
may be i am wrong
waiting for the oa
( but honestly i don`t like this problem at all, to me it is poorly constructed-my own opinion)