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dolorumsimilique

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16 Dec 2025, 07:32

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I like the solution - it’s helpful.

commodiad

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21 Dec 2025, 09:59

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I did not quite understand the solution. Please show how to get 2^8 from 256

Bunuel

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21 Dec 2025, 10:05

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commodiad

I did not quite understand the solution. Please show how to get 2^8 from 256

Not sure how to show this but 2^8 = 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 = 256.

CaroK

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22 Dec 2025, 11:11

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I like the solution - it’s helpful.

BigBacBrother333

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10 Jan 2026, 16:08

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bruhidk

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17 Jan 2026, 02:26

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I like the solution - it’s helpful.

sunshineeee

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31 Jan 2026, 02:48

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can u elaborate more why we need to make n = 0, is it because of the m already covered ?

Bunuel

Here is an explanation:

Since both $2^{90}$ and $m$ are integers, then $\frac{n}{2^8}$ must also be an integer. Considering the constraint $0 \le n \le 4$, this is only possible when $n = 0$.

So, n can only be 0, 1, 2, 3, or 4. Only if n is 0 do we get an integer value for n/2^8.

Bunuel

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31 Jan 2026, 05:13

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sunshineeee

can u elaborate more why we need to make n = 0, is it because of the m already covered ?

We have: $2^{90} = m + \frac{n}{2^8}$.

Since 2^90 and m are integers, n/2^8 must also be an integer. But n is restricted to 0, 1, 2, 3, 4. None of those divided by 2^8 gives an integer, except 0. So the only possible value for n is 0.

blanditiisoptio

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09 Feb 2026, 02:43

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I like the solution - it’s helpful.

Omomomo

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Updated on: 17 Feb 2026, 01:25

Originally posted by Omomomo on 13 Feb 2026, 03:18.
Last edited by Omomomo on 17 Feb 2026, 01:25, edited 1 time in total.

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the question is worded in a confusing manner. they should've used brackets to specify whether its (256*m)+n or if its 256*(m+n), or they should've simply worded it as 256m+n.

Bunuel

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13 Feb 2026, 03:52

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Omomomo

I don’t quite agree with the solution. the question should be worded in a confusing manner. they should've used brackets to specify whether its (256*m)+n or if its 256*(m+n), or they should've simply worded it as 256m+n.

You are wrong. Mathematically, 256m + n can only mean (256m) + n, nothing more. If it were 256(m + n), it would have been written that way. So there is no ambiguity there whatsoever.

elgatocosmico

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16 Mar 2026, 13:02

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I like the solution - it’s helpful.

parthk24

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24 Mar 2026, 06:33

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Here's how I did it, but this method is better applied to just this question instead of all questions of this type-

If n can be 0, try to check whether the equation can be solved just in terms of m. If you remove n, you get 2^98 = 256∗m

Since 256 = 2^8, you can rewrite this as 2^98=2^8 * m. Hence m will have to be 2^90.

n = 0 is sufficient.

Golabhai

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25 Mar 2026, 20:30

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I like the solution - it’s helpful. Since m can only take a value 2^90, anything less than this will not result in 2^98, as n can take a max value of 4, so m= 2^90 and n= 0 is the only possibility.

dignissimosminima

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08 May 2026, 03:02

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2^98 = 2^8 m + n
Dividend = devisor + remainder

2^8 will divide 2^98 hence leave 0 reminder.

Bunuel

If $2^{98} = 256*m + n$, where $m$ and $n$ are integers and $0 \le n \le 4$, what is the value of $n$?

A. 0
B. 1
C. 2
D. 3
E. 4

TautHorn

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15 May 2026, 11:09

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I am not sure if its more helpful but just for a tricky worded problem like this , I didn't see a whole lot of value given in the constraint other than it being such. I just separated as I knew 2 to power of something was 256. so separated into 2^8 x 2^90 = 256(m) + n . so I assumed 2^90 would be m and there for no need for n to equal anything. this may be some sort of rule for quotient expression but not sure but that is how I justified an answer of 0. Mind you I did on whiteboard do 2*2*2*2*2*2*2*2 to make sure it was 256 but I would imagine the gmat knows this and does offer a few different elegant ways. Probably why it wouldn't give too odd of a number but still works for most fairly easily.