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can we post answers for these question with solutions under the posts?
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cryuss
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\(x=125^1/3\) how do i give cube root sign

x=125^1/3

nvm i got it
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A-singh
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Hey can anyone please help me with this question
Q- N is a positive interger such that N^4 is divisible by 96. If N is divided by 96, the remainder obtained has
Options
a) 5 possible values
b) 6 possible values

A-singh
Hey can anyone please help me with this question
Q- N is a positive interger such that N^4 is divisible by 96. If N is divided by 96, the remainder obtained has
Options
a) 5 possible values
b) 6 possible values
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c)7 possible values
d) 8 possible values
e) 9 possible values
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A-singh
c)7 possible values
d) 8 possible values
e) 9 possible values
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is it c) 7 possible values
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A-singh
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It is D 8 possible values
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yeah correct should be 8
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Could you help me with the solution
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To determine the remainder when \( N \) is divided by 96, given that \( N^4 \) is divisible by 96, we start by analyzing the prime factorization of 96:

\[
96 = 2^5 \times 3^1
\]

For \( N^4 \) to be divisible by \( 96 \), it must be divisible by both \( 2^5 \) and \( 3^1 \).

### Step 1: Divisibility by \( 2^5 \)

Let \( N \) be expressed in terms of its prime factors:

\[
N = 2^a \times 3^b \times k
\]

where \( k \) is an integer not divisible by 2 or 3. Then,

\[
N^4 = (2^a \times 3^b \times k)^4 = 2^{4a} \times 3^{4b} \times k^4
\]

For \( N^4 \) to be divisible by \( 2^5 \), we need:

\[
4a \geq 5 \implies a \geq \frac{5}{4} \implies a \geq 2
\]

Thus, \( a \) must be at least 2.

### Step 2: Divisibility by \( 3^1 \)

For \( N^4 \) to be divisible by \( 3^1 \), we need:

\[
4b \geq 1 \implies b \geq \frac{1}{4} \implies b \geq 1
\]

Thus, \( b \) must be at least 1.

### Step 3: Form of \( N \)

From the above conditions, we can conclude that:

\[
N = 2^a \times 3^b \times k
\]

where \( a \geq 2 \) and \( b \geq 1 \). The smallest values satisfying these conditions are \( a = 2 \) and \( b = 1 \). Therefore, we can express \( N \) as:

\[
N = 2^2 \times 3^1 \times k = 12k
\]

### Step 4: Finding the Remainder of \( N \) when Divided by 96

Now, we need to find the remainder of \( N = 12k \) when divided by 96. We can express this as:

\[
N \mod 96 = (12k) \mod 96
\]

To find the possible values of \( k \), we note that \( k \) can be any positive integer. The values of \( 12k \) modulo 96 will depend on \( k \):

- If \( k = 1 \), \( N = 12 \)
- If \( k = 2 \), \( N = 24 \)
- If \( k = 3 \), \( N = 36 \)
- If \( k = 4 \), \( N = 48 \)
- If \( k = 5 \), \( N = 60 \)
- If \( k = 6 \), \( N = 72 \)
- If \( k = 7 \), \( N = 84 \)
- If \( k = 8 \), \( N = 96 \) (which gives a remainder of 0)

### Step 5: Possible Remainders

The possible remainders when \( N \) is divided by 96 are:

\[
12, 24, 36, 48, 60, 72, 84, 0
\]

### Conclusion

Thus, the possible remainders when \( N \) is divided by 96 are:

\[
\{0, 12, 24, 36, 48, 60, 72, 84\}
\]
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so dies 0 count as a remainder?

making there 8 remainders and not 7?
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how does 0 count as a remainder

if it is perfectly divisible 96 with k=8

kingbucky
Yes. It does.
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will it be true for all gmat question in such context?
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cryuss
how does 0 count as a remainder

if it is perfectly divisible 96 with k=8
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­Positive integer \(a\) divided by positive integer \(d\) yields a reminder of \(r\) can always be expressed as \(a=qd+r\), where \(q\) is called a quotient and \(r\) is called a remainder, note here that \(0\leq{r}<d\) (remainder is non-negative integer and always less than divisor).

For example, when dividing by 3, there are three possible remainders: 0, 1, and 2. A remainder of 0 would imply that a number is divisible by 3.­
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Bunuel
Positive integer \(a\) divided by positive integer \(d\) yields a reminder of \(r\) can always be expressed as \(a=qd+r\), where \(q\) is called a quotient and \(r\) is called a remainder, note here that \(0\leq{r}<d\) (remainder is non-negative integer and always less than divisor).

For example, when dividing by 3, there are three possible remainders: 0, 1, and 2. A remainder of 0 would imply that a number is divisible by 3.
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thnx, just checked the same thrgh GMAT club math book
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Hi @Bunuel,
can we find all official questions from Quant official pack (costing 29.99 USD)
on gmat club? or atleast a majority of them? same qs for DI official question pack as well
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+1

SKDEV
Hi @Bunuel,
can we find all official questions from Quant official pack (costing 29.99 USD)
on gmat club? or atleast a majority of them? same qs for DI official question pack as well
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also did they remove og category from the gmat club forum quizzes
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cryuss
also did they remove og category from the gmat club forum quizzes
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Yes, OG category is removed from GC Forum quiz.

SKDEV
Hi @Bunuel,
can we find all official questions from Quant official pack (costing 29.99 USD)
on gmat club? or atleast a majority of them? same qs for DI official question pack as well
Show more
Exact questions may not be available but similar type are avl. You will get enough questions for quant to practice from Gmat club.
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