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# Quant Question of the Day Chat

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Math Expert
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Math Expert
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Posts: 95429
Own Kudos [?]: 657483 [0]
Given Kudos: 87241
Math Expert
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Intern
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Re: Quant Question of the Day Chat [#permalink]
can we post answers for these question with solutions under the posts?
Intern
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Re: Quant Question of the Day Chat [#permalink]
$$x=125^1/3$$ how do i give cube root sign

x=125^1/3

nvm i got it
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Re: Quant Question of the Day Chat [#permalink]
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 wrote:
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

c)7 possible values
d) 8 possible values
e) 9 possible values
Intern
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Re: Quant Question of the Day Chat [#permalink]
A-singh wrote:
c)7 possible values
d) 8 possible values
e) 9 possible values

is it c) 7 possible values
Intern
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Re: Quant Question of the Day Chat [#permalink]
It is D 8 possible values
Intern
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Re: Quant Question of the Day Chat [#permalink]
yeah correct should be 8
Intern
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Re: Quant Question of the Day Chat [#permalink]
Could you help me with the solution
Manager
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Re: Quant Question of the Day Chat [#permalink]
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:

$\{0, 12, 24, 36, 48, 60, 72, 84\}$
Intern
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Re: Quant Question of the Day Chat [#permalink]
so dies 0 count as a remainder?

making there 8 remainders and not 7?
Intern
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Re: Quant Question of the Day Chat [#permalink]
how does 0 count as a remainder

if it is perfectly divisible 96 with k=8

kingbucky wrote:
Yes. It does.

will it be true for all gmat question in such context?
Math Expert
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Re: Quant Question of the Day Chat [#permalink]
cryuss wrote:
how does 0 count as a remainder

if it is perfectly divisible 96 with k=8

­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.­
Intern
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Re: Quant Question of the Day Chat [#permalink]
Bunuel wrote:
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.

thnx, just checked the same thrgh GMAT club math book
Manager
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Re: Quant Question of the Day Chat [#permalink]
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
Intern
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Re: Quant Question of the Day Chat [#permalink]
+1

SKDEV wrote:
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

also did they remove og category from the gmat club forum quizzes
Manager
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Re: Quant Question of the Day Chat [#permalink]
cryuss wrote:
also did they remove og category from the gmat club forum quizzes

Yes, OG category is removed from GC Forum quiz.

SKDEV wrote:
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

Exact questions may not be available but similar type are avl. You will get enough questions for quant to practice from Gmat club.
Math Expert
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Re: Quant Question of the Day Chat [#permalink]
 Problem Solving Butler: August 2024 August 15 PS 1 PS 2
­
Math Expert
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Re: Quant Question of the Day Chat [#permalink]