Re: Quant Question of the Day Chat
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14 Aug 2024, 10:15
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:
\[
\{0, 12, 24, 36, 48, 60, 72, 84\}
\]