If `52272 = p^2xxq^3xxr^4`,
where p, q and r are prime numbers then the value of `(2p+q-r) is:

Options

A. 22
B. 23
C. 29
D. 21

Answer

To solve the equation \( 52272 = p^2 \times q^3 \times r^4 \), where \( p, q, \) and \( r \) are prime numbers, we will first factor \( 52272 \) into its prime factors.

### Step 1: Factor 52272
We start by dividing \( 52272 \) by the smallest prime number, which is \( 2 \).

\[ 52272 \div 2 = 26136 \]
\[ 26136 \div 2 = 13068 \]
\[ 13068 \div 2 = 6534 \]
\[ 6534 \div 2 = 3267 \]

Now \( 3267 \) is not divisible by \( 2 \), so we try the next prime number, \( 3 \).

\[ 3267 \div 3 = 1089 \]
\[ 1089 \div 3 = 363 \]
\[ 363 \div 3 = 121 \]

Next, we factor \( 121 \) which is \( 11 \times 11 \) or \( 11^2 \).

Putting it all together, we have:

\[ 52272 = 2^4 \times 3^3 \times 11^2 \]

### Step 2: Identify p, q, and r
From the factorization \( 52272 = 2^4 \times 3^3 \times 11^2 \), we can assign the prime numbers to \( p, q, \) and \( r \) based on their powers:

- \( p = 2 \) (since it has the power of 4)
- \( q = 3 \) (since it has the power of 3)
- \( r = 11 \) (since it has the power of 2)

### Step 3: Calculate \( 2p + q - r \)
Now we substitute the values of \( p, q, \) and \( r \) into the expression \( 2p + q - r \):

\[ 2p + q - r = 2(2) + 3 - 11 \]
\[ = 4 + 3 - 11 \]
\[ = 7 - 11 \]
\[ = -4 \]

### Final Answer
The value of \( 2p + q - r \) is \( -4 \).

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