Question

Find the sum of all positive divisors of 242 except 1 and itself.
A.156
B.242
C.342
D.399

Answer 1

Hint: First obtain the factors of 242, then apply the sum of all positive divisors of a number to obtain the sum. Then subtract 242 and 1 form the obtained result for solution.

Formula Used:
If a number A have 2 factors a, b with multiplicity 3 then the sum of all divisors of A is
$\left( {{a^0} + {a^1} + {a^2} + {a^3}} \right) \times \left( {{b^0} + {b^1} + {b^2} + {b^3}} \right)$ .

Complete step by step solution:
The given number is 242.
The given number can be written as $242 = 2 \times 11 \times 11$
Therefore, 2 have multiplicity 1 and 11 have multiplicity 2.
Hence, the sum of divisor is
$\left( {{2^0} + {2^1}} \right) \times \left( {{{11}^0} + {{11}^1} + {{11}^2}} \right)$
$=\left( {1 + 2} \right)\left( {1 + 11 + 121} \right)$
$=3 \times 133$
$=399$
Now, as per the question, subtract 242 and 1 from 399 to obtain the required answer.
$399 - 242 - 1 = 156$

Option ‘A’ is correct

Note: We call a number perfect if it is the sum of all its positive divisors except itself. Sometime students wrote the answer as 399 and did not do any subtractions, but if the question was to find the sum all positive divisor then the answer will be 399 but here we are asked to find the sum of all positive divisor except 1 and itself, so we have to subtract 242 and 1 from 399 to obtain the required answer.