This question was previously asked in

Territorial Army Paper I : Official Practice Test Paper - 3

Option 4 : 3719

**Concept used**

N = a^{p }b^{q }c^{r}

Sum of factors = \(\frac{{\left( {{a^{p + 1}} - 1} \right)\left( {{b^{q + 1}} - 1} \right)\left( {{c^{r + 1}} - 1} \right)}}{{\left( {a - 1} \right)\left( {b - 1} \right)\left( {c - 1} \right)}}\)

**Calculation**

1520 = 2^{4} × 5^{1} × 19^{1}

⇒ \(\frac{{\left( {{2^{4 + 1}} - 1} \right)\left( {{5^{1 + 1}} - 1} \right)\left( {{{19}^{1 + 1}} - 1} \right)}}{{\left( {2 - 1} \right)\left( {5 - 1} \right)\left( {19 - 1} \right)}}\)

⇒ \(\frac{{\left( {{2^5} - 1} \right)\left( {{5^2} - 1} \right)\left( {{{19}^2} - 1} \right)}}{{72}}\)

⇒ \(\frac{{31 \times 24 \times 360}}{{72}}\) = 3720

Sum of factors (except unity)

⇒ 3720 - 1 = 3719