Problems on Numbers Quiz Set 007

The digit at units place = (last digit of 7575) × (last digit of 1390) × (last digit of 9505) = 5 × 0 × 5 = 0, which = 0.

Let the numbers be 2n - 3, 2n - 1 and 2n + 1. The sum is 3(2n - 1) = 3 x middle. We are given 3 x middle = -2313, ⇒ middle = $-2313/3$, i.e., middle = -771.

The digit at units place = (last digit of 2840) × (last digit of 2492) × (last digit of 6032) = 0 × 2 × 2 = 0, which = 0.

Let the number be ab. When it is reversed and added to itself we get (10a + b) + (10b + a) = 11 × (a + b). We are given 154 = 11 × (a + b) ⇒ $a + b = 154 / 11 = 14$, so the digits are $a$ and $14 - a$. We are given their product as a × (14 - a) = 49, which is a quadratic expression that can be simplified to $(a - 7) × (7 - a) = 0$. So the number could be 77 or 77.

Let the number be n. From the given conditions we have $3n + 13 = 13n - 67$. Rearranging we have $67 + 13 = 13n - 3n$, ⇒ $n = {67 + 13}/{13 - 3}$. So n = 8.