Question

What is the sum of all the 2-digit numbers which leave a remainder of 6 when divided by 8?

• A. 612
• B. 594
• C. 324
• D. 872

Hint:

When a condition involves remainders, use modular arithmetic and apply arithmetic progression summation.

Answer 1

The Correct Option is B

We want 2-digit numbers \( x \) such that \( x \equiv 6 \pmod{8} \).
Smallest 2-digit number: 10, largest: 99.
First number satisfying condition: \( x = 14 \) (since \( 14 \mod 8 = 6 \))
Next terms: \( 22, 30, 38, \ldots \) till \( \leq 98 \).
This forms an AP with first term \( a = 14 \), common difference \( d = 8 \).
Let’s find number of terms: \[ a_n = 14 + (n-1) \cdot 8 \leq 98 \Rightarrow (n-1) \cdot 8 \leq 84 \Rightarrow n-1 \leq 10.5 \Rightarrow n = 11 \] Now sum of AP: \[ S = \frac{n}{2}(2a + (n-1)d) = \frac{11}{2}(28 + 80) = \frac{11}{2} \cdot 108 = \boxed{594} \]