Question

If \( x = 16^3 + 17^3 + 18^3 + 19^3 \), then \( x \) divided by 70 leaves a remainder of...

• A. 0
• B. 1
• C. 69
• D. 35

Hint:

When given cubes of consecutive integers, look for patterns or try actual substitution and simplify modulo.

Answer 1

The Correct Option is A

Let us denote: \[ x = 16^3 + 17^3 + 18^3 + 19^3 \] This is of the form \( a^3 + b^3 + c^3 + d^3 \), and if you try to calculate this modulo 70 directly: Instead use modulo arithmetic simplification. Note: \[ x = 16^3 + 17^3 + 18^3 + 19^3 = \text{sum of four consecutive cubes} \] We can observe: \[ 16 + 19 = 35,\quad 17 + 18 = 35 \Rightarrow \text{mean symmetry around } 17.5 \] Calculating actual values: \[ 16^3 = 4096,\quad 17^3 = 4913,\quad 18^3 = 5832,\quad 19^3 = 6859 \] \[ x = 4096 + 4913 + 5832 + 6859 = 21700 \] Now divide 21700 by 70: \[ 21700 \div 70 = 310 \Rightarrow \text{Remainder} = 0 \] \fbox{Final Answer: (A) 0}