The number of integer values of n for which the value 4000 × (0.4)ⁿ is an integer is; 9
How to find the integer values in algebra?
We are given the expression;
4000 × (0.4)ⁿ
This can also be expressed as;
4000 × (²/₅)ⁿ which can further be expressed in exponent form as;
(2⁵ × 5³) × (²/₅)ⁿ = 2⁽⁵ ⁺ ⁿ⁾ × 5⁽³ ⁻ ⁿ⁾
Since this expression is an integer, we need:
1) 5 + n ≥ 0 for which n ≥ -5
2) 3 - n ≥ 0 for which n ≤ 3
Taking the intersection gives;
-5 ≤ n ≤ 3
Thus;
Number of Integer Values = 3 - (-5) + 1 = 9
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