Question

If (81√3x³ + 625√5y³) = (Ax + By) (Cx² + Dy² + Exy), then find the ten's place digit of (A × B × E) is:
1. 5
2. 6
3. 7
4. 4

Answer 1

Correct Answer - Option 3 : 7

Given:

(81√3x3 + 625√5y3) = (Ax + By) (Cx2 + Dy2 + Exy)

Formula used:

x³ + y³ = (x + y)(x² + y² - xy)

Calculation:

(81√3x3 + 625√5y3) = (Ax + By) (Cx2 + Dy2 + Exy)

⇒ (3√3x)³ + (5√5y)³ = (Ax + By) (Cx2 + Dy2 + Exy)

⇒ (3√3x + 5√5y) [(3√3x)² + (5√5y)² - (3√3x × 5√5y)] = (Ax + By) (Cx2 + Dy2 + Exy)

⇒ (3√3x + 5√5y) (27x² + 125y² - 15√15xy) = (Ax + By) (Cx2 + Dy2 + Exy)

On comparing both sides, we get 

A = 3√3, B = 5√5, C = 27, D = 125 and E = -15√15

According to the question, we have to find the value of ABE

A × B × E = 3√3 × 5√5 × (-15 √15)

⇒ A × B × E = -(225 × 15)

⇒ A × B × E = -3375

Now, we can see that the ten's place digit of (A × B × E) = 7

∴ The ten's place digit of (A × B × E) is 7.