An algebra problem by Naren Bhandari

Algebra Level 3

{ A × 4 = E B ÷ 4 = E C + 4 = E D − 4 = E A + B + C + D = 100 \begin{cases} A × 4 = E \\ B ÷ 4 = E \\ C + 4 = E \\ D - 4 = E \\ A + B + C + D = 100 \end{cases}

If real numbers A A , B B , C C , D D and E E satisfy the system of equations above, find the product A B C D ABCD .


The answer is 61440.

Naren Bhandari by Naren Bhandari , 21, India

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1 solution

4 A = E 4A = E or A = E 4 A = \frac{E}{4}

B 4 = E \frac{B}{4} = E or B = 4 E B = 4E

C + 4 = E C + 4 = E or C = E − 4 C = E - 4

D − 4 = E D - 4 = E or D = E + 4 D = E + 4

Substitute A , B , C A,B,C and D D in the fifth equation.

A + B + C + D = 100 A + B + C + D = 100

E 4 + 4 E + E − 4 + E + 4 = 100 \frac{E}{4} + 4E + E - 4 + E + 4 = 100

E = 16 E = 16

solving for A,B,C and D, we have

A = 16 4 = 4 A = \frac{16}{4} = 4

B = 4 ( 16 ) = 64 B = 4(16)=64

C = 16 − 4 = 12 C = 16 - 4 = 12

D = 16 + 4 = 20 D = 16 + 4 = 20

A B C D = ( 4 ) ( 64 ) ( 12 ) ( 20 ) = ABCD = (4)(64)(12)(20) = 61440 \boxed{61440}

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