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A(2^3) + B(2^2) + C(2^1) + D(2^0) = 17
A(8) + B(4) + C(2) + D(1) = 17

Let A have the maximum value among all 4 that is nearest to 17; 8*2 = 16
therefore, A = 2, then only D can be 1 because its constant is 1,
so, 2(8) + B(4) + C(2) + 1(1) = 17
where B and C has to be zero,
so the product of ABCD can be 0

Let A,B,C have the equal value,
therefore, A = 1 = B = C , then only D can be 3 because its constant is 1,
so, 1(8) + 1(4) + 1(2) + D(1) = 17
where D has to be 3,
so the product of ABCD can be 1*1*1*3 = 3

the values of ABCD can be 0 as well as 3. Statement 1 and 3 are correct,
we have only option E with the validation of 1 and 3 together, so we don't have to calculate for statement 2.

Answer should be E
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We have 8a + 4b + 2c + d = 17
If a=b=c=1
8 + 4 + 2 + d = 17
d = 3
abcd = 3

If a = b = 1 and c = 2
8 + 4 + 2*2 + d = 17
d = 1
abcd = 2

If a = b = c = 0
0 + 0 + 0 + d = 17
d=17
abcd = 0

Therefore answer is E
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Hello from the GMAT Club BumpBot!

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