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# If a, b, c, and d are non-negative integers and a(2^3) + b(2^2) + c(2^

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If a, b, c, and d are non-negative integers and a(2^3) + b(2^2) + c(2^  [#permalink]

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23 Jun 2020, 02:59
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55% (hard)

Question Stats:

60% (02:13) correct 40% (01:51) wrong based on 43 sessions

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If a, b, c, and d are non-negative integers and $$a(2^3)+b(2^2)+c(2^1)+d(2^0)=17$$, which of the following could be the product of $$abcd$$ ?

I. 0
II. 2
III. 3

A. I only
B. II only
C. I and II only
D. II and III only
E. I, II, and III

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Re: If a, b, c, and d are non-negative integers and a(2^3) + b(2^2) + c(2^  [#permalink]

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23 Jun 2020, 03:06
8a + 4b + 2c + d = 17

Case 1: a = 2, b = 0 , c = 0, d = 1
=> abcd = 0

Case 2: abcd = 2
=> One of them is 2 and all others are 1

The values are:
a = 1, b = 1, c = 2, d = 1

Case 3: abcd = 3
=> One of them is 3, and all others are 1.

Possible values: a = 1, b = 1, c = 1, d = 3

=> All statements could be true.

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Re: If a, b, c, and d are non-negative integers and a(2^3) + b(2^2) + c(2^  [#permalink]

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23 Jun 2020, 03:11
Bunuel wrote:
If a, b, c, and d are non-negative integers and $$a(2^3)+b(2^2)+c(2^1)+d(2^0)=17$$, which of the following could be the product of $$abcd$$ ?

I. 0
II. 2
III. 3

A. I only
B. II only
C. I and II only
D. II and III only
E. I, II, and III

First of all, Non-Negative Integer Includes Zero
(Zero is neither positive, nor negative)

We have a(8) + b(4) + c(2) + d(1) = 17
suppose a=b=c=1; we'll have 8 + 4 + 2 + d = 17
therefore d = 3
Product = 3

Suppose a = b = 1 and c = 2
8 + 4 + 2*2 + d = 17
We'll have d = 1
Product = 2

Suppose a = b = c = 0 "(given: Non-Negative ) and d = 17
We'll have product = 0

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Re: If a, b, c, and d are non-negative integers and a(2^3) + b(2^2) + c(2^  [#permalink]

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23 Jun 2020, 05:23
1
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.

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Re: If a, b, c, and d are non-negative integers and a(2^3) + b(2^2) + c(2^  [#permalink]

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02 Jul 2020, 12:12
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

Re: If a, b, c, and d are non-negative integers and a(2^3) + b(2^2) + c(2^   [#permalink] 02 Jul 2020, 12:12