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14 Sep 2018, 13:57
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C
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A four-digit number with consecutive positive integers is a product of 2x and x^2. What is the sum of these two factors?
A. 80
B. 99
C. 168
D. 224
E. 288
A. 80
B. 99
C. 168
D. 224
E. 288
14 Sep 2018, 21:37
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gracie
Number is 4-digit and =2x*x^2=2x^3, thus an even number..
Cases
1) 1234..
\(2x^3=1234....x^3=617... Between 8^3--9^3\) , so no integer value possible
2) 3456..
\(2x^3=3456......x^3=1728......x=12\), so we have a possible value
We are looking for \(2x+x^2=2*12+12^2=24+144=168\)
C
18 Sep 2020, 07:09
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Let the 4 consecutive digits be a, a+1, a+2 and a+3
The number is therefore denoted as 1000(a) + 100(a+1) + 10(a+2) + (a + 3)
= 1000a + 100a + 100 + 10a + 20 + a + 3
= 1111a + 123
This value is equal to the product of 2x and \(x^2 = 2x * x^2 = 2x^3\). This value is even as Even * Odd = Even
We have \(1111a + 123 = 2x^3\). Since 1111 and 123 are Odd, and we want the sum to be Even, then \(a\) has to be Odd.
(Odd * Odd) + Odd = Odd + Odd = Even
When a = 1, \((1111 * 1) + 123 = 1234 = 2x^3\).
\(x^3 = \frac{1234}{2} = 617\) which is prime, and therefore is not a perfect cube
When a = 3, \((1111*3) + 123 = 3333 + 123 = 3456 = 2x^3\).
\(x^3 = \frac{3456}{2} = 1728 = 2^3 * 2^3 * 3^3\)
Therefore x = 2 * 2 * 3 = 12
\(2x + x^2 = 2(12) + 12^2 = 24 + 144 = 168\)
Option C
Arun Kumar
The number is therefore denoted as 1000(a) + 100(a+1) + 10(a+2) + (a + 3)
= 1000a + 100a + 100 + 10a + 20 + a + 3
= 1111a + 123
This value is equal to the product of 2x and \(x^2 = 2x * x^2 = 2x^3\). This value is even as Even * Odd = Even
We have \(1111a + 123 = 2x^3\). Since 1111 and 123 are Odd, and we want the sum to be Even, then \(a\) has to be Odd.
(Odd * Odd) + Odd = Odd + Odd = Even
When a = 1, \((1111 * 1) + 123 = 1234 = 2x^3\).
\(x^3 = \frac{1234}{2} = 617\) which is prime, and therefore is not a perfect cube
When a = 3, \((1111*3) + 123 = 3333 + 123 = 3456 = 2x^3\).
\(x^3 = \frac{3456}{2} = 1728 = 2^3 * 2^3 * 3^3\)
Therefore x = 2 * 2 * 3 = 12
\(2x + x^2 = 2(12) + 12^2 = 24 + 144 = 168\)
Option C
Arun Kumar
