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5^3x+5^2y+5z+p=264, where x, y, z, p are all positive
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30 Jan 2022, 19:08
30 Jan 2022, 19:08
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Question Stats:
66% (02:46) correct
33% (01:39) wrong based on 3 sessions
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\((5^3)x+(5^2)y+5z+p=264\), where \(x\), \(y\), \(z\), and \(p\) are all positive integers less than or equal to 5, what is the value of \(x+y+z+p\)?
A. 8
B. 10
C. 12
D. 14
E. 16
Posted from my mobile device
A. 8
B. 10
C. 12
D. 14
E. 16
Posted from my mobile device
Re: 5^3x+5^2y+5z+p=264, where x, y, z, p are all positive
[#permalink]
31 Jan 2022, 08:05
31 Jan 2022, 08:05
1
\(5^3x+5^2y+5z+p=264\)
\(125x+25y+5z+p=264\)
\(125x+25y+5z=264-p\)
Since \(p\) is less than or equal to \(5\) and all the \(L.H.S\) elements are multiples of \(5\), to convert \(R.H.S\) into multiple of 5, let's assume \(p=4\)
\(125x+25y+5z=264–4\)
\(125x+25y+5z=260\)
Divide by \(5\)
\(25x+5y+z=52\)
Now, similarly to convert \(R.H.S\) into multiple of \(5\), let's assume that \(z=2\)
\(25x+5y= 52–2\)
\(25x+5y=50\)
\(5x+y=10\)
Now, similarly to convert \(R.H.S\) into multiple of \(5\), let's assume that \(y=5\)
\(5x=5\)
\(x=1\)
So, \(x+y+z+p= 1+5+2+4 = 6+6 =12\)
Answer C
\(125x+25y+5z+p=264\)
\(125x+25y+5z=264-p\)
Since \(p\) is less than or equal to \(5\) and all the \(L.H.S\) elements are multiples of \(5\), to convert \(R.H.S\) into multiple of 5, let's assume \(p=4\)
\(125x+25y+5z=264–4\)
\(125x+25y+5z=260\)
Divide by \(5\)
\(25x+5y+z=52\)
Now, similarly to convert \(R.H.S\) into multiple of \(5\), let's assume that \(z=2\)
\(25x+5y= 52–2\)
\(25x+5y=50\)
\(5x+y=10\)
Now, similarly to convert \(R.H.S\) into multiple of \(5\), let's assume that \(y=5\)
\(5x=5\)
\(x=1\)
So, \(x+y+z+p= 1+5+2+4 = 6+6 =12\)
Answer C
gmatclubot
Re: 5^3x+5^2y+5z+p=264, where x, y, z, p are all positive [#permalink]
31 Jan 2022, 08:05
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