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GRE Prep Club Team Member
Joined: 20 Feb 2017
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x = 10^10 - z, where z is a two-digit integer. If the sum of the digit
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04 May 2021, 03:19
04 May 2021, 03:19
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\(x = 10^{10} - z,\) where \(z\) is a two-digit integer. If the sum of the digits of \(x\) equals 84, how many values for \(z\) are possible?
A. 3
B. 4
C. 5
D. 6
E. 7
A. 3
B. 4
C. 5
D. 6
E. 7
Re: x = 10^10 - z, where z is a two-digit integer. If the sum of the digit
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04 May 2021, 22:14
04 May 2021, 22:14
need help!
x = 10^10 - z, where z is a two-digit integer. If the sum of the digit
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05 May 2021, 00:53
05 May 2021, 00:53
2
This is a very nice question.
\(x = 10^{10} - z\) & \(z\) - two digit integer
Now \(10^3 - \) any two digit no = \(9ab\)
\(10^4 - \)any two digit no = \(99ab\)
Similarly
\(10^{10} - \) any two digit no = \(99999999ab\)
It's given that the sum of digits of \(x\) = \(84\)
We already have \(8\)times \(9\) = \(72\)
So sum of \(ab = 12\)
So we need \(ab\) in which the sum of the digits = \(12\)
\(39 , 48 , 57 , 66 , 75 , 84 , 93\)
Now if we put ab = 93 , \(z\) will be \(7\) , but z is a two digit integer
So total 6 nos.
Answer D
\(x = 10^{10} - z\) & \(z\) - two digit integer
Now \(10^3 - \) any two digit no = \(9ab\)
\(10^4 - \)any two digit no = \(99ab\)
Similarly
\(10^{10} - \) any two digit no = \(99999999ab\)
It's given that the sum of digits of \(x\) = \(84\)
We already have \(8\)times \(9\) = \(72\)
So sum of \(ab = 12\)
So we need \(ab\) in which the sum of the digits = \(12\)
\(39 , 48 , 57 , 66 , 75 , 84 , 93\)
Now if we put ab = 93 , \(z\) will be \(7\) , but z is a two digit integer
So total 6 nos.
Answer D
