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If [m]Z = 10^x + 10^y + 10^z [/m], where [m]x, y, z [/m]are distinct p

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If [m]Z = 10^x + 10^y + 10^z [/m], where [m]x, y, z [/m]are distinct p  [#permalink]

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New post 29 Sep 2018, 10:29
10
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A
B
C
D
E

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  45% (medium)

Question Stats:

60% (01:37) correct 40% (01:36) wrong based on 30 sessions

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Weekly Quant Quiz Question -7



If \(Z = 10^x + 10^y + 10^z\), where \(x, y, z\)are distinct positive integers, How many different value of \(Z\) is there if \(Z < 10^9\) ?
a) 24
b) 34
c) 56
d) 84
e) 120

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Re: If [m]Z = 10^x + 10^y + 10^z [/m], where [m]x, y, z [/m]are distinct p  [#permalink]

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New post 29 Sep 2018, 10:38
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IMO the answer should be option C. Please find my answer in image below.

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Re: If [m]Z = 10^x + 10^y + 10^z [/m], where [m]x, y, z [/m]are distinct p  [#permalink]

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New post 29 Sep 2018, 10:37
(8*7*6)/6!= 56 Ans C
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Re: If [m]Z = 10^x + 10^y + 10^z [/m], where [m]x, y, z [/m]are distinct p  [#permalink]

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New post 29 Sep 2018, 10:39
Answer is E

as x can be anything between 0 and 8
y can be anything between 0 to 7
z can be anything between 0 to 7
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Re: If [m]Z = 10^x + 10^y + 10^z [/m], where [m]x, y, z [/m]are distinct p  [#permalink]

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New post 29 Sep 2018, 10:42
If Z=10x+10y+10zZ=10x+10y+10z, where x,y,zx,y,zare distinct positive integers, How many different value of ZZ is there if Z<109Z<109 ?
a) 24
b) 34
c) 56
d) 84
e) 120

x,y,z can take values from 1 to 8 only

total combination will be 8*7*6=336
but as all are powers of 10 and we are asked to calculate different values of Z
divide this value by 3! i.e. 6
therefore answer=336/6=56
C
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Re: If [m]Z = 10^x + 10^y + 10^z [/m], where [m]x, y, z [/m]are distinct p  [#permalink]

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New post 29 Sep 2018, 10:42
Z<10^9 implies that 10's can only have max power upto 8
8*7*1=56
Since we can have 1st positive 8 integer values for the variable x
7 integer values for the y( as 1 used for x already)
and only one integer value for z as that is based on x and y
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Re: If [m]Z = 10^x + 10^y + 10^z [/m], where [m]x, y, z [/m]are distinct p  [#permalink]

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New post 29 Sep 2018, 10:44
As x,y,z are distinct, x,y,z can take 1 to 8 because once if one of the x,y,z is 9 then the condition is invalid.

If x,y,z can take any of the 8 integers and be distinct, then no. of possibilities are 8*7*6. But we need unique Z values, so it would be (8*7*6)/6 = 56. Each set of three integers have 6 combinations which give rise to the same Z value. Hence the number of possibilities is divided by 6.

Answer C.
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Re: If [m]Z = 10^x + 10^y + 10^z [/m], where [m]x, y, z [/m]are distinct p  [#permalink]

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New post 29 Sep 2018, 12:33
X, Y , Z are distinct and positive integers. And x or y or z is less than 9 and greater than 0

(Less than 9 and equation is less than (10) raised to the power 9.

Thus maximum value each of x, y and z can take is 8

For the number, if x can take 8 values, then y will take 7 values and z will take 6 ( because they are distinct).

Plus, as the order in which they take the value is not important, that is the combination of x*y*z can be 1*2*3 or 3*2*1 or 2*1*3 or so on (6 combinations are possible for 3!), we will divide 8*7*6 by 3! Thus, (8*7*6)/(6) = 56

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Re: If [m]Z = 10^x + 10^y + 10^z [/m], where [m]x, y, z [/m]are distinct p &nbs [#permalink] 29 Sep 2018, 12:33
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