If Z = 10^x + 10^y + 10^z, where x, y, z are distinct positive integer

Question

If  A = 10^x + 10^y + 10^z , where x, y, z are distinct positive integers, How many different value of A is there if  A < 10^9  ?

A. 24
B. 34
C. 56
D. 84
E. 120

Gladiator59 · Accepted Answer

IMO the answer should be option C. Please find my answer in image below.

Best,
Gladi

honneeey · Answer

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

u1983 · Answer

(8*7*6)/6!= 56  Ans C

saukrit · Answer

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

Manvi_nagpal · Answer

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
 C

akshaykotha · Answer

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.

mongmat27 · Answer

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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gmatcafe · Answer

In order to satisfy the given condition, the largest of x, y and z should be 8 at most. There are 8*7*6=336 ways of choosing such numbers. We need to divide by 3! because we don't care about the order. So 336/6=56. Answer is C)

chetanraghav1705 · Answer

Min value 1 & max value 8. Choose any three of 8 i.e. 8c3 or 56 or C is the ans.

Posted from my mobile device

SirSanguine · Answer

Bunuel  
Shouldn't a different symbol be used for the second Z? Otherwise the question is crazy tough!

Bunuel · Answer

Changed Z with A. Thank you!

anish777 · Answer

I did not understand about the ordering part ( why we are dividing the combination by 3!). Can anyone explain to me in detail?

Bunuel · Answer

That's because, for example, if x = 5, y = 2, and z = 1, the value of A will be the same as when x = 2, y = 5, and z = 1, or any other permutation of (5, 2, 1). Since there are 3! = 6 possible permutations of these three values, we divide the total number of combinations by 6 to avoid counting the same value of A multiple times.

If Z = 10^x + 10^y + 10^z, where x, y, z are distinct positive integer

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If Z = 10^x + 10^y + 10^z, where x, y, z are distinct positive integer

Prep Toolkit

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