10^x + 10^y + 10^z = n, where x, y, and z are positive integ

[hasibul.alam]

10^x + 10^y + 10^z = n, where x, y, and z are positive integers

Which of the following could be the number of zeroes, to the left of the decimal point, contained in n?

Indicate all such answers

A. x + y
B. y - z
C. z

[tommywallach]

Hey Hasibul,

In general, we prefer if you post a question ALONG with your question, rather than just ask us to solve. Tell us where the question confused you, and how you tried to solve it, and where it went wrong.

To solve, let's look at some samples:

x, y, z = 1
10 + 10 + 10 = 30
ONE zero --> This meets the condition for C

x = 1 z = 1 y = 2
10 + 10 + 100 = 120
ONE zero --> This meets the condition for B

x = 2 y = 2 z = 2
100 + 100 + 100 = 300

We can create a rule now. We can only get as many zeroes as whichever variable is largest (x or y or z). So we could never get as many zeroes as the sum of two of them.

The answer is B and C.

-t

[hasibul.alam]

thank you very much, actually the wording of the question was vague to me

[tommywallach]

Gotcha. Well I hope it's clearer now!

-t

[ash2rash]

If z = 5 x = 2 y =2 then
10^z = 100000
10^x=100
10^y= 100
so 10^x +10^y +10^y = 100200

then total zeros = 4 = x+y

Hence option 1 is also correct

[tommywallach]

Oh dear! Interesting! You're right that wording implies that your example would also work. I was assuming it would be number of zeroes DIRECTLY to the left of the decimal point, but you're absolutely right!

-t