Problem Solving

P_mashru wrote:The function f is defined for each positive three-digit integer n by f(n) = 2(x).3(y).5(z), where x, y & z are the hundreds, tens, and units digits of n, respectively.

There is something wrong with the wording of this question, for a start. I wonder whether you mean 2^x * 3^y * 5^z? It looks suspiciously like a prime factorization, which is why I'd expect x, y and z to be exponents here. I'll do both questions below- first I'll assume it's a product, and after, I'll assume x, y and z are exponents. Either way, there is something wrong with the answer choices.

I'll first assume that the question is supposed to be as you've written it- that we are multiplying x, y and z.

Then

f(n) = 30xyz

Let's let m = abc, v = xyz, where a, b, c, x, y and z are digits.

We know
f(m) = 9f(v)
30abc = 9*30xyz
abc = 9xyz

That is, the product of m's digits is 9 times the product of v's digits. That could happen in quite a few different ways; the answer choices may tell us what we're looking for. We want m-v to be a two digit number. So, why not try making the second digit of m = 9 times the second digit of v, and let the other digits remain equal? That is, let the middle digit of v be 1, and the middle digit of m be 9. We'd then have numbers like:

m = 191
v = 111

or

m = 492
v = 412

and in each case, m-v = 80.

So m-v could be 80. Could it be any of the other answer choices? Yes, in fact; it could be all of them. If we allow one digit to be 0, something which is not excluded in the question, we can get any two-digit value for m-v that we like. e.g.

m = 220
v = 202
m-v = 18

Notice the product of m's digits is still 9 times the product of v's digits, since 9*0 = 0, so in the above case, f(m) = 9*f(v). So, as the question is worded, all five answers are correct. Either x, y and z are intended to be exponents, or the question is incorrectly worded. I'm curious- where is this question from?

I'd note, finally, that if x, y and z are intended to be exponents, (which I suspect is what's really supposed to be happening here) then we would know (again letting m = abc, v = xyz):

f(m) = 9*f(v)
2^a * 3^b * 5^c = 3^2 * 2^x * 3^y * 5^z
2^a * 3^b * 5^c = 2^x * 3^(y+2) * 5^z

These are prime factorizations of integers, so a=x, b=y+2, c=z. Thus m and v only differ in their second digit; the second digit of m is two larger than the second digit of v. The difference must be 20, which isn't in the five answer choices you've listed above.

P_mashru wrote: If m an v are three - digit positive integers such that
f(m)=9f(v), then what could be the value of (m-v)=?

a) 18
b) 19
c) 15
d) 21
e) 80

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