Problem Solving

(replaced squares and triangles with letters to make this easier to type)

YZ
x(ZY)
-------

The product of the two-digit numbers above is the three-digit number YCZ, where Y,C,Z are three different nonzero digits. If YZ <10, what is the two digit number YZ?

a) 11
b) 12
c) 13
d) 21
e) 31

the book has an explanation but ive read it over and over and it makes absolutely no sense to me.

thanks for the help.

YZ x ZY = (10Y + Z) x (10Z + Y) = 100YZ + 10YY + 10 ZZ + ZY
YZ x ZY = YCZ
So we have
100YZ + 10YY + 10 ZZ + ZY = 100Y + 10C + Z
YZ < 10 therefor YZ has to be equal to Z (100YZ, 10YY, 10ZZ, 100Y, 10C, all have at least one zero at the end)
YZ = Z => Y = 1

Now we have
100Z + 10 + 10ZZ + Z = 100 + 10C + Z
100Z + 10 + 10ZZ = 100 + 10C
10Z + 1 + ZZ = 10 + C
From now in my opinion it's much faster to plug in 11, 12, 13 or follow your intuition. The answer is 11.

yankees660 wrote:(replaced squares and triangles with letters to make this easier to type)

YZ
x(ZY)
-------

The product of the two-digit numbers above is the three-digit number YCZ, where Y,C,Z are three different nonzero digits. If YZ <10, what is the two digit number YZ?

How can the correct answer be YZ=11 when the question states Y, C, and Z are three different non-zero digits?

yankees660 wrote:(replaced squares and triangles with letters to make this easier to type)

YZ
x(ZY)
-------

The product of the two-digit numbers above is the three-digit number YCZ, where Y,C,Z are three different nonzero digits. If YZ <10, what is the two digit number YZ?

a) 11
b) 12
c) 13
d) 21
e) 31

the book has an explanation but ive read it over and over and it makes absolutely no sense to me.

thanks for the help.

It's much easier to backsolve this question after applying a bit of logic.

First, we know that z and y are different non-zero digits. If we make z and y as small as possile, i.e. 1 and 2 (in either order), then we'll get a product of at least 200 (12*21 > 200). Therefore, we know that Y > 1 (since our product is YCZ). Eliminate (a), (b) and (c).

Now we could use fancy algebra to figure out the answer OR we could just plug in either (d) or (e) and be done. I choose door #2.

21*12 = 252, which is YCY, therefore incorrect.

Now, if this were the actual test, I'd just choose (E) and move on. However, just to be sure:

31*13 = 403, which is.. umm.. also not what we want. So, either the question is wrong or you've misposted it (sorry, I'm not in the office and don't have my Green book handy).

11*11 = 121, which does work except for the "different non-zero digits" issue raised by nytivofan.

In fact, if we think about it, none of the answers make any sense.

We've already shown that y must be greater than 1. However, we could also prove that y must equal 1, since the only way two distinct non-zero digits give you a produce ending in one of the digits is if one of them is 1. 1*(non-1 digit) = (non-1 digit), so z can't be the "1". By default, y=1.

So, we've proven that:

Y > 1
and
Y = 1.

Methinks there's a problem!

BIG PROBLEM

question is wrong here, the question actually says the result of YZ x ZY as YCY NOT YCZ

:roll:

realanoop wrote:BIG PROBLEM

question is wrong here, the question actually says the result of YZ x ZY as YCY NOT YCZ

:roll:

OK, then per my post above, (D) is the correct answer

Stuart..
1 query ...if yz <10 as per stem, then how the value of Yz>10

stuart, thanks for the explanation..makes a lot more sense than the one posted in the book and apologies for posting the question incorrectly

vivek.kapoor83 wrote:Stuart..
1 query ...if yz <10 as per stem, then how the value of Yz>10

The stem says yz < 10, i.e. y*z < 10. That's not to be confused with the 2 digit integer "yz" (although I can see how it could be confusing). It's impossible for a positive 2 digit integer to be less than 10, since 10 is the smallest postive 2 digit integer.

So, if the 2 digit integer "yz" = "21", then y*z = 2*1 = 2, which is indeed less than 10.