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### ELITE EIGHT: East & West Regions

EAST:

Game G:

H: Xavier, Yvonne, and Zack all have the same birthday, but were born in different years, so every "when", "was", "is", or "will be" refers to their common birthday in some year, when every age is a whole number. Now read carefully and try not to get confused!

In 2000, Yvonne was 14 years older than Zack will be when Xavier is half as old as Yvonne was when Zack was twice as old as Xavier.

In 2003, Zack was 15 years younger than Yvonne was when Xavier was one-third the age Zack was when Yvonne was twice as old as Xavier will be when Zack is 35.

Now, in 2006, Xavier is 16 years older than Zack was when Yvonne was three times as old as Xavier was when Zack was one-third the age Yvonne will be when Xavier is 9 years younger than Zack will be when his age is the difference between those of the other two.

So, the question is, how old will Yvonne be when Xavier is 3 years younger than Zack was when Xavier and Yvonne's ages combined equalled 12 less than three times Zack's age?

L: The age that the actress that has her picture on one of the cards of the "All in the Cards" contest from GAMES will turn when, according to a famous sci-fi series android, Ireland is reunified

WEST:

Game G:

H:

1. Multiply these two numbers together:

* Number of US states that share no point of common latitude with any point in Canada or Mexico

* Number of US Presidents who, for whatever reason, never resided in the White House with a wife

2. If the result is less than 25, add 50.

3. If the result is more than 99, subtract 50.

4. If necessary, repeat Step 3 until the result is 2 digits.

L: Ricardo is as old as Michael Douglas’s wife used to be when Ricardo was as old as she is now.

How old is Ricardo?

## Comments

Am I missing something? The first part of the question implies that Ricardo is younger than Mrs. Michael Douglas, but the second part implies he is older.

Posted by: Will Nediger | Mar 25, 2006 10:46:24 AM

I think I found an answer that works. Clever.

Posted by: Bob Lodge | Mar 25, 2006 7:24:27 PM

I just got it too. Very clever.

Posted by: Will Nediger | Mar 26, 2006 12:00:50 PM

I've got two possible leads for the Ricardo/Michael Douglas's wife, but neither of them is absolutely convincing. Both Bob and Will seem convinced of their finite solution. That's (March) maddening!

Posted by: Jay Winter | Mar 28, 2006 2:33:54 PM

Any chance that someone would be kind enough to scan the All in the Cards contest? We recently moved, and my wife convinced me to discard a number of things to lighten the load, including back issues of GAMES. So, I cut out some of the better contests and articles and such, and kept only the issues that we have not yet seen the results published for. Since the All in the Cards results came out recently, and I did not keep that contest, those pages are being recycled somewhere. I have the issue with the contest results, which I may be able to use, but it would be easier (I think) with the original contest pages.

Or, would you rather see me penalized for not standing up for my right to keep my GAMES back issues???

Posted by: Jay Winter | Mar 29, 2006 8:52:10 AM

I can understand asking a question about something in archives on line. But back issues of a magazine? If you don't have them anymore, you're doomed. Call it a day. Game over. It's not really right.

Posted by: Ross | Mar 29, 2006 2:56:30 PM

If you read the current issue, it should be at least able to narrow the list of potential actresses down to small number (from there it may be a matter of comparing the other score to determine if it could possible be higher). Most magazines (although not GAMES) are available via fiche in most libraries. If you are really, really stuck a hint can be obtained via e-mail...but only if every other avenue has been exhausted.

Posted by: JmSR | Mar 29, 2006 3:08:00 PM

OK. Message received.

Posted by: Jay Winter | Mar 29, 2006 6:03:28 PM

Just finished the East, Game G, High Seed. Not for the algebraically weak at heart! I'm confident that Bob Lodge was at the heart of this one.

Posted by: Jay Winter | Mar 31, 2006 10:04:40 AM

Just to be sure -

In 2000, Yvonne was 14 years older than Zack will be when Xavier is half as old as Yvonne was when Zack was twice as old as Xavier.

Does that mean Zack was twice as old as Xavier WAS or twice as old as Xavier IS now?

Posted by: Paula | Apr 7, 2006 12:52:51 PM

"...when Zack was twice as old as Xavier" means the year that Zack's age was twice Xaviers age, then (at the same time).

Posted by: Bob Lodge | Apr 7, 2006 2:52:41 PM

Aha! I knew it was yours!

Posted by: Jay Winter | Apr 7, 2006 4:32:14 PM

**White House: Clarification** -- It has come to my attention that the White House was primarily named that by Theodore Roosevelt, before whom it was usually referred to as the Executive Mansion or some other appellation. This was not an intended consideration in solving this question, and the reference is to any U.S. President who lived in the building now called the White House with his wife, regardless of what it was called at the time.

Posted by: The Author | Apr 18, 2006 4:58:03 AM

EAST

Game G:

H: Declarative State 61 W

L: University Cantbeserious 50

H: Answer: 61

Solution:

Surely more than one approach is possible, but the shifting time frames make most algebraic attempts rather awkward and invite error. Personally, I have found it more reassuring to home in on the years of birth, which remain constant, rather than continuously changing ages.

Let X, Y, Z represent the YEAR OF BIRTH of each person. Use T where needed as a temporary facilitating variable. For each of the three statements, start at the end and work BACKWARDS one step at a time.

In 2000, Yvonne was 14 years older than Zack will be when Xavier is half as old as Yvonne was when Zack was twice as old as Xavier.

Let T = the year Zack was twice the age of Xavier

T-Z = 2(T-X)

T-Z = 2T - 2X

T = 2X-Z

Yvonne's age that year = 2X-Z-Y

half that age = (2X-Z-Y)/2

year Xavier is that age: (2X-Z-Y)/2 + X

Zack's age that year: (2X-Z-Y)/2 + X - Z

14 years older: (2X-Z-Y)/2 + X - Z + 14

= Yvonne's age in 2000, so

2000 - Y = (2X-Z-Y)/2 + X - Z + 14

1986 = (2X-Z-Y)/2 + X - Z + Y

3972 = 2X - Z - Y + 2X - 2Z + 2Y

3972 = 4X + Y - 3Z

In 2003, Zack was 15 years younger than Yvonne was when Xavier was one-third the age Zack was when Yvonne was twice as old as Xavier will be when Zack is 35.

The year Zack is 35 = Z + 35

Xavier's age that year = Z + 35 - X

Twice that age = 2Z + 70 - 2X

Year Yvonne was that age = Y + 2Z + 70 - 2X

Zack's age then = Y + 2Z + 70 - 2X - Z

combining & simplifying: 70 - 2X + Y + Z

one-third of that (70-2X+Y+Z)/3

Year Xavier was that age: X + (70-2X+Y+Z)/3

Yvonne's age that year: X + (70-2X+Y+Z)/3 - Y

15 years younger than that: X + (70-2X+Y+Z)/3 - Y - 15

= Zack's age in 2003, so

2003 - Z = X + (70-2X+Y+Z)/3 - Y - 15

6009 - 3Z = 3X + 70 - 2X + Y + Z - 3Y - 45

5984 = X - 2Y + 4Z

Now, in 2006, Xavier is 16 years older than Zack was when Yvonne was three times as old as Xavier was when Zack was one-third the age Yvonne will be when Xavier is 9 years younger than Zack will be when his age is the difference between those of the other two.

From the first statement we know that Zack is older than Xavier, therefore to make this last statement work, Yvonne must be the oldest. Let T = the year Zack's age is Yvonne's age minus Xavier's age.

Then T - Z = T - Y - (T - X)

T - Z = T - Y - T + X

T = X - Y + Z

Zack's age that year: X - Y

9 years younger: X - Y - 9

year Xavier is that age: 2X - Y - 9

Yvonne's age that year: 2X - 2Y - 9

one-third of that: (2X - 2Y - 9)/3

year Zack was that age: (2X - 2Y - 9)/3 + Z

Xavier's age that year: (2X - 2Y - 9)/3 + Z - X

three times that amount: 2X - 2Y - 9 + 3Z - 3X

= 3Z - X - 2Y - 9

year Yvonne was that age: 3Z - X - Y - 9

Zack's age then: 2Z - X - Y - 9

16 years older: 2Z - X - Y + 7

year Xavier was that age: 2Z - Y + 7

So, 2006 = 2Z - Y + 7

1999 = 2Z - Y

We now have three simultaneous equations in three unknowns:

3972 = 4X + Y - 3Z

5984 = X - 2Y + 4Z

1999 = 2Z - Y

add 1st and 3rd: 5971 = 4X - Z

add 2nd and 2x 1st: 13928 = 9X - 2Z

double new 1st: 11942 = 8X - 2Z

subtract 1986 = X

subst in 1st: 5971 = 7944 - Z

Z = 1973

subst in 3rd above: 1999 = 3946 - Y

Y = 1947

How old will Yvonne be when Xavier is 3 years younger than Zack was when Xavier and Yvonne's ages combined equalled 12 less than three times Zack's age?

In a given year T, their ages are T-X, T-Y, and T-Z

The year Xavier and Yvonne combined = 12 less than 3 x Zack

T - X + T - Y = 3(T-Z) - 12

2T - X - Y = 3T - 3Z - 12

T = 12 - X - Y + 3Z

= 12 - 1986 - 1947 + 5919

= 1998

In 1998, Xavier was 12, Yvonne was 51, and Zack was 25.

12 + 51 = 3 x 25 - 12 checks!

How old will Yvonne be when Xavier is 3 years younger than Zack was...

3 years younger than 25 is 22

Xavier is 22 in 2008

In 2008 Yvonne will be 61 !

Final solution is 61. Bob Lodge, writer.

L: The actress is Lark Voorhies from the TV show "Saved By The Bell".

She was born on March 25, 1974 & would turn 50 in the year 2024.. Jim from Minnesota & JonMichael Rasmus, writers.

WEST

Game G:

H: University of Catan 91 W

L: Clown College 48* (See below)

H: Solution:

The latitude of the southernmost point of Canada passes through these states, eliminating them and all farther north.

CA NV UT WY NE IA IL IN OH PA NY CT RI MA

The latitude of the northernmost point of Mexico passes through these states, eliminating them and all farther south.

CA AZ NM TX LA AL GA SC MS

Only these 13 states remain totally in between Canada and Mexico, overlapping no point of either.

CO KS OK MO AR KY TN NC VA WV MD DE NJ = 13

---------

Most of the 42 men who have been President of the US lived in the White House with their wives, but there were some exceptions:

When George Washington was President, the capital was in Philadelphia and the White House had not been built.

James Buchanan was a bachelor.

Wm H Harrison died one month after assuming office in 1841. Although she had planned to join him soon, his wife had not yet moved to Washington DC before he died.

Four presidents, Jefferson, Jackson, Van Buren, and Arthur, entered the White House as widowers. All had been married but their wives had died before their presidency.

So, altogether 7 Presidents never lived in the White House with their wives.

13 x 7 = 91, the answer to the procedure. Bob Lodge, writer.

L: After closer investigation, there are 3 possible ways to interpret

the question -

with three different answers.They are of varied "cleverness"...

A)48 is Ricardo's age. This is what I submitted. I believe I missed the word "twice" and it should read "ricardo is twice as old as Michael Douglas's wife..."

B)36 is Ricardo's age. This would fit as he is as old as she used to be and as old as he used be.(They are the same)

C) 42 is Ricardo's age. Michael Douglas had two wives. WifeA is now 48 and WifeB is 36. When Ricardo was 36, Wife A was 42. Therefore Ricardo is 42.

Since any of these don't change the result, all of them are correct. Paula Stevens, writer.

Posted by: JmSR | May 1, 2006 8:52:36 AM

This whole contest was great fun! But there's still one answer I'm just not getting. Would someone be willing to explain Ricardo's age again? What was the intended answer?

Posted by: Ross | May 2, 2006 6:10:25 AM

So, it turns out the word "twice" was inadvertently omitted from the Ricardo puzzle (which should have been caught when Will first posted his query) but that just happened to convert it to a much cleverer and still solvable puzzle, using TWO Mrs. Michael Douglases! I found Version three, which I believe is the only one that works with the omitted word as published. Here's my wording, Ross:

Ricardo is **42**.

Michael Douglas's wife (Catherine Zeta-Jones) is now 36.

Ricardo was 36 six years ago, in 2000.

In 2000, Michael Douglas's wife was Diandra Luker, who was then… (ta daaa!) … **42** !

Posted by: Bob Lodge | May 2, 2006 1:02:31 PM

PS -- The above depends partly on when Diandra Luker's birthday is, i.e., has she had one yet this year? Despite considerable searching, I never did find it, only the year she was born. But nothing else made ANY sense!

Posted by: Bob Lodge | May 2, 2006 1:07:22 PM

I spent way too much time looking at Michael Keaton's wives, Mike Douglas's wives, and Michael Douglas characters' wives (particularly using the movie The In-Laws) than I'd care to admit to.

Posted by: Jay Winter | May 2, 2006 2:07:57 PM

Then the question doesn't read so well. Doesn't "she" sound like the same person as "MD's wife"?? If it had said "MD's wife" twice (instead of "she"), I could hear the answer.

Posted by: Ross | May 2, 2006 2:22:18 PM

My thought exactly when I first solved it, Ross. I'm sure Paula would have thought of that, too, but remember, this is an unintended alteration of her original intent, which just happened to make a most interesting teaser!

Posted by: Bob Lodge | May 2, 2006 6:33:36 PM

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