Think Tank

Real Estate

Four entrepreneurs – Amber, Beryl, Carrie and Denise – each purchased a rectangular lot to build a factory. Each side of each lot was an exact number of feet. They found, to their surprise, that the diagonal of each lot was exactly 697 feet.

Amber and Denise owned the 2 larger allotments. Amber found that her lot was larger than Denise's by 38,520 square feet.

Similarly, Beryl found that her lot was larger than Carrie's by 20,280 square feet.

What were the dimensions of each allotment of the four companies?

CrossNumber 55 (All answers are numbers. No number is repeated, none begins with zero, “prime” is a prime number and “root” means “square root.”)

Across

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1. See 11 Across
5. Square
8. Square
10. 2 x Root of 7 Down
11. 4 x Root of 2 Across
12. 13 x Root of 15 Down
14. Prime
16. 7 x 24 Down
17. Root of 6 Down x Root of 8 Down
20. 3 x 24 Down
22. Prime
23. Prime
24. See 1 Down
26. Square
28. Square

Down

1. Root of 24 Across
3. 3 x 1 Down
4. 2 x Root of 2 Across
6. See 17 Across
7. See 10 Across
8. See 17 Across
9. Root of 8 Down x Root of 20 Down
12. 13 x 25 Down
13. (8 Down + 5) / 2
15. Square
18. 15 Down – Root of 20 Down
19. 26 Across + Root of 20 Down
20. See 18 Down
21. 23 x Root of 8 Down
24. Prime
25. Root of 28 Across
27. One-third of 13 Down

Contest Rules: All entries must be received by Oct. 10, 2000. The winners of the word puzzle and CrossNumber (can be the same person) will be drawn from all correct entries. Fax all entries to 847-634-4240. All entries must include name, complete address, company affiliation and daytime phone number in order to be considered. Employees of PennWell and their immediate families, agents contracted by PennWell and their immediate families, employees of PennWell and its subsidiary companies and their immediate families, and members of the Advanced Packaging Advisory Board and their immediate families, are not eligible to compete.

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Solutions to June/July's Think Tank Puzzles CrossNumber 53

The School Break

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Answer: 10Ω = 20; 100Ω = 15; 10KΩ = 10; 100KΩ = 11; Total = 112

Solution: Let the resistors be indicated by the first letter of the following colors assigned to them, and the other by X (i.e., 10Ω = Red; 100Ω = Brown; 10KΩ = Yellow and 100KΩ = Green).

Let the amount needed for SP be R + B + G + Y; let the spares be 2b + r + 2g + y.

What Tom used for his model train (MT) = r + b +g + y.

What Dad used for his Burglar Alarm (BA) = b + g + X.

Let total = 4T = 2b + r + 2g + y + R + B + G + Y + X ———(1) b + r + g + y = X + b + g = T —- (2) [they used the same amount, 1/4 of total]

I.e., X = r + y —- (3); also (b + g + X)/2 = X [Half of Dad's share in other resistors] I.e., b + g =X — (4); also b = g —- (5) [Same number of 100s and 100Ks] From (4), b = X/2 = g ; y – g = 1 —- (7) [Tom used 1 more 10K than 100K for MT]

y + g = B — (8) [ Tom's 10Ks and 100Ks = 100s for SP]

From (7) and (8), y = (B+1)/2; from (7), y = g + 1 = X/2 + 1 —– (i)

From (8), B = X/2 + 1 + X/2 = X + 1 —– (ii)

2g + G = B + Y —- (9) [Total of 100Ks = total of 10s and 10Ks for SP]

G + Y = 3b —- (10) [Total of 10K and 100K for SP = 3 times 100s Dad used]

B + 2b + r + R = 5G —- (11) [Total of 10s and 100s Dad bought = 5 times 100Ks for SP] (9) – (10), 2g – Y = B + Y – 3b. I.e., 2g + 3b = B + 2Y

Substituting from (6) and (ii), X + 3X/2 = X + 1 + 2Y. I.e., 4Y = 3X – 2

From (10), G = 3b – Y = 3X/2 – (3X – 2)/4 = (3X + 2)/4 — (iii)

Substituting in (11), r + R = 5G – 2b – B = 5(3x + 2)/4 – (X + 1) – X = (7X + 6)/4

From (3), X = r + y, I.e., r = X – (X/2 + 1) = (X – 2)/2 —- (iv)

R = (7X + 6)/4 – (X – 2)/2 = (5X + 10)/4 —- (v); substituting in (1), 4T = (31X + 14)/4.

From (2), T = b + g + X = 2X. I.e. T = (31X + 14)/16 = 2X. I.e., X = 14; T = 28; 4T = 112; b = g = 7; r = 6; y =8; B = 15; R = 20; G = 11 and Y = 10.

Grand Prize Winners

Thanks to all who faxed in their responses to our June/July puzzles. Congratulations to Norbert Samek of NS Consulting Services, who is now the proud owner of an Advanced Packaging T-shirt for winning CrossNumber 53. Sadly, no one was able to solve the School Break challenge. Good luck next time and don't forget to fax in your solutions.

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