# Week 1 Solutions

## How hard were these week problems?

## Week 1 Solutions

Dear users of this forum,

Congratulations on another great week of problems. Here are the official answers from Mr. Stephens.

"The greatest digital sum date won't occur till almost the end of this century: 09/29/99, which has digital sum 38. "

"Fifty miles is 5280 x 50 = 264,000 feet. So, a circle of radius 50 miles has approximately pi x (264,000)2 = (3.14)(69,696,000,000) = 218,845,440,000, or almost 219 billion, square feet of area. Since between 7 and 19 inches of snow fell in various areas, it's reasonable to assume that all areas received an average of (7 + 19) / 2 = 13 inches. And that layer of snow, we found, fell on 219 billion square feet of area. In total, that's 13 x (219 billion x 144) = 4.09 x 1014 cubic inches of snow. We were told that all of this snow weighed 5 billion tons, which is (5 x 109) x 2000 = 1.0 x 1013 pounds. Consequently, one cubic inch of snow weighs (1.0 x 1013) / 4.09 x 1014 = 0.024 pounds. Since there are 123 = 1728 cubic inches in a cubic foot, a cubic foot of this snow weighs 0.024 x 1728 = 41.5 pounds."

"Every fourth year will be a leap year, except that three of the four years ending in double zero will not have leap days. Hence, every 400 years, there will be 400/4 - 3 = 97 leap days. From 2001 to 4000, there are 2000 years, so there will be 2000/400 x 97 = 485 leap days. As a fraction, that's 485/2000 = 97/400."

"The same date in successive non-leap years occurs one day later in the week, so 2/28/1910 was a Monday and 2/28/1911 was a Tuesday. In a leap year, it's two days later, so 2/28/1912 was a Thursday. That means that 2/28/1913 was a Friday. Consequently, five days early, 23 February 1913, was the last Sunday in February that year. "

"Use A, B and C to identify the number of votes received by each candidate. By using the information, B is greater than or equal to C/5, and is less than or equal to A/3. And, since B+C is greater than or equal to 600 and B is greater than or equal to C/5, 6C/5 is greater than or equal to 600. Therefore, C is greater than or equal to 500. By using the inequality above, B is greater than or equal to C/5, or 100. Since the least value of B is 100, the least value of A is 300 "

I hope you will soon be ready for next week's problems!

Archis

Congratulations on another great week of problems. Here are the official answers from Mr. Stephens.

**Problem 1:**"The greatest digital sum date won't occur till almost the end of this century: 09/29/99, which has digital sum 38. "

**Problem 2:**"Fifty miles is 5280 x 50 = 264,000 feet. So, a circle of radius 50 miles has approximately pi x (264,000)2 = (3.14)(69,696,000,000) = 218,845,440,000, or almost 219 billion, square feet of area. Since between 7 and 19 inches of snow fell in various areas, it's reasonable to assume that all areas received an average of (7 + 19) / 2 = 13 inches. And that layer of snow, we found, fell on 219 billion square feet of area. In total, that's 13 x (219 billion x 144) = 4.09 x 1014 cubic inches of snow. We were told that all of this snow weighed 5 billion tons, which is (5 x 109) x 2000 = 1.0 x 1013 pounds. Consequently, one cubic inch of snow weighs (1.0 x 1013) / 4.09 x 1014 = 0.024 pounds. Since there are 123 = 1728 cubic inches in a cubic foot, a cubic foot of this snow weighs 0.024 x 1728 = 41.5 pounds."

Problem 3:Problem 3:

"Every fourth year will be a leap year, except that three of the four years ending in double zero will not have leap days. Hence, every 400 years, there will be 400/4 - 3 = 97 leap days. From 2001 to 4000, there are 2000 years, so there will be 2000/400 x 97 = 485 leap days. As a fraction, that's 485/2000 = 97/400."

**Problem 4:**"The same date in successive non-leap years occurs one day later in the week, so 2/28/1910 was a Monday and 2/28/1911 was a Tuesday. In a leap year, it's two days later, so 2/28/1912 was a Thursday. That means that 2/28/1913 was a Friday. Consequently, five days early, 23 February 1913, was the last Sunday in February that year. "

**Problem 5:**"Use A, B and C to identify the number of votes received by each candidate. By using the information, B is greater than or equal to C/5, and is less than or equal to A/3. And, since B+C is greater than or equal to 600 and B is greater than or equal to C/5, 6C/5 is greater than or equal to 600. Therefore, C is greater than or equal to 500. By using the inequality above, B is greater than or equal to C/5, or 100. Since the least value of B is 100, the least value of A is 300 "

I hope you will soon be ready for next week's problems!

Archis

**Archis**- Admin
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## Re: Week 1 Solutions

Wait did you come up with those solutions archis or did mr. stephens send it to you?

**Steve**- The Cool Cube
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## Re: Week 1 Solutions

Dear Steve,

Mr. Stephens sent it to me, as well as next week's problems. The setup is after a week alloted for Week 1, he emails me the Week 1 Solutions and Week 2 problems.

It follows like that for all instances.

Sincerely,

Archis

Mr. Stephens sent it to me, as well as next week's problems. The setup is after a week alloted for Week 1, he emails me the Week 1 Solutions and Week 2 problems.

It follows like that for all instances.

Sincerely,

Archis

**Archis**- Admin
- Posts : 78

Points : 38

Reputation : 4

Join date : 2008-09-29

## Re: Week 1 Solutions

thats a good system!

**Steve**- The Cool Cube
- Posts : 85

Points : 4

Reputation : 4

Join date : 2008-10-07

Age : 22

Location : ...here

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