Download 4A Strategy: List the single-digit primes. The single

4A Strategy: List the single-digit primes.
The single-digit prime numbers are 2, 3, 5, and 7. Select the 3 greatest numbers from this list and write
them from largest to smallest. The last time before noon when all 3 digits are prime is 7:53.
4B Strategy: Make an organized list.
List all the ways four different numbers can add to 15. Starting with the largest number reduces the
number of trials necessary. Because 3 + 2 + 1 = 6 is the least possible sum for 3 of the numbers, the
greatest number can't exceed 15 — 6 = 9.
15 = 9 + 6 = 9 + 3 + 2 + 1
15 = 8 + 7 = 8 + 4 + 2 + 1
15 = 7 + 8 = 7 + 5 + 2 + 1 or 7 + 4 + 3 + 1
15 = 6 + 9 = 6 + 5 + 3 + 1 or 6 + 4 + 3 + 2
15 can be written as the sum of four different counting numbers in 6 different ways.
4C METHOD 1 : Strategy: Count horizontally, from the top layer down.
Make a table that counts cubes separately for each layer. In each case, add the
number of hidden cubes to the number of visible cubes.
1 +3 + 6 + 10 = 20 cubes are used to form the tower.
METHOD 2 : Strategy: Count vertically, stack by stack.
This table counts cubes separately for each stack (column), from the shortest to the tallest. Both hidden
and visible cubes are counted.
A total of 20 cubes are used to form the tower.
4D Strategy: Consider the clues one at a time, starting with the most restrictive clue.
Clue 3: The only prime factor of the 4-digit number is 11, so the number = 11 x 11 or 11 x 11 x 11 or
11 x 11 x 11 x 11 , etc. Of these, only 11 x 11 x 11 = 1331 has 4 digits, so the 6-digit number is __1331__
Clue 1: The number reads the same right to left, so the first and last digits are the same. Call the number
A1331A.
Clue 2: The number is a multiple of 9, so the sum of its digits is a multiple of 9. A +1+3+3 +1 +A = A
+ 8 + A must equal 9 or 18. No digit A satisfies A + 8 + A = 9, but if A + 8 +A =18, A = 5. Hannah's
number is 513 315.
4E Strategy: Make the figure as compact as possible.
The area of one L-shape is 3 sq cm. A figure made up of 5 L-shapes has an area of 15 sq
cm. The fact that all the angles in each shape are right angles suggests trying to pack the
5 shapes into a square of area 16 sq cm. This can be done as shown in the figure.
The least possible perimeter is 4 + 4 + 3 +3 + 1 + 1 = 16 cm.
4A Strategy: Use the property of additive inverses (opposites).
Write the sum as (-2007+-2006+ … + +2006 + +2007)+ +2008 + +2009. The sum in parentheses is 0.
Thus the sum of all the integers from -2007 through 2009 is +2008 + +2009 = 4017.
4B Strategy: Draw a picture and examine the possibilities.
There are two possible cases. Either
is 55 cm.
The first case
is not possible. If the sum of 8 and three widths is 55 cm, then the
width is not an integer.
The second case
is possible. If the sum of 16 and three widths is 55 cm, then
the width is 13 cm. This makes the length 21 cm, and the perimeter of the rectangle 68 cm.
4C Strategy: Use the definition of average speed.
The average speed is the total distance traveled divided by the total time elapsed.
The graph shows that Jen traveled 10 miles from home before she rested and then
traveled 10 miles back. The total time was 4 hours, so her average speed was 20
mi ÷ 4 hrs = 5 mph.
4D Strategy: Count in an organized way.
METHOD 1: Strategy: Examine all the ones and tens digits separately.
The ones digit of 8: It occurs 1 time in every set of 10 consecutive numbers. From 1 through 384, there
are 38 complete sets of 10 and therefore the digit 8 appears 38 times as a ones digit.
The tens digits of 8: It occurs 10 times in every set of 100 consecutive numbers. From 1 through 384,
there are 3 complete sets of 100 (1-100, 101-199, 200-299); the digit 8 appears 30 times as a tens digit.
In addition, the numbers 380-384 contain 5 more tens digits of 8. In all, the digit 8 appears a total of 73
times.
METHOD 2: Strategy: Count by intervals, treating the 80s as a special case.
Add the numbers in the second row. The digit 8 appears a total of 73 times.
4E Strategy: Examine the prime factors.
Since 180 x N is a perfect cube, it is the product of three equal factors. Each of these factors is prime
or is the product of prime numbers. Because 180 = 2 x 2 x 3 x 3 x 5, the perfect cube, 180 x N, must
represent the product of 23 , 33 , and 53 . N must represent the missing additional factors 2, 3, 5, and 5.
The least possible value of N is 2 x 3 x 5 x 5 = 150. As a check, 180 x 150 = 27,000, which is 303 .
Alternately, (23 x 33 x 53) ÷ 180 = 27,000 ÷ 180 = 150.