7th Graders, please click on the following link for practice problems for Friday's test:

https://sites.google.com/site/lassmath/my-forms


Problem of the Week

10/23/17

10 Digit Number

Write a ten-digit number so that the first digit indicates how many 0s are in the number, the second digit indicates how many 1s are in the number, the third digit indicates the number of 2s, etc. 


10/16/17

Pluto Math

Pluto's inhabitants use the same mathematical operators that we do (+, −, etc.). They also use an operator, @, that we do not know. The following are true for any real numbers x and y.

x @ 0 = x

x @ y = y @ x

x @ y = ((y–1) @ x) + (x+1) = (y-2) @ x + 2(x+1) = ((y–3) @ x) + 3(x+1)….

What is the value of 12 @ 5?

Solution

77. We have:

12 @ 5 = 5 @ 12
= (4 @ 12) + 13
= (3 @ 12) + 13 + 13
= (3 @ 12) + 26.

Continuing,

(3 @ 12) + 26 = (2 @ 12 + 39)
= (1 @ 12) + 52
= (0 @ 12) + 65
= (12 @ 0) + 65
 = 77

.

10/9/17

Path Distance

Jeremy walks along a spiral path (as shown). If the path is 2 meters wide, how far does he walk?

Solution

97 meters. By superimposing a grid of 2X2 squares, one can see that the path Jeremy travels can be broken into 13 sections of lengths 13, 12, 12, 10, 10, 8, 8, 6, 6, 4, 4, 2, and 2 meters.


10/2/17

Chicken Nuggets

Chicken nuggets come in packets of 6, 9, or 20. What is the largest number of nuggets that you cannot buy when combining various packets?

Solution

43

Any multiple of 3 that is greater than 3 can be obtained from packets of 6 and 9 nuggets. Since 36 = 9 + 9 + 9 + 9, 38 = 20 + 9 + 9, and 40 = 20 + 20, any even number ≥ 36 can be achieved by adding 6s to each of these. Similarly, by adding another 9, any odd number ≥ 45 can be achieved. But 43 is not yet guaranteed, so we need to examine the possibility of combinations that yield 43 nuggets. Since 43 is not a multiple of 3, likewise, 43 – 20 = 23 is not a multiple of 3, and 43 – 2 • 20 = 3 is too small to achieve. Consequently, 43 cannot be obtained.


9/25/17

Rainy Days

It rained on exactly 7 of the days during Jane's trip. On each day that it rained, it rained either in the morning or in the afternoon but not both. There were exactly 5 afternoons when it did not rain and exactly 6 mornings when it did not rain. How many days did the trip last?

Solution

9 days. Let M be the number of days it rained in the morning; let A be the number of days it rained in the afternoon; and let N be the number of days when it did not rain. We have M + A = 7 (days when it rained); M + N = 5 (afternoons when it did not rain); and A + N = 6 (mornings when it did not rain). Adding the equations we get 2(M + A + N) = 18, M + A + N = 9.


9/18/17

Painted Cube 

A cube has an edge length of 4 in. The cube is painted red all over and then cut into 64 cubes of edge length 1 in. How many of these cubes have exactly one face painted red?

Solution:

24. When the original cube is divided into 64 smaller cubes as shown, each face has only 4 cubes in the center with exactly 1 face painted red. (The cubes on the interior of the original cube will have no painted faces, and the other exterior cubes will have at least 2 painted faces.) This will result in 6 faces with 4 cubes each, or 4 × 6 = 24 cubes with exactly 1 face painted red.


9/11/17

No Fives

How many six-digit whole numbers are there which DO NOT contain a 5?

Solution

Let the six digits number be abcdef, which a, b, c, d ,e, f represent a digit respectively. 

For a, neither 0 nor 5 could place in it, thus, 8 digits are available here (1,2,3,4,6,7,8,9)

For b, c, d, e and f, they can't contain 5, hence, 9 digits are available for them (0,1,2,3,4,6,7,8,9)

Therefore, the number of six digits number which does not contain any 5 is

8 * 9 * 9 * 9 * 9 *9 = 472,392 .