AMC8 1991

Write the problem vertically and compute the difference: $1,000,000,000,000 - 777,777,777,777 = 222,222,222,223$ OR What must be added to $777,777,777,777$ to get $1,000,000,000,000$? From the right, one adds a final digit of 3 and then eleven 2’s.

Using the standard order of operations, first simplify the numerator and then the denominator. Finally compute the quotient: $(16 + 8) / (4 - 2) = 24 / 2 = 12$. Note. Keying $16 \div 8 + 4 - 2$ on the calculator will give an incorrect answer for this problem. The problem means $(16 + 8) \div (4 - 2) = 24 \div 2 = 12$.

Using arithmetic notation: $200,000 \times 200,000 = 40,000,000,000$ OR Using scientific notation $(2 \times 10^5)(2 \times 10^5) = 4 \times 10^{10} = 40 \times 10^9 = 40$ billion. OR Two hundred times two hundred is forty thousand. A thousand thousands is a million. The answer is forty thousand millions, or forty billion.

Answer (E): Each of the five numbers on the left side of the equation is approximately equal to 1,000. Thus $N$ can be found by computing the difference between 1,000 and each number, so $N = 9 + 7 + 5 + 3 + 1 = 25$.

A collection of non-overlapping dominoes must cover an even number of squares. Since checkerboard (B) has an odd number of squares, it follows that it cannot be covered as required. A little experimentation shows how the other checkerboards can be covered.

First mark the largest number in each column. Only 7 is the smallest in its row. (Note: One could also begin by finding the smallest number in each row. Then a check of the columns yields the answer 7.)

Rounding each number to one significant digit (highest place value) yields $(500,000)(10,000,000) + (10,000,000)(500,000) / (20,000)(.05)$ which equals $(500,000)(10,000,000 + 10,000,000) / 1,000$ which equals $(500)(20,000,000) = 10,000,000,000$.

The largest quotient would be a positive number. To obtain a positive quotient either both numbers must be positive or both must be negative. Using two positive numbers, the largest quotient is $8 / 1 = 8$. Using two negative numbers, the largest quotient is $-24 / -2 = 12$.

A number is divisible by 3 if it is a multiple of 3, and it is divisible by 5 if it is a multiple of 5. There are 15 multiples of 3, and 9 multiples of 5 which are whole numbers less than 46. However, 3 numbers (15, 30 and 45) which are divisible by both 3 and 5 have been counted twice. Thus the total number which are divisible by either 3 or 5 or both is $15 + 9 - 3 = 21$.

The parallelogram rests on the horizontal axis. Since the coordinates of point C are (4, 2), it follows that the height of the parallelogram is 2. Since point B is (0, 2), it follows that the length of the base BC is 4. The area of a parallelogram is base times height. Thus the area is $4 \times 2 = 8$.

After the 5 is selected, a sum of 10 is needed. There are four pairs that yield 10: 9+1, 8+2, 7+3, 6+4. Thus there are four 3-element subsets which include 5 and whose sum is 15.

Any fraction of the form $((k-1) + k + (k+1)) / k$ equals 3, since $(k-1) + k + (k+1) = 3k$ and $3k / k = 3$. The denominator of the fraction must equal the middle term of the numerator. Thus $N = 1991$.

Since $2 \times 5 = 10$, each zero at the end of the product comes from a product of 2 and 5 in the prime factorization. Since $25 = 5 \times 5$ and $8 = 2 \times 2 \times 2$, it follows that there are fourteen factors of 5 and nine factors of 2. This yields 9 pairs of $2 \times 5$ and results in 9 zeros at the end of the product.

If one student earns $5 + 5 + 5 = 15$ points, no other student can earn more than $3 + 3 + 3 = 9$ points. If one student earns $5 + 5 + 3 = 13$ points, no other student can earn more than $3 + 3 + 5 = 11$ points. However, if one student earns $5 + 3 + 3 = 11$ or $5 + 5 + 1 = 11$ points, some other student can earn $3 + 5 + 5 = 13$ or $3 + 3 + 5 = 11$ points. Thus 13 points is the smallest number of points a student must earn to be guaranteed of earning more points than any other student.

When the one-foot cube is removed, three square feet of surface area are “removed”, but three new square feet of surface area are “uncovered”. Thus the original surface area is unchanged.

A fold from the top leaves #9-16 on the bottom. A fold from the bottom leaves #9-12 on the bottom. A fold from the right leaves #9 and #10 on the bottom. A fold from left leaves #10 on the bottom with number #9 moving to the top.

The first row has 10 seats, so 5 students can sit in row 1. The second row has 11 seats, so 6 students can sit in row 2. The third row has 12 seats, so 6 students... The sum is $5 + 6 + 6 + 7 + 7 + 8 + 8 + 9 + 9 + 10 + 15 + 14 + 14 + 13 + 13 + 12 + 12 + 11 + 11 + 10 = 200$.

Regardless of the scale on the vertical axis, 9 X's out of 30 X's represent employees who have worked 5 years or more. This is $9/30$ or 30%.

Since the mean is 10, it follows that the sum of the numbers is $10 \times 10 = 100$. Taking the smallest possible value for the 9 smaller numbers would give the largest possible value of the tenth number. Thus, the largest possible number is $100 - (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9) = 100 - 45 = 55$.

Answer (A): For the sum to be 300, $A=2$ in the hundreds’ place, since $A=1$ gives numbers too small and $A=3,4,\ldots$ makes the sum too large. If $A=2$ then $B=7$, since $A=2$ and $B=8$ or $9$ would be too large for the ones’ place and $A=2$ and $B=6,5,\ldots$ would not be enough to carry a 1 from the tens’ to the hundreds’ place. Thus, if $A=2$ and $B=7$ and $A+B+C=10$ in the ones’ place, then $C=1$.

The rate of change is 4 cubic centimeters per 3°. The temperature change is $32° - 20° = 12°$. The corresponding change in volume is $12 \times (4/3) = 16$ cubic centimeters. Thus the initial volume of the gas was $24 - 16 = 8$.

Answer (D): The only way to get an odd number for the product of two numbers is to multiply an odd number times an odd number. This happens if one spins 1 or 3 on the first spinner (2 chances out of 3) and 5 on the second spinner (1 chance out of 3). Thus, the probability of an odd product is $\frac{2}{3}\times\frac{1}{3}=\frac{2}{9}$. If one does not get an odd product, then the product is even. Hence the probability of an even product is $1-\frac{2}{9}=\frac{7}{9}$.

Answer (A): There area 100 females in the band, 80 in the orchestra, and 60 in both. Thus, there are $(100+80)-60=120$ females in at least one of the groups. Since the total is 230, then there are $230-120=110$ males in at least one of the groups. There are 80 males in band and 100 males in orchestra, thus to find the number of males in both, $(80+100)-?=110$. There are 70 in both. Finally, the number of males in band who are not in orchestra is $80-70=10$.

Since the edge of each smaller cube must be a whole number, the smaller cubes must be $1\times1\times1$ or $2\times2\times2$ cubes. There can be only one smaller cube of edge 2, so the rest of the smaller cubes have edge 1. Since the volume of the original cube was $3 \times 3 \times 3 = 27$ cubic cm, and the volume of the cube of edge 2 is $2 \times 2 \times 2 = 8$ cubic cm, then there must be $27 - 8 = 19$ cubes of edge 1 (volume = 1 cubic cm). There are a total of 20 cubes so $N = 20$.

Answer (C): Since $1/4$ turns white, it follows that $3/4$ remains black after the first change. After the second change, $3/4$ of the remaining $3/4$ stays black or $\frac{3}{4}\times\frac{3}{4}$ remains black. Thus, after the fifth change, the amount remaining black is $\frac{3}{4}\times\frac{3}{4}\times\frac{3}{4}\times\frac{3}{4}\times\frac{3}{4}=\frac{243}{1024}$ of the original area.