Grade 6 AMO Practice Paper 3

Answer: B

The ones, tens, hundreds, thousands and ten thousands place digits each consist of 1, 3, 5, 7 and 9, in some order. So, using 1 + 3 + 5 + 7 + 9 = 25 in each place value, the problem becomes 25 + 250 + 2500 + 25000 + 250000, which equals 277,775 and sum of its digits is 7 × 5 = 35.

Answer: C

Make lists and eliminate duplicates. Multiples of 2: 42, 44, 46, 48 and 50. Multiples of 3: 42, 45 and 48. Multiples of 5: 45, 50. There are 6 unique numbers that satisfy the requirements.

Answer: D

123456789 + 987654321 = 1111111110. Using the respective Divisibility Rules, 1111111110 is divisible by 2, 3, 5 and 9. It is also divisible by 6 since it is divisible by 2 and 3. From the Divisibility Rules of 4, 7 and 8, 1111111110 is not divisible by 4, 7 or 8. Obviously, 1111111110 is also divisible by 1. Hence there are 6 possible digits.

Answer: A

Every triangle consists of three vertices shown here in alphabetical order: △ABC, △ABD, △ABE, △ACD, △ACE, △ADE, △BCD, △BCE, △BDE and △CDE. These are the 10 possible triangles with vertices on the indicated points.

Answer: B

The page numbers facing him must be two consecutive whole numbers. 462 is more than 20² = 400. We can check that it is also more than 21² = 441 and less than 22² = 484. Multiplying 21 by 22 gives the desired product 462. Hence the sum is 21 + 22 = 43.

Answer: E

The numbers 1 through 38 are possible to obtain. The examples of how to obtain the first 23 numbers are shown below. The numbers 24 through 38 can be obtained similarly. 1, 1+1, 1+1+1, 4, 4+1, 4+1+1, 4+1+1+1, 4+4, 9, 9+1, 9+1+1, 9+1+1+1, 9+4, 9+4+1, 9+4+1+1, 9+4+1+1+1, 16, 16+1, 16+1+1, 16+1+1+1, 16+4, 16+4+1, 16+4+1+1. Thus, the least whole number score that is impossible to obtain is 39.

Answer: B

Let the age of the youngest brother be 1 unit, then the ages of the two older brothers are (2 units − 3) and 2 units. Hence 1 unit + (2 units − 3) + 2 units = 32 5 units − 3 = 32 5 units = 32 + 3 = 35 1 unit = 35 ÷ 5 = 7

Answer: C

The difference between the first term in the first group and the last term in the second group is 115 − 105 = 10. The difference between the second term of the first group and the second last term of the second group is also 114 − 104 = 10. This pattern continues for each of the rest of the term pairs. There are five pairs whose difference is 10, therefore the value is 5 × 10 = 50.

Answer: D

The single-digit prime numbers are 2, 3, 5 and 7. The single-digit perfect squares are 0, 1, 4 and 9, leaving 6 and 8 as neither prime nor perfect square. Choose the largest digit in each group and arrange from the greatest to least to get the largest possible number, 987. But 987 is divisible by 3. The greatest possible value of Larry's number is 985.

Answer: 94

Prime factorize 4096. 4096 = 2 × 2 × ... × 2 × 2 (12 factors) = 2¹². 2¹² can be expressed as (1) 2¹² (2) (2×2)⁶ = 4⁶ (3) (2×2×2)⁴ = 8⁴ (4) (2×2×2×2)³ = 16³ (5) (2×2×2×2×2×2)² = 64² So, all possible values of a are 2, 4, 8, 16, 64, 4096. The sum is 2 + 4 + 8 + 16 + 64 = 94.

Answer: C

If each dice shows 1, the total is 10. If each dice shows 6, the total is 60. All whole number sums from 10 to 60 inclusive are possible. These are all the whole numbers up to 60, except for 1 through 9. Then 60 − 9 = 51 different sums are possible.

Answer: A

If the hundreds digit is 2, then there is only 1 number 210. If the hundreds digit is 3, then there are only 3 numbers: 321, 320 and 310. If the hundreds digit is 4, then there are only 6 numbers: 432, 431, 430, 421, 420 and 410. If the hundreds digit is 5, then there are only 10 numbers: 543, 542, 541, 540, 532, 531, 530, 521, 520 and 510. The numbers 1, 3, 6 and 10 form the following pattern: 1, 3, 6, 10, 15, 21, 28, 36 (with differences +2, +3, +4, +5, +6, +7, +8). So, if the hundreds digit is 6, then there are only 15 numbers and so on for the rest of the hundreds digits. Thus, there are 1+3+6+10+15+21+28+36 = 120 such 3-digit numbers.

Answer: D

The pattern of the last digits of the powers of 2 is 2, 4, 8 and 6. Hence the last digit of 2¹⁰ is 4. The pattern of the last digits of the powers of 3 is 3, 9, 7 and 1. Hence the last digit of 3⁹ is 3. The ones digit of the sum is 4 + 3 = 7.

Answer: D

The middle 2 small cubes on each edge of the big cube have 2 faces painted each. A small cube on each vertex of the big cube has 3 faces painted. Since there are 12 edges and 8 vertices, the desired number is 12 × 2 + 8 = 32.

Answer: C

Let us count the number of unique games first. Each team plays with the other 9 teams. Therefore, there are 10 × 9 = 90 games. However, each game between any two teams was counted twice. For example, the games between A and B, AB and BA were counted in 90 games though it is the same game. Hence there are 90 ÷ 2 = 45 unique games. Since each unique game is played 3 times, the total number of games is 45 × 3 = 135.