Grade 7 AMO Practice Paper 3

Answer: 79

The six-digit palindrome has the form 47__74. Checking the units digit of the product of any 2 consecutive integers (1×2=2, 2×3=6, 3×4=2, 4×5=0, 5×6=0, 6×7=2, 7×8=6, 8×9=2, 9×0=0) shows it is never 4, so 2 consecutive integers cannot work. Considering 3 consecutive integers, the units digit of the product is 4 only for the cases 2×3×4 and 7×8×9 (by the same unit-digit check). Testing candidates near the target size: 72×73×74 = 388944 (too small to start with 47), 82×83×84 = 571704 (too big), and 77×78×79 = 474474, which is indeed a six-digit palindrome beginning with 47. Therefore, the largest of the three consecutive integers is 79.

Answer: 25

Let N be the required number, with quotients a, b, c and common remainder R, so 266 = Na + R, 516 = Nb + R, and 741 = Nc + R. Subtracting pairs of these equations eliminates R: 741 − 516 = 225 = N(c − b); 741 − 266 = 475 = N(c − a); 516 − 266 = 250 = N(b − a). Since N must divide each of these differences, N is the greatest common factor of 225, 250, and 475. Since 225 = 3²×5², 250 = 2×5³, and 475 = 5²×19, the greatest common factor is 5² = 25. So N = 25.

Answer: 161

Let d be the distance from A to B, x be the distance from the car's current position to A, and S be Lean's biking speed. Since Lean is 5/17 of the way from A to B, he is (5/17)d from A and (12/17)d from B. Equal travel times give: going back to A, x/391 = [(5/17)d]/S, so S = 391 × [(5/17)d]/x; going forward to B, (x+d)/391 = [(12/17)d]/S, so S = 391 × [(12/17)d]/(x+d). Setting these equal and simplifying gives 7x = 5d. Substituting x = 5d/7 into the first expression: S = 391 × [(5/17)d]/(5d/7) = 391 × 7/17 = 161. So Lean's biking speed is 161 km/h.

Answer: 3

Since the car and train move in the same direction, the car closes the length of the train at the relative speed of 22 − 17 = 5 km/h. The train's length is 250 m = 1/4 km. The time to close this gap is (1/4 km) ÷ (5 km/h) = 1/20 hour = (1/20) × 60 minutes = 3 minutes. So M = 3.

Answer: 12

Consider two cases based on the sign of N − 3. Case I: If N − 3 ≥ 0, then 2 < N − 3 < 5, so 5 < N < 8, giving integer values N = 6, 7. Case II: If N − 3 < 0, then 2 < −(N − 3) < 5, i.e., 2 < 3 − N < 5, so −2 < N < 1, giving integer values N = −1, 0. The sum of these 4 values is 6 + 7 + 0 + (−1) = 12.

Answer: 7

For 2/N, 3/N and 5/N to all be in lowest terms, N cannot share a common factor with 2, 3, or 5, so N cannot be divisible by 2, 3, or 5. Since N must also be composite, N must be a product of two or more primes from {7, 11, 13, 17, 19, ...}. The composites formed from these primes that fall between 20 and 150 are 7×7=49, 7×11=77, 7×13=91, 7×17=119, 7×19=133, 11×11=121, and 11×13=143 (7×23=161 and 13×13=169 both exceed 150). There are 7 possible values of N.

Answer: 513

Clue 3: The only prime factor of the 4-digit number is 11, so the number is a power of 11: 11¹, 11², 11³, etc. Of these, only 11³ = 1331 has 4 digits, so the 6-digit number has the form _1331_. Clue 1: The number reads the same right to left, so the first and last digits are equal; call the number A1331A. Clue 2: The number is a multiple of 9, so its digit sum A + 1 + 3 + 3 + 1 + A = 2A + 8 must be a multiple of 9. No digit A gives 2A + 8 = 9, but 2A + 8 = 18 gives A = 5. So Gina's postal code is 513315, and the first 3 digits are 513.

Answer: 4

Let the two-digit number be 10a + b, with tens digit a and ones digit b. The first condition gives 10a + b = 7(a + b) + 6, which simplifies to a = 2b + 2. Since a is a digit from 1 to 9 and b is a digit from 0 to 9, and a = 2b + 2 ≤ 9 means b ≤ 3.5, b can be 0, 1, 2, or 3, giving (a, b) = (2, 0), (4, 1), (6, 2), (8, 3) — the numbers 20, 41, 62, and 83. There are 4 possible answers.

Answer: 65

Since the piles are unordered, fix the smallest pile's size and count the ways to split the remaining marbles into two odd piles, each at least as large as the smallest pile. Case 1: smallest pile has 1 marble: (1,51), (3,49), ..., (25,27) — 13 ways. Case 2: smallest pile has 3: (3,47), (5,45), ..., (25,25) — 12 ways. Case 3: smallest pile has 5: (5,43), (7,41), ..., (23,25) — 10 ways. Case 4: smallest pile has 7: (7,39), (9,37), ..., (23,23) — 9 ways. Case 5: smallest pile has 9: (9,35), (11,33), ..., (21,23) — 7 ways. Case 6: smallest pile has 11: (11,31), (13,29), ..., (21,21) — 6 ways. Case 7: smallest pile has 13: (13,27), (15,25), ..., (19,21) — 4 ways. Case 8: smallest pile has 15: (15,23), (17,21), (19,19) — 3 ways. Case 9: smallest pile has 17: (17,19) — 1 way. Total: 13+12+10+9+7+6+4+3+1 = 65 ways.

Answer: 12

Every two-digit prime number is odd (since even two-digit numbers greater than 2 are not prime), so its ones digit is automatically odd. It remains to find, among the two-digit primes, those whose tens digit is also odd (1, 3, 5, 7, or 9): disregarding all two-digit primes with an even tens digit leaves 11, 13, 17, 19, 31, 37, 53, 59, 71, 73, 79, and 97. There are 12 such prime numbers.