Grade 6 math and coding help for Ontario students

Source: Ontario’s Grade 6 mathematics curriculum (2020), Grade 6 strands and specific expectations, pulled on 22 October 2026 from the data service that feeds the ministry’s curriculum website. We teach independently and are not linked to the ministry or to any board.

1. Write each rule in words first. 2: the last digit is even. 3: the digit sum divides by 3. 4: the last two digits divide by 4. 5: the last digit is 0 or 5. 6: it passes both 2 and 3. 8: the last three digits divide by 8. 9: the digit sum divides by 9. 10: the last digit is 0.
2. Spot the work that repeats. The digit sum is needed twice, for 3 and for 9, so it becomes its own small function.
3. Draw the flow chart. One decision diamond per rule, each with a yes branch that adds the divisor to a list.
4. Find the free answer. The test for 6 needs no new arithmetic at all: if 2 and 3 are already on the list, 6 goes on too. That reuse is the efficiency.
5. Code it and run it on three numbers chosen to behave differently: one that passes everything, one that passes some, one that passes few.
6. Check the rules against real division. We had the computer compare our rule-based list with plain division for every whole number from 1 to 10,000. There were no mismatches.

# Grade 6: which divisibility tests does a number pass? (ours)
def digit_sum(n):
    total = 0
    for d in str(n):
        total = total + int(d)
    return total

def passes(n):
    tests = []
    last = n % 10
    if last % 2 == 0:
        tests.append(2)
    if digit_sum(n) % 3 == 0:
        tests.append(3)
    if n % 100 % 4 == 0:              # last two digits
        tests.append(4)
    if last == 0 or last == 5:
        tests.append(5)
    if 2 in tests and 3 in tests:      # 6 needs both: no new work
        tests.append(6)
    if n % 1000 % 8 == 0:             # last three digits
        tests.append(8)
    if digit_sum(n) % 9 == 0:
        tests.append(9)
    if last == 0:
        tests.append(10)
    return tests

for n in [2520, 7416, 1155]:
    print(n, "passes", passes(n))

1. Yes, yes, yes. The digit sum is 7 + 4 + 1 + 6 = 18, which divides by 3 and by 9. For 8, look at the last three digits: 416 = 8 × 52. (Our worked program above gives the same list.)
2. 36. 10% of 240 is 24. 5% is half of that, 12. 15% is 24 + 12 = 36.
3. 12 degrees. From −7 up to 0 is 7, and from 0 up to 5 is 5 more. 7 + 5 = 12. On a vertical number line, count the steps.
4. 4.6 and 4.64. For tenths, look at the hundredths digit, 3, which is less than 5, so round down. For hundredths, look at the thousandths digit, 8, so round up.
5. 11/12. The lowest common denominator of 3 and 4 is 12. 2/3 = 8/12 and 1/4 = 3/12, so the sum is 11/12.
6. 2 × 2 × 2 × 3 × 3 × 5, or 2³ × 3² × 5. One tree: 360 = 36 × 10, then 36 = 4 × 9 and 10 = 2 × 5, then 4 = 2 × 2 and 9 = 3 × 3.
7. Linear; 3n + 1; 61. The pattern grows by 3 every time, the same amount, so it is linear. Term 1 is 3 × 1 + 1 = 4, which checks. Term 20 is 3 × 20 + 1 = 61.
8. x = 0, 1, 2, 3, 4, 5, 6, 7 or 8. Subtract 5 from both sides: 3x < 27, so x < 9. Check the edge: 3 × 8 + 5 = 29, which is less than 32, and 3 × 9 + 5 = 32, which is not.
9. 8. It counts the multiples of 6 from 1 to 50: 6, 12, 18, 24, 30, 36, 42 and 48. The if asks the divisibility question once per number.
10. The first loop runs 50 times; the second runs 8 times. Both print 8. The second is more efficient because it jumps straight from one multiple of 6 to the next instead of testing every number. This is the Grade 6 C3.2 question in miniature: same outcome, less work.
11. 8. The loop keeps adding 1 to x while 3x + 5 is still less than 32. It stops when x reaches 9, the first value that fails, so the largest solution is one less: 8. That matches Question 8, worked by code instead of algebra.
12. (2, −2), in Quadrant IV. Right 5 changes x from −3 to 2; down 4 changes y from 2 to −2. Positive x and negative y is the fourth quadrant.

Bonus answers. Six of the 36 equally likely outcomes give a sum of 7, so the theoretical probability is 6/36 = 1/6, about 0.17. The experiment gives 40/200 = 1/5 = 0.2. Experiments vary from the theory, especially with few trials; more rolls usually bring them closer. For the data set, sorted it is 9, 9, 14, 16, 20, 22: the mean is 90 ÷ 6 = 15, the median is halfway between 14 and 16, so 15, the mode is 9 and the range is 22 − 9 = 13. The trapezoid’s area is (8 + 12) ÷ 2 × 5 = 50 cm².