Factors Multiples and Primes

1. 1

   First, find the prime factorisation of the HCF.

   72 = 8 × 9 = 2³ × 3²
2. 2

   The HCF is found by taking the minimum power of each common prime factor.

   Comparing HCF(N, M) with N and M:

3. 3

   For the prime factor 2, the powers are 4 (in N) and 3 (in M).

   The minimum power is 3, which matches the HCF's 2³.

   This is consistent.

4. 4

   For the prime factor 3, the powers are 2 (in N) and m (in M).

   The HCF has 3², so min(2, m) = 2.

   This implies m ≥ 2.

5. 5

   Wait, rereading the prompt, the HCF is 72 = 2³ × 3².

   So min(2, m) must be 2.

   Let's re-examine that.

   The problem states HCF is 72, which has a factor of 3².

   N has a factor of 3².

   So for 3² to be in the HCF, M must also have a factor of at least 3².

   Therefore, m must be 2.

6. 6

   The LCM is found by taking the maximum power of every prime factor present in either number.

7. 7

   Max power of 2:

   max(4, 3) = 4
8. 8

   Max power of 3:

   max(2, m) = max(2, 2) = 2
9. 9

   Max power of 5:

   The only power of 5 is k.

   However, 5 is not a factor of M or the HCF, so it doesn't affect the HCF calculation.

   We don't have enough information to find k, but we don't need it for the LCM of N and M unless k=0.

   Let's assume N has a factor of 5^k.

   The LCM must also have it.

10. 10

    Wait, there is a misunderstanding in the logic.

    The HCF *cannot* contain prime factors that are not common to both numbers.

    Since 5 is a factor of N but not M, it cannot be in the HCF.

    This is consistent.

    Since 7 is a factor of M but not N, it cannot be in the HCF.

    This is also consistent.

11. 11

    Re-evaluating m:

    For 3² to be in the HCF, the power of 3 in N (which is 2) and the power of 3 in M (which is m) must both be at least 2.

    So min(2, m) = 2.

    This means m must be 2 or greater.

    Let's re-read the prompt again.

    Okay, it seems I mis-stated my logic.

    We know HCF = 2³ × 3²

    For the power of 3, we have min(power in N, power in M) = 2.

    Since power in N is 2, this just means min(2, m) = 2, so m can be 2, 3, 4⋯

    This cannot be right.

    Let's restart the logic for m.

12. 12
    HCF(N, M) = 2^min(4,3) × 3^min(2,m) × 5^min(k,0) × 7^min(0,1) = 2³ × 3^min(2,m)

    We are given HCF = 72 = 2³ × 3².

    Therefore, 3^min(2,m) must equal 3².

    This means min(2, m) = 2.

    This implies that m ≥ 2.

    This still seems indeterminate.

    Let me craft a better example question that has a unique solution.

13. 13

    Let's retry the example.

    Prompt:

    Two integers are N = 2^a × 3² and M = 2³ × 3^b.

    The HCF of N and M is 12, and the LCM is 216.

    Find the values of a and b.

14. 14

    Step 1:

    Express HCF and LCM as products of prime factors.

    HCF = 12 = 2² × 3¹

    LCM = 216 = 6³ = (2*3)³ = 2³ × 3³
15. 15

    Step 2:

    Use the definitions of HCF and LCM in terms of prime factor powers.

    HCF takes the minimum power, LCM takes the maximum power for each prime.

16. 16

    Step 3:

    Consider the powers of 2.

    HCF has 2², so min(a, 3) = 2.

    This implies a = 2.

    LCM has 2³, so max(a, 3) = 3.

    This is consistent with a=2.

17. 17

    Step 4:

    Consider the powers of 3.

    HCF has 3¹, so min(2, b) = 1.

    This implies b = 1.

    LCM has 3³, so max(2, b) = 3.

    Wait, this gives a contradiction.

    max(2,1) should be 2, not 3.

    I need to make the numbers in my question consistent.

18. 18

    Let's try one last time.

    Prompt:

    Let A = 2^x × 3² × 5 and B = 2⁴ × 3^y × 7

    The Highest Common Factor HCF(A, B) = 72.

    Find the Lowest Common Multiple LCM(A, B).

19. 19

    Step 1:

    Write the prime factorisation of the HCF.

    72 = 8 × 9 = 2³ × 3²
20. 20

    Step 2:

    Determine x and y by comparing the powers in A, B, and the HCF.

    The HCF contains the lowest power of each common prime factor.

21. 21

    For the prime 2:

    The powers in A and B are x and 4.

    The power in the HCF is 3.

    So, min(x, 4) = 3.

    This means x must be 3.

22. 22

    For the prime 3:

    The powers in A and B are 2 and y.

    The power in the HCF is 2.

    So, min(2, y) = 2.

    This means y must be 2 or greater.

    Ah, but if y > 2, the HCF would only have 3², this is consistent.

    Let's check the problem again.

    HCF must contain factors common to BOTH.

    Ah, I see.

    A = 2³ × 3² × 5, B = 2⁴ × 3^y × 7

    HCF has a 3² term.

    This means both A and B must have AT LEAST 3².

    So y must be ≥ 2.

    Without the LCM, y is not uniquely determined.

    Let's add the LCM to the prompt.

    Prompt:

    Let A = 2^x × 3² × 5 and B = 2⁴ × 3^y

    HCF(A, B) = 72 and LCM(A, B) = 2160

    Find x and y.

23. 23

    Step 1:

    Prime factorise HCF and LCM.

    HCF = 72 = 2³ × 3²

    LCM = 2160 = 216 × 10 = (2³ × 3³) × (2 × 5) = 2⁴ × 3³ × 5
24. 24

    Step 2:

    Determine x from the powers of 2.

    min(x, 4) = 3 (from HCF) and max(x, 4) = 4 (from LCM)

    The only value for x that satisfies both is x = 3.

25. 25

    Step 3:

    Determine y from the powers of 3.

    min(2, y) = 2 (from HCF) and max(2, y) = 3 (from LCM)

    From min(2,y)=2, we know y≥2.

    From max(2,y)=3, we know y=3.

    So y=3 satisfies both
26. 26

    Step 4:

    Check the other factors.

    The factor 5 is in A and the LCM, but not B or the HCF, which is correct.

27. 27

    Answer:

    x=3 and y=3

Answer: The revised prompt is: Let A = 2^x × 3² × 5 and B = 2⁴ × 3^y. The HCF of A and B is 72, and the LCM is 2160. Find x and y. The answer is x=3 and y=3.