Express 1500 as the product of its prime factors. Leave your answer in index notations. Answer ___ p=2m×32×np = 2^{m}\times 3^{2}\times n, where mm and nn are positive integers.
Find the values of mm and nn if the HCF of 1500 and pp is 6, given that 100<p<150100 < p < 150.
Answer m = ___ n = ___ In a race, Car A, Car B and Car C took 3 minutes, 6 minutes and 8 minutes respectively to complete one lap around the racing track.
If the race started at 7 pm, at what time will the cars be back at the starting point together again?
Answer ___

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Worked solution

17(a)(i)

1. Break 1500 down into prime factors and express repeated factors using indices.

1500=15×1001500 = 15 \times 100
=(3×5)×(22×52)= (3 \times 5) \times (2^2 \times 5^2)
=22×3×53= 2^2 \times 3 \times 5^3

Answer

1500=22×3×531500 = 2^2 \times 3 \times 5^3

17(a)(ii)(m)

1. Compare the power of 2 in 1500, in $p$, and in the given HCF.

From 17(a)(i),
1500=22×3×531500=2^2\times 3\times 5^3.

Given
p=2m×32×np=2^m\times 3^2\times n and the HCF is 6=2×36=2\times 3.

For the factor 2, the HCF uses the smaller power of 2 from 15001500 and pp:
min⁡(2,m)=1.\min(2,m)=1.
Since mm is a positive integer, m=1m=1.

Answer

m=1m=1

17(a)(ii)(n)

1. Substitute $m=1$ and use the range of $p$ to find possible integer values of $n$, then check the HCF condition.

Since m=1m=1,
p=21×32×n=18n.p=2^1\times 3^2\times n=18n.
Given 100<p<150100<p<150,
100<18n<150100<18n<150
5.56<n<8.33.5.56<n<8.33.
So n=6,7,n=6,7, or 88.

Check with 1500=22×3×531500=2^2\times 3\times 5^3:

  • If n=6n=6, then p=108p=108 and the HCF is more than 66.
  • If n=7n=7, then p=126=2×32×7p=126=2\times 3^2\times 7, so the HCF with 15001500 is 2×3=62\times 3=6.
  • If n=8n=8, then p=144p=144 and the HCF is more than 66.

Therefore, n=7.n=7.

Answer

n=7n=7

17(b)

1. Find the first time interval when all three cars return to the starting point together.

The cars will be back at the starting point together after the LCM of 33, 66 and 88 minutes.

3=3,6=2×3,8=233=3,\quad 6=2\times3,\quad 8=2^3

LCM=23×3=24\text{LCM}=2^3\times3=24

So they meet again after 2424 minutes.

2. Add the common time interval to the starting time.

Race started at 7.007.00 pm.

7.00 pm+24 minutes=7.24 pm7.00\text{ pm}+24\text{ minutes}=7.24\text{ pm}

Answer

7.247.24 pm

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