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Find the GCF(Greatest Common Factor) of 78 and 86.(1 point)

Responses

2
2

1
1

86
86

3

9 answers

To find the GCF (Greatest Common Factor) of 78 and 86, we can start with the prime factorization of each number:

Prime factorization of 78:

  • 78 = 2 × 39
  • 39 = 3 × 13
  • Therefore, 78 = 2 × 3 × 13

Prime factorization of 86:

  • 86 = 2 × 43
  • Therefore, 86 = 2 × 43

Now we can identify the common factors:

  • The only common factor of 78 and 86 is 2.

Thus, the GCF of 78 and 86 is 2.

So the answer is:

2

What is the greatest common factor of 45 and 54?(1 point)
Responses

9
9

1
1

45
45

3

To find the greatest common factor (GCF) of 45 and 54, we can start with the prime factorization of each number:

Prime factorization of 45:

  • 45 = 5 × 9
  • 9 = 3 × 3
  • Therefore, 45 = 3^2 × 5

Prime factorization of 54:

  • 54 = 2 × 27
  • 27 = 3 × 9
  • 9 = 3 × 3
  • Therefore, 54 = 2 × 3^3

Now we can identify the common factors:

  • The prime factors of 45 are 3^2 and 5.
  • The prime factors of 54 are 2 and 3^3.
  • The common prime factor is 3, and the lowest power of 3 in both factorizations is 3^2 (which is 9).

Thus, the GCF of 45 and 54 is 9.

So the answer is:

9

Find the greatest common factor of 17 and 51.(1 point)
Responses

3
3

51
51

17
17

1

To find the greatest common factor (GCF) of 17 and 51, we can use the fact that 17 is a prime number.

Prime factorization:

  • 17 is prime, so its only factors are 1 and 17.
  • 51 can be factored as 3 × 17.

The common factors of 17 and 51 are:

  • 1 and 17.

Therefore, the GCF of 17 and 51 is 17.

So the answer is:

17

A rabbit is packing lunches for his children. He has 90 carrots and 72 leaves of lettuce. If he is giving each bunny the same number of carrots and the same number of leaves of lettuce, how many baby bunnies must he have? He will have no extra vegetables after he packs the lunches.(1 point)
Responses

18
18

9
9

6
6

3

To determine how many baby bunnies there are, we need to find the greatest common factor (GCF) of the number of carrots (90) and leaves of lettuce (72). The GCF will tell us how many bunnies he can evenly distribute the carrots and lettuce among, with none left over.

Finding the GCF:

  1. Prime factorization of 90:

    • \(90 = 2 \times 3^2 \times 5\)

  2. Prime factorization of 72:

    • \(72 = 2^3 \times 3^2\)

  3. Identifying the common factors:

    • For \(2\): the minimum power is \(2^1\) (from 90).
    • For \(3\): the minimum power is \(3^2\) (both have \(3^2\)).
    • For \(5\): it only appears in 90, so it is not counted.

  4. Calculating the GCF:

    • The GCF = \(2^1 \times 3^2 = 2 \times 9 = 18\).

Thus, the rabbit can pack the lunches evenly among 18 baby bunnies.

So the answer is:

18

A carnival game gives variety bags as prizes. The game operator uses 100 balloons and 68 stickers to put into a certain number of bags. How many balloons and how many stickers can go in each bag if they make the greatest number of bags possible so that each bag has the same number of balloons and the same number of stickers?(1 point)
Responses

25 balloons and 17 stickers
25 balloons and 17 stickers

17 balloons and 25 stickers
17 balloons and 25 stickers

50 balloons and 34 stickers
50 balloons and 34 stickers

4 balloons and 4 stickers

To determine how many balloons and how many stickers can go in each bag while maximizing the number of bags, we need to find the greatest common factor (GCF) of the number of balloons (100) and the number of stickers (68). The GCF will indicate the largest number of bags possible, with each bag containing the same number of balloons and stickers.

Finding the GCF:

  1. Prime factorization of 100:

    • \(100 = 2^2 \times 5^2\)

  2. Prime factorization of 68:

    • \(68 = 2^2 \times 17\)

  3. Identifying the common factors:

    • For \(2\): the minimum power is \(2^2\) (from both).
    • There are no other common factors.

  4. Calculating the GCF:

    • The GCF = \(2^2 = 4\).

Now that we know the GCF is 4, we can calculate how many balloons and stickers can go into each bag:

  • Number of balloons per bag: \[ \frac{100}{4} = 25 \text{ balloons} \]

  • Number of stickers per bag: \[ \frac{68}{4} = 17 \text{ stickers} \]

Thus, each bag can contain 25 balloons and 17 stickers.

So the answer is:

25 balloons and 17 stickers.