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frosja888 [35]
2 years ago
11

Francis works at Carlos Bakery and is making cookie trays. She has 48 chocolate chip cookies, 64 rainbow cookies, and 120 oatmea

l cookies to put on the trays. Part A: How many trays can Francis make under the following condition? Part B: How many of each type of cookie would fit on each of the trays? Justify your answer.
48 chocolate chip cookies = 1 batch

64 rainbow cookies = 1 batch

120 oatmeal cookies = 1 batch
What are the smallest number of batches of each type of cookie she would need to bake so that Francis has the same amount of chocolate chip, rainbow, and oatmeal cookies?
Mathematics
1 answer:
amm18122 years ago
4 0

The number of cookies and trays are illustrations of greatest common factors.

  • The number of trays is 8
  • 6 chocolate chips, 8 rainbows and 15 oatmeal cookies would fit each tray

The given parameters are:

\mathbf{Chocolate\ chip=48}

\mathbf{Rainbow=64}

\mathbf{Oatmeal=120}

<u>(a) The number of trays</u>

To do this, we simply calculate the greatest common factor of 48, 64 and 120

Factorize the numbers, as follows:

\mathbf{48 = 2 \times 2 \times 2 \times 2 \times 3}

\mathbf{64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2}

\mathbf{120 = 2 \times 2 \times 2 \times 3 \times 5}

So, the GCF is:

\mathbf{GCF= 2 \times 2 \times 2}

\mathbf{GCF= 8}

Hence, the number of tray is 8

<u>(b) The number of each type of cookie</u>

We have

\mathbf{Chocolate\ chip=48}

\mathbf{Rainbow=64}

\mathbf{Oatmeal=120}

Divide each cookie by the number of trays

So, we have:

\mathbf{Chocolate\ chip = \frac{48}{8} = 6}

\mathbf{Rainbow = \frac{64}{8} = 8}

\mathbf{Oatmeal = \frac{150}{8} = 15}

Hence, 6 chocolate chips, 8 rainbows and 15 oatmeal cookies would fit each tray

Read more about greatest common factors at:

brainly.com/question/11221202

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If a positive integer is equal to the following product: 25b3c425b3c4, where b and c are distinct prime numbers greater than 2,
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Answer: 64 distinct even factors

Step-by-step explanation:

let the 2 distinct numbers be b and c and the integer is expected to be 25b3c425b3c4

since b, c > 2 and prime numbers,

potential options include 3, 5, and 7

hence the likelihoods are (b = 3, c = 5), (b = 5, c = 3), (b = 3, c = 7), (b = 7, c = 3), (b = 5, c = 7), (b = 7, c = 5)

Possibility 1 (b = 3, c = 5)

integer is now 253354253354

the distinct even factors = 2, 202, 262, 1934, 19802, 26462, 195334, 253354, 2000002, 2594062, 19148534, 25588754, 262000262, 1934001934, 2508457954, 253354253354

number of distinct even factors = 16

Possibility 2 (b = 5, c = 3)

integer is now 255334255334

the distinct even factors = 2, 86, 202, 5938, 8686, 19802, 253354, 599738, 851486, 2000002, 25788734, 58792138, 25588754, 86000086, 2528061934, 1934001934, 5938005938, 253354253354

number of distinct even factors = 18

Possibility 3 (b = 3, c = 7)

integer is now 253374253374

the distinct even factors = 2, 6, 22, 66, 202, 242, 606, 698, 726, 2094, 2222, 6666, 7678, 19802, 23034, 24442, 59406, 70498, 73326, 84458, 211494, 217822, 253374, 653466, 775478, 2000002, 2326434, 2396042, 6000006, 6910898, 7188126, 8530258, 20732694, 22000022, 25590774, 66000066, 76019878, 228059634, 242000242, 698000698, 726000726, 836218658, 2094002094, 2508655974, 7678007678, 23034023034, 84458084458,253354253354

number of distinct even factors = 48

Possibility 4 (b = 7, c = 3)

integer is now 257334257334

the distinct even factors = 2, 6, 14, 22, 42, 66, 154, 202, 462, 606, 1114, 1414, 2222, 3342, 4242, 6666, 7798, 12254, 15554, 19802, 23394, 36762, 46662, 59406, 85778, 112514, 138614, 217822, 257334, 337542, 415842, 653466, 787598, 1237654, 1524754, 2000002, 2362794, 3712962, 4574262, 6000006, 8663578, 11029714, 14000014, 22000022, 25990734, 33089142, 42000042, 66000066, 77207998, 121326854, 154000154, 231623994, 363980562, 462000462, 849287978, 1114001114, 2547863934, 3342003342, 7798007798, 12254012254, 23394023394, 36762036762, 85778085778, 257334257334

number of distinct even factors = 64

Possibility 5 (b = 5, c = 7)

integer is now 255374255374

the distinct even factors = 2, 14, 34, 58, 74, 202, 238, 406, 518, 986, 1258, 1414, 2146, 3434, 5858, 6902, 7474, 8806, 15022, 19802, 24038, 36482, 41006, 52318, 99586, 127058, 138614, 216746, 255374, 336634, 574258, 697102, 732674, 889406, 1517222, 2000002, 2356438, 3684682, 4019806, 5128718, 9762386, 12455458, 14000014, 21247546, 25792774, 34000034, 58000058, 68336702, 74000074, 87188206, 148732822, 238000238, 361208282, 406000406, 518000518, 986000986, 1258001258, 2146002146, 2528457974, 6902006902, 8806008806, 15022015022, 36482036482, 255374255374

number of distinct even factors = 64

Possibility 6 (b = 7, c = 5)

integer is now 257354257354

the distinct even factors = 2, 202, 19802, 257354, 2000002, 25992754, 2548061954, 257354257354

number of distinct even factors = 8

From the six possibilities, the highest number of likely distinct even factors is 64

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