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Question

Difficulty: HardFactors, Multiples, and Prime Factorization

An event coordinator is preparing identical welcome packets for a conference. She has 7272 promotional pens, 108108 notebooks, and 130130 keychains. She plans to distribute all of the pens and notebooks into the maximum number of identical packets such that no pens or notebooks are left over. She also wants to distribute the keychains equally among these packets. Finally, the coordinator requires that the total number of items (pens, notebooks, and keychains combined) in each packet is a multiple of 66. To meet these conditions, she must purchase additional keychains. What is the minimum number of additional keychains she needs to buy?

  1. A
    14
  2. B
    50
  3. C
    86
  4. 122Answer
  5. E
    230

Answer

122
To maximize the number of identical packets using all 72 pens and 108 notebooks without leftovers, we find the greatest common factor (GCF) of 72 and 108, which is 36. This means 36 packets are created, with each packet containing 2 pens and 3 notebooks. If each packet contains q keychains, the total number of keychains is 36q. Since she starts with 130 keychains, we must have 36qgeq13036q \\geq 130, which means qgeq4q \\geq 4. The total number of items in each packet is 2+3+q=5+q2 + 3 + q = 5 + q. For this total to be a multiple of 6, and given qgeq4q \\geq 4, the smallest possible integer value for q is 7 (since 5+7=125 + 7 = 12). With 7 keychains per packet, the total number of keychains needed is 36times7=25236 \\times 7 = 252. Subtracting the 130 keychains she already has, she must buy a minimum of 122 keychains.

Step-by-Step Solution

1
Find the maximum number of packets that can be created using all 72 pens and 108 notebooks.
The number of packets must be the greatest common factor (GCF) of 72 and 108, which is 36.
Since all pens and notebooks must be distributed equally with none left over, the number of packets must divide both 72 and 108. To maximize the packets, we find their GCF.
2
Calculate the number of pens and notebooks in each of the 36 packets.
Each packet contains 2 pens (72div3672 \\div 36) and 3 notebooks (108div36108 \\div 36).
Dividing the total quantities of each item by the number of packets gives the quantity of that item per packet.
3
Set up the inequality for the number of keychains per packet, q.
Since she already has 130 keychains, the total keychains needed is 36qgeq13036q \\geq 130, which simplifies to qgeq3.61q \\geq 3.61. Since q must be an integer, qgeq4q \\geq 4.
The keychains must be distributed equally, meaning each packet gets an integer number of keychains q. She must buy additional keychains, so the total keychains must be at least 130.
4
Apply the constraint that the total items per packet must be a multiple of 6 to find the minimum value of q.
The total items per packet is 2+3+q=5+q2 + 3 + q = 5 + q. The smallest integer qgeq4q \\geq 4 that makes 5+q5 + q a multiple of 6 is q=7q = 7 (since 5+7=125 + 7 = 12, which is a multiple of 6).
This satisfies the divisibility condition for the total packet size while respecting the minimum quantity of keychains.
5
Calculate the number of additional keychains to buy.
Total keychains needed is 36times7=25236 \\times 7 = 252. The number of additional keychains to buy is 252−130=122252 - 130 = 122.
Subtracting the starting quantity of keychains from the total required quantity gives the number of additional keychains to purchase.

Key Concept

Using the Greatest Common Factor (GCF) to solve division and optimization problems with multiple constraints.

Alternative Method

Alternatively, a student can test the answer choices by working backwards. Add each choice to 130 to find the total keychains, check if the result is divisible by 36 (the GCF of 72 and 108), and verify if the resulting number of keychains per packet makes the total number of items per packet a multiple of 6.
Estimated Time:2m 0s
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Topics

Factors, Multiples, and Prime Factorization

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