A logistics manager is arranging shipping containers into equal stacks. When the container… — State PSC Exam

A logistics manager is arranging shipping containers into equal stacks. When the containers are stacked in groups of $18$ , $24$ , or $32$ , there are always exactly $11$ containers left over. However, when they are stacked in groups of $35$ , there are no containers left over. What is the minimum total number of containers the manager could have?

Answer: 875 containers

Answer

875

The correct answer is derived by recognizing that the total number of containers must take the form of

LCM(18, 24, 32) \times k + 11

. The LCM is

288

, so the number is

288k + 11

. We then find the smallest integer

k

such that

(288k + 11)

is perfectly divisible by

35

. By evaluating the expression for sequential values of

k

(or using modular arithmetic,

8k \equiv 24 \pmod{35}

), we find that

k = 3

is the smallest valid multiplier. Substituting this back yields

288(3) + 11 = 875

Step-by-Step Solution

Calculate the Least Common Multiple (LCM) of the initial stack group sizes (

18

24

, and

32

The prime factorizations are

18 = 2 \times 3^2

24 = 2^3 \times 3

, and

32 = 2^5

. The LCM is

2^5 \times 3^2 = 32 \times 9 = 288

The base cycle for the stacks without remainders requires finding the smallest number perfectly divisible by all three group sizes.

Express the total number of containers algebraically.

Let the total number of containers be

N

. We can write

N = 288k + 11

, where

k

is a positive integer.

Since there is always a remainder of

11

containers when divided by these numbers, the total must be

11

more than a multiple of their LCM.

Apply the secondary divisibility condition.

The problem states that

N

is exactly divisible by

35

. Therefore,

288k + 11 \equiv 0 \pmod{35}

When the containers are grouped by

35

, there is no remainder.

Solve the congruence to find the smallest integer

k

First, simplify

288 \pmod{35}

: since

35 \times 8 = 280

, we have

288 \equiv 8 \pmod{35}

. The equation becomes

8k + 11 \equiv 0 \pmod{35}

, or

8k \equiv -11 \equiv 24 \pmod{35}

. Dividing both sides by

8

gives

k = 3

Finding the smallest valid positive integer

k

will yield the minimum total number of containers.

Calculate the final number of containers using

k = 3

N = 288(3) + 11 = 864 + 11 = 875

Substitute the multiplier back into the original algebraic expression.

Key Concept

Solving for an unknown quantity using the Least Common Multiple (LCM) combined with constant remainder logic and a secondary divisibility condition.

Alternative Method

Instead of using modular arithmetic to solve

288k + 11 \equiv 0 \pmod{35}

, you can manually test integer values for

k

in the formula

N = 288k + 11

. For

k=1

N = 288(1) + 11 = 299

(

299 \div 35

leaves remainder

19

). For

k=2

N = 288(2) + 11 = 587

(

587 \div 35

leaves remainder

27

). For

k=3

N = 288(3) + 11 = 875

(

875 \div 35 = 25

with no remainder). Therefore,

875

is the smallest valid number.

Estimated Time:2m 30s

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

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