An event organizer is arranging chairs for a large conference. When the chairs are arrange… — State PSC Exam

An event organizer is arranging chairs for a large conference. When the chairs are arranged in rows of $18$ , $24$ , or $36$ , there are always exactly $5$ chairs left over. However, when the chairs are arranged in rows of $13$ , all chairs are perfectly accommodated with none left over. What is the minimum possible total number of chairs the organizer has?

Answer: 221

Answer

The minimum possible total number of chairs is 221.

The correct answer is derived by first establishing that any number leaving a remainder of

5

when divided by

18

24

, and

36

must be of the form

72k + 5

, where

72

is the LCM of the divisors. By systematically checking values of

k

, we find that

k=3

is the smallest integer that makes the expression

(72k + 5)

perfectly divisible by

13

, resulting in

72(3) + 5 = 221

Step-by-Step Solution

Set up the conditions for the total number of chairs mathematically.

Let the total number of chairs be

N

. We are given

N \equiv 5 \pmod{18}

N \equiv 5 \pmod{24}

N \equiv 5 \pmod{36}

, and

N \equiv 0 \pmod{13}

Translating the word problem into modular arithmetic helps systematically apply the concepts of LCM and divisibility.

Find the Least Common Multiple (LCM) of the first set of divisors.

The divisors are

18

24

, and

36

. Their prime factorizations are

18 = 2 \times 3^2

24 = 2^3 \times 3

, and

36 = 2^2 \times 3^2

. The LCM is

2^3 \times 3^2 = 8 \times 9 = 72

Any number that leaves the same remainder when divided by multiple divisors must be a multiple of their LCM plus that remainder.

Express

N

using the LCM and the common remainder.

Since

N

leaves a remainder of

5

when divided by

18

24

, or

36

, we can write

N = 72k + 5

, where

k

is a non-negative integer (

k = 0, 1, 2, \dots

This general formula captures all possible numbers of chairs that satisfy the first condition.

Apply the final divisibility condition to find

k

We require

N

to be perfectly divisible by

13

, meaning

72k + 5 \equiv 0 \pmod{13}

This guarantees the solution satisfies the second condition where arranging chairs in rows of 13 leaves no remainder.

Simplify the congruence modulo 13 and solve for

k

Divide

72

13

to find the remainder:

72 = 13 \times 5 + 7

. So,

72k \equiv 7k \pmod{13}

. The equation becomes

7k + 5 \equiv 0 \pmod{13}

. Testing values for

k

: if

k=1

7(1)+5=12

(not divisible); if

k=2

7(2)+5=19

(not divisible); if

k=3

7(3)+5=26

(divisible by

13

, since

26 = 13 \times 2

). The smallest valid

k

3

Finding the smallest non-negative integer

k

ensures we find the minimum possible number of chairs.

Calculate the final value of

N

N = 72(3) + 5 = 216 + 5 = 221

Substituting

k=3

back into our general formula gives the final answer.

Key Concept

Solving simultaneous remainder and divisibility conditions using the Least Common Multiple (LCM).

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