When the positive integer N is divided by 15, 20, and 24, it leaves a remainder of 8 in ea… — State PSC Exam

When the positive integer $N$ is divided by $15$ , $20$ , and $24$ , it leaves a remainder of $8$ in each case. If $N$ is the smallest such number that is also exactly divisible by $13$ , what is the value of $N$ ?

A
832
B
128
728Answer
D
1664

Answer

728

The Lowest Common Multiple (LCM) of 15, 20, and 24 is 120. Any number leaving a remainder of 8 when divided by these numbers can be expressed in the form 120k + 8. By testing integer values for k, we find that when k = 6, the number is 120(6) + 8 = 728. Since 728 is perfectly divisible by 13 (728 / 13 = 56), it satisfies all conditions and is the correct answer.

Step-by-Step Solution

Find the Lowest Common Multiple (LCM) of the given divisors: 15, 20, and 24.

LCM(15, 20, 24) = 120.

The base number without the remainder must be exactly divisible by all three numbers.

Set up a general algebraic expression for the number

N

N = 120k + 8, where k is a positive integer.

Because the number leaves a remainder of 8 in each case, it must be 8 more than a multiple of their LCM.

Apply the final divisibility condition to find the smallest valid

k

Test values of k in the expression (120k + 8) to see which result is divisible by 13.

The problem specifies that the final number

N

must be exactly divisible by 13.

Evaluate

120k + 8

for

k = 1, 2, 3...

until a multiple of 13 is reached.

For k = 6, N = 120(6) + 8 = 728. Checking divisibility: 728 / 13 = 56 (an exact integer).

Sequential testing guarantees we find the smallest possible value for

N

Key Concept

Applications of LCM and Modular Arithmetic

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