When the positive integer n is divided by 7, the remainder is 3, and when n is divided by … — GRE General Test

When the positive integer $n$ is divided by 7, the remainder is 3, and when $n$ is divided by 11, the remainder is 5. If $n$ is a three-digit integer less than 200 that is divisible by 6, what is the value of $n$ ?

Answer: 192

Answer

The value of

n

is 192.

To find

n

, combine the remainder requirements

n \equiv 3 \pmod 7

and

n \equiv 5 \pmod{11}

. The smallest positive integer solution is

38

. Because

\text{lcm}(7, 11) = 77

, all valid integers take the form

n = 77m + 38

. Restricting

n

to three-digit numbers less than 200 gives candidates

n = 115

(

m=1

) and

n = 192

(

m=2

). Among these, only 192 is divisible by 6 (since

192 = 6 \times 32

Step-by-Step Solution

Formulate remainder congruences for the given conditions

n \equiv 3 \pmod{7}

and

n \equiv 5 \pmod{11}

Dividing

n

by 7 leaves remainder 3, and dividing by 11 leaves remainder 5.

Find the smallest positive integer satisfying both congruences

n = 38

Checking values

11k + 5

5, 16, 27, 38

38 \div 7 = 5

remainder 3, so 38 satisfies both conditions.

Determine the general solution for

n

using the Chinese Remainder Theorem logic

n = 77m + 38

for integer

m \ge 0

Since 7 and 11 are coprime, the solutions repeat every

\text{lcm}(7, 11) = 77

Evaluate candidate values for

n

such that

100 \le n < 200

For

m=1

n = 115

; for

m=2

n = 192

These are the only three-digit integers less than 200 of the form

77m + 38

Apply the final constraint that

n

must be divisible by 6

192

is divisible by 6 (

192 = 6 \times 32

)

115

is odd, so it is not divisible by 6.

192

is even and the sum of its digits (

1+9+2=12

) is a multiple of 3, so it is divisible by 6.

Key Concept

Simultaneous congruences and combined divisibility rules (Chinese Remainder Theorem)

Estimated Time:2m 0s

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