Test: Divisibility - 1

Given: 108 = 2² × 3³

Since 8t is divisible by 96, we may write

8t = 96k, where k is a positive integer

• 2³t = 2⁵ × 3× k
• t = 2² × 3× k
• t³ = 2⁶ × 3³ × k³
• t³ = ( 2² × 3³ )( 2⁴  × k³)
• t³ = 108( 2⁴  × k³)

This shows that t³ is completely divisible by 108, implying that the remainder is 0.

Answer: Option (A)

Step 1: Question statement and Inferences

32455 × 3145208 × K² is divisible by 3

• Either 32455 or 3145208 or K² must be divisible by 3

Step 2: Finding required values

Given:

A number is divisible by 3 when sum of its digits are divisible by 3

• Sum of the digits of 32455 = 3+2+4+5+5 = 19 not divisible by 3
• Sum of the digits of 3145208 = 3+1+4+5+2+0+8 = 23 not divisible by 3
• Therefore 32455 and 3145208 are not divisible by 3.

Hence, K² should be divisible by 3

• K should be divisible by 3 (3 is prime, Kⁿ and K will have same prime factors)

Step 3: Calculating the final answer

Checking for all the options:

• Sum of the digits of 6000209 = 6+0+0+0+2+0+9 = 17, not divisible by 3
• Sum of the digits of 6111209 = 6+1+1+1+2+0+9 = 20, not divisible by 3
• Sum of the digits of 6111309 = 6+1+1+1+3+0+9 = 21, divisible by 3
• Sum of the digits of 6111109 = 6+1+1+1+1+0+9 = 19, not divisible by 3
• Sum of the digits of 6111809 = 6+1+1+1+8+0+9 = 26, not divisible by 3

Only 6111309 is divisible by 3

Answer: Option (C)

Steps 1 & 2: Understand Question and Draw Inferences

t² + 1 can be evenly divided by 10 if t² + 1 = 10 *k, where k is a positive integer

• We need to check if the last digit of t² + 1 is 0

Step 3: Analyze Statement 1

91⁶ × t leaves a remainder of 1 when divided by 2

• t is odd
• last digit of t can be 1, 3, 5, 7 or 9

Not Sufficient.

Step 4: Analyze Statement 2

91⁶ × t leaves a remainder of 2 when divided by 5

• last digit of 91⁶ × t can be 2 or 7
• last digit of t can be 2 or 7

Not Sufficient

Step 5: Analyze Both Statements Together (if needed)

Inference from statement 1: last digit of t can be 1, 3, 5, 7 or 9

Inference from statement 2: last digit of t can be 2 or 7

Inference from statement 1 and statement 2: last digit of t can be 7

--> Last digit of t² + 1 = Last digit of (9 + 1) = 0

Hence, t² + 1 can be evenly divided by 10

Statement 1 and Statement 2 together are sufficient to answer the question.

 Answer: Option (C)