Test Level 1: Number System - 1 Solved MCQs CAT

The units digit in this case would obviously be ‘0’ because the given expression has a pair of 2 and 5 in it’s prime factors.

The two numbers should be factors of 405. A factor search will yield the factors. (look only for 2 digit factors of 405 with sum of digits between 1 to 19).
Also 405 = 5 × 3⁴. Hence: 15 × 27
45 × 9 are the only two options. 
From these factors pairs only the second pair gives us the desired result. 
i.e. Number × sum of digits = 405.
Hence, the answer is 45.

• it's a very simple question,
• Let D=0.abababab        (say - 1)
• Multiply both sides by 100 i.e
• 100D = ab.abababab     (say - 2)
• now subtract 1 from 2. That gives
• 99D = ab
• D = ab/99.
• hence out of all the options if we multiply 198 by ab/99 we get
• = (ab / 99)x 198 = 2ab
• hence we got an integer.
• So option B is the right answer.

For 381A to be divisible by 9, the sum of the digits 3 + 8 + 1 + A should be divisible by 9.
For that to happen A should be 6.
Option (d) is correct.

LCM of 5, 15 and 20 = 60.
HCF of 5, 15 and 20 = 5.
The required ratio is 60:5 = 12:1

Only the first option can be verified to be true in this case.
If A is even, 3A would always be divisible by 6 as it would be divisible by both 2 and 3.
Options b and c can be seen to be incorrect by assuming the value of A as 4.

For a number to be divisible by 6, it must be divisible by both 2 and 3.
Therefore, For the number 15B to be divisible by 6. it must satisfy two conditions:

1. Divisibility by 2: The number must be even.
2. Divisibility by 3: The sum of its digits must be divisible by 3.

Since 15B represents a number with digits 1, 5, and B, let’s analyze each option for B:

• For divisibility by 2, B must be even (i.e., 0, 2, 4, 6, or 8) because an even B makes 15B an even number.
• For divisibility by 3, the sum 1+5+B=6+B must be divisible by 3.

Since 6 is already divisible by 3, any even B (0, 2, 4, 6, or 8) will keep the sum divisible by 3.

The correct answer is: (d) Both (a) and (c)

Also, the value of B is divisible by 6. Therefore, both options are correct, B will be even and B will be divisible by 6.

Let’s re‐evaluate each term’s units digit carefully:

1. 25⁶²⁵¹
   Any power of a number ending in 5 ends in 5 (once past the first power).
   ⇒ units digit = 5.

2. 36⁵²⁸
   Any power of a number ending in 6 ends in 6.
   ⇒ units digit = 6.

3. 73⁵⁴
   We only care about the base’s units digit, 3. The cycle for 3ⁿ mod 10 is length 4:

   • 3¹ ≡ 3

   • 3² ≡ 9

   • 3³ ≡ 7

   • 3⁴ ≡ 1
     Then it repeats every 4.
     Compute 54 mod 4 = 2.
     So 3⁵⁴ has the same units digit as 3², namely 9.

Now add the three units digits: 5 + 6 + 9 = 20 ⇒ units digit = 0.
Hence the correct answer is 0.

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The number of zeroes would be given by adding the quotients when we successively divide 1090 by 5 : 1090/5 + 218/5 + 43/5 + 8/5 = 218 + 43 + 8 + 1 = 270. Option (a) is correct.