Test Level 1: Number System - 1 - CAT MCQ

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.

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.

As we know an even number is divisible by 3 then it will always be divisible by 6.

Thus, it is clear that B will be an even number.

Now applying rule for divisibility by 3, as if sum of digits is divisible by 3 then the number will also be divisible by 3.

i.e. 1+5+B should be divisible by 3.

As B is an even number thus on equating B=  satisfies the divisibility rule of 3.

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

The units digit would be given by 5 + 6 + 9 (numbers ending in 5 and 6 would always end in 5 and 6 irrespective of the power and 3⁵⁴ will give a units digit equivalent to 3⁴ⁿ⁺² which would give us a unit digit of 3² i.e.9).

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.