What’s the CATch- 10?

In this article certain questions have been randomly taken from different topics of the quantitative section of CAT. Each question has been selected keeping in mind certain challenges you need to manage as a test taker. These challenges may range from concept basics to application orientation and test taking strategies.
How many numbers with distinct digits are possible product of whose digits is 28?
1. 4
2. 6
3. 8
4. None of these
Topic :Numbers
Approach :
Two digit numbers:
The two digits can be 4 and 7: Two possibilities 47 and 74.
Three-digit numbers:
The three digits can be 1, 4 and 7, so a total of 3! or 6 possibilities.
We cannot have three digits as (2, 2, 7) as the digits have to be distinct.
We cannot have numbers with 4 digits or more without repeating the digits.
So, there are 8 such numbers.
Answer: Option 3
Challenge: Understanding multiple possibilities in the product of numbers
What are the last two digits of the number 747?
1. 07
2. 49
3. 43
4. 01
Topic : Numbers
Approach :The last two digits of 71 are 07.
The last two digits of 72 are 49.
The last two digits of 73 are 43.
The last two digits of 74 are 01.
The last two digits of powers of 7 go in a cycle - 07, 49, 43, 01.
So, to find the last two digits of 747, divide the index 47 by 4; check the remainder which is 3, at which level the actual remainder is 43
Answer : Option 3
Challenge :Understanding cyclicity and symmetry in numbers
x2 + 5x + 6 is a multiple of 6. x is natural number less than 100. How many values can x take?
1. 33
2. 66
3. 55
4. None of these
Topic : Numbers
Approach: x2 + 5x + 6  = (x + 2) (x + 3)
Now (x + 2) and (x + 3) are two consecutive numbers. One of (x + 2) or (x + 3) has to be even.
We need x such that (x + 2) (x + 3) is a multiple of 3; which means one out of (x + 2) or (x + 3) should be a multiple of 3. Thus x leaves a remainder of 0 or 1 when divided by 3.
So x could be 1, 3, 4, 6, 7, 9, 10, 12, 13, 15, 16, 18 ....
So, from among the first 99 natural numbers, x cannot take 2, 5, 8, 11, 14…….98.
There are 33 numbers in this list –  (3 × 0 + 2, 3 × 1 + 2, 3 × 3 + 2, ….3 × 32 + 2).
So, x can take 99 – 33 = 66 values.
Answer : Option 2
Challenge: Understanding multiples of numbers and exploring different possibilities there
x is an 86 digit positive integer (in the decimal number system). All digits except the 50th  digit (from the left) are 2s. If x is divisible by 13, find the 50th digit?
1. 6
2. 8
3. 4
4. 7
Topic : Numbers
Approach :Any number of the form abcabc is a multiple of 1001. Now 1001 is 7 * 11 * 13. So, any number of the form abcabc is a multiple of 13.
Consider a number formed from the first 48 2s of the given number. This number would be a multiple of 13 (8 sets of 222222), so would a number comprising 36 2's.
Now we are left with a two digit number 2n, where n is the 50th digit. 26 is a multiple of 13, so the 50th digit should be 6.
Answer: Option 1
Challenge: Application of the concept of divisibility in numbers
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