CAT Quantitative Aptitude Questions on Divisibility

This article shows the application of divisibility to some complex problems that are used to form questions in examinations like CAT.  We first show a problem and its detailed solution to clarify that approach used in this problem and also help you develop the logic for such problem solving. The second problem is of the form that you may encounter in CAT.

Problem 1:
How many five digit multiples of 11 are there, if the five digits be 7, 6, 5, 4 and 3 in some order?

Solution:
We know from divisibility rules that for any five digit number to be a multiple of 11, if the number is of the form: abcde, then should be equal to either 0 or a multiple of 11.

In the present question, this difference between (a+c+e) and (b+d) should be 0, 11 or -11.

In case, 

[(a+c+e)] - (b+d)]=0, then

(a+c+e)= (b+d)

Then, if we take the sum of all digits of this five digit number abcde, we get

(a+c+e)+(b+d)

(b+d)+(b+d) = 2(b+d)

or (a+b+c+d+e) = 7+6+5+4+3=25 which is odd.

But 2(b + d) will always be even. This implies that the difference will never be zero.

In case,

[(a+c+e)] - (b+d)]= 11, then

(a+b+c+d+e) - 11= 2 (b+d)

so that (b+d) = 7

This is only possible, when b and d are 3 and 4 in some order; and a, c, e are 7, 6 & 5 in some order.

There will be 2 × 3! = 12 numbers satisfying this.

In case

[(a+c+e)] - (b+d)]= 11, then

(a+b+c+d+e) - 11= 2 (b+d), so that

(b+d)=18 which again is not possible

Therefore, there will only be 12 numbers satisfying the conditions of the problem.

Problem 2:

If then x divided by 70 leaves a remainder of:

[1] 0
[2] 1
[3] 69
[4] 35

Solution:

This problem appeared in CAT 2005. Just a second to soak in the problem.
One way to solve this problem is to find the cubes of the numbers and then sum them and then divide by 70 to get the solution. Of course, this can be done…but will require a lot of calculations and is surely not comfortable. An alternative way to tackle the problem is shown below.

Inspection of the problem shows that (16+19) and (17+18) will both give a sum of 35…which is a multiple of 70...so there is some lead to go by.

Rearranging the terms, we can write x as

x= (16³+19³+17³+18³), and now breaking the terms further for easy handling

x= (16+9) [(16²+19²)-(16x19)] + (17+18) [17²+18²)-(17x18)]

Note: if the terms in the bracket are a multiple of 2 then the problem is solved.
So, let us now look only at the terms in the brackets.
We have, terms that will end in:

16² will be end in 6

19² will be end in 1

Therefore, there sum will have 7 as its last (unit's) digit.

Also, 16x19 will have 4 as its units digit...which leads us to the conclusion that we will have 3 as the last (unit's) digit of this term

Similarly, we can come to the conclusion that the solution to the other set of terms will have 7 as the unit’s digit.

And when we will take a sum of the two, the final sum will have 0 as its unit’s digit…which means we will have an even number.

Therefore,

x is of the form 35 [2(some number)] which will be completely divisible by 70 (and will leave 0 as remainder).

Divisibility is a concept to be understood and applied in many questions in CAT Examination directly and indirectly.Read and practice as many questions as you can.