Problems on Number Theory: CAT Quantitative Aptitude

Let us go through the following problems to understand the application of underlying concepts of Number Theory. The solutions given here are deliberately detailed so as to make the solution steps comprehensible to an average student. Do take care that in the actual exam you will not have the time to work out detailed solutions and much of these steps will need to be worked out without putting pen to paper.

Example:

Arnab was working with a number system to a certain base. He multiplied 41 and 53 to obtain 3133. What will he obtain if he were to multiply 14 and 35?

[1] 350

[2] 420

[3] 556

[4] 3313

Solution:

Let the base of the system in which Arnab is working ben.

We can write 41 x 53 = 5133 in base n as:

Now, solving for n, we get n = 7

i.e., (41)₇ x (53)₇ = (3133)₇

(41)₇ x (35)₇ = ?

Therefore changing base of L.H.S. of (i), we have:

[(1.7) + 4] x [(3.7 + 5) ] = 11 x 26

                                     = 286

Again, converting base of 286 back to base 7 (from base 10 that we normally use): Divide 286 by 7, and keep dividing the resulting quotient till the quotient becomes less than the divisor... as shown below:

                  Therefore the number for (286)₁₀ is (556)₇

Therefore, the correct option is [3].

Example:

Along a road, on a side,lie an odd number of stones placed at intervals of 10 m. These stones have to be assembled around the middle stone. A person can carry only one stone at a time. A man carried out the job starting with the stone in the middle, carrying stones in succession, thereby covering a distance of 4.8 km. Then the number of stones is:

[1] 35

[2] 15

[3] 29

[4] 31

Solution:

Take a deep breath and read the question slowly,once again, visualizing the situation as you read the question! Since, there are an odd number of stones, let this be (2n + 1) stones; i.e. n stones on each side of the middle stone.

The man will have to walk 20 m to pick up the first stone and return, 40 m for the second stone

                                    60 m for the third stone and so on…

This gives us a series of respective distances (in metres)walked/travelled by the man to pick each stone. This will be:

20, 40, 60, 80,..   [the sum of the series will give us the total distance

travelled by the man to collect all, (2n + 1) the stones]

The series above is an AP, with the first term a = 20 and common difference, d = 20

Therefore, the sum of the series,

That is, the man walks [10n(n + 1)] metres to pick up the stones on one side, and hence 20n(n + 1) m to pick up all the
stones.

therefore 20n(n + 1) = 4800, or n = 15

therefore There are 2n + 1 = 31 stones.

Therefore, the correct option is [4].