Number System is an all-time favorite with the CAT exam setters as well as the students. The topic is known for intriguing conceptual problems that test the best brains. For this topic, you need to use a combination of pure mathematical knowledge and logic skills. In fact, this is one topic that begins with the absolute basics of mathematics, that is, numbers. It explains you the different types of numbers and then extends the application of this knowledge to a variety of domains.
Number system is such a vast area that the syllabus for the same cannot be defined exactly. Keeping in mind the concepts from which questions can appear in the exam, the main list of topics for this area is: unit's digit, remainders, digital root, divisibility rules, integers and number of zeroes. Besides these topics, questions are also based on real numbers, prime numbers, natural numbers, and whole numbers. In short, questions in number system may be based on basic definitions as well as advanced concepts.
(i) Every number has the same unit’s digit at its fifth power as it has at its first power, thus the standard method that can be followed is to divide the power given by 4, find the remaining power and check the unit’s digit in that number. This short cut can be applied because you will always get the same unit’s digit as otherwise.
(ii) For calculating number of zeroes at the end of factorial of a number, you should divide the number by 5, the quotient obtained is again divided by 5 and so on till the last quotient obtained is smaller than 5. The sum of all the quotients is the number of 5s, which then becomes the number of zeroes in the given number.
(iii) The digital root of a number is the sum of the digits, over and over again, till it becomes a single digit number. For example, the digital root of 87984 will be 8 + 7 + 9 + 8 + 4 ⇒ 36 = 3 + 6 ⇒ 9.
(iv)When the concept of Euler number is used and the dividend and divisor happen to be co-prime, the remainder questions become very easy.
(v) The product of 3 consecutive natural numbers is divisible by 6.
(vi) The product of 3 consecutive natural numbers, the first of which is an even number is divisible by 24.
(vii) The sum of a two-digit number and a number formed by reversing its digits is divisible by 11. E.g. 28 + 82 = 110, this is divisible by 11. At the same time, the difference between those numbers will be divisible by 9. e.g. 82 – 28 = 54, this is divisible by 9.
(viii) ∑n = n(n+1)/2, ∑n is the sum of first n natural numbers.
(ix) ∑n2 = n(n+1)(n+2)/6, ∑n2 is the sum of first n perfect squares.
(x) ∑n3 = n2(n+1)2/4 = (∑n)2, ∑n3 is the sum of first n perfect cubes.
(xi) xn + yn = (x + y) (xn-1 - xn-2.y + xn-3.y2 - ... +yn-1) when n is odd. Therefore, when n is odd, xn + yn is divisible by x + y. e.g. 33 + 23 = 35 and is divisible by 5, which is (3 + 2).
(xii) xn - yn = (x + y) (xn-1 - xn-2.y + ... yn-1) when n is even. Therefore, when n is even, xn - yn is divisible by x + y. e.g. 72 - 32 = 40, is divisible by 10, which is (7+3).
(xiii) xn - yn = (x - y) (xn-1 + xn-2.y + .... + yn-1) for both odd and even n. Therefore, xn - yn is divisible by x - y. For example: 94 - 24 = 6545, is divisible by 7, which is (9 – 2).
Example 1: N = (18n2 + 9n + 8)/n; where N belongs to integer. How many integral solutions of N are possible?
Solution: The given expression can be broken as: 18n2/n + 9n/n + 8/n. This gives us: 18n + 9 + 8/n. Now we can see that whatever the value of ‘n’, 18n + 9 will always give an integral value. Therefore, it now depends upon 8/n only ⇒ n can have any integral value, which is a factor of 8. The integers, which will satisfy this condition are ±1, ±2, ±3, ±4. Thus, in total,n can take 8 values.
Example 2: N = 960. Find total number of factors of N.
Solution: The number of divisors of a composite number: If D is a composite number in the form D = ap × bq × cr, where a, b, c are primes, then the no. of divisors of D, represented by n is given by n = (p+1)(q+1)(r +1). Following the same, after dividing 960 into prime factors: 26 × 31 × 51, we can calculate the total number of factors as: (6+1).(1+1).(1+1) = 28.
Example 3: Find the unit digit of the following expression: (123)34 × (876)456 × (45)86.
Solution: Whenever an even unit digit and a 5 at the unit digit are present, they will always give a 0 at the unit digit, no matter if any other number is present or not. Hence, with this approach, this question proves to be a sitter.
Example 4: What will be the number of zeroes at the end of the product of the first 100 natural numbers?
Solution: In this kind of questions, you need to find greatest power of 5, which can divide the product of the first 100 natural numbers. Remember, a multiple of 5 multiplied by any even number, gives you a zero. Now divide 100 by 5 and take 20 as quotient. Then divide 20 (the quotient) by 5 and get the new quotient 4, which further cannot be divided by 5. The sum of all such quotient gives you the greatest power of 5, which can divide that number. The sum is 24 and this is the number of zeroes at the end of the product of the first 100 natural numbers.
Example 5: Which letter should replace the $ in the number 2347$98, so that it becomes a multiple of 9?
Solution: As you know that if the sum of all the digits is divisible by 9, then the number is divisible by 9. Now sum of the given digits is 2 + 3 + 4 + 7 + 9 + 8 = 33 + $. Now think the next multiple of 9 after 33 i.e. 36. This means you add 3 in this. The value of $ is 3.
Example 6: In a party there are 20 persons present. If each of them shakes hand with all the other persons,in total how many handshakes will take place?
Solution: Out of 20 persons, the first person will shake hand with 19 persons. The second person will shake hand with 18 persons (because he has already shaken hand with first person). The third person will shake hand with 17 persons and so on. The second last person shakes hand with only one person. And last will shake hand with none (because he has already shaken hand with all persons). In order to find the total number of handshakes you have to add all the natural numbers from 1 to 19 i.e. ∑ 19. ∑19 = 19 x 20/2 = 190 handshakes.