1994
1995
Qu. 001: On dividing a number by 3, 4 and 7, the remainders are 2, 1 and 4 respectively. If the same number is divided by 84 then the remainder is: (1) 80 (2) 76 (3) 53 (4) None of these. (CAT 2002)
Method I: The question of this type can be solved using Chinese Remainder Theorem. But this may take some additional time.
Second method is by hit and trial. And I claim that this is the easiest method. In fact, the final remainder should leave the same remainders i.e. 2, 1 and 4 when this final remainder is divided by 3, 4 and 7 respectively.
The first answer is 80. Obviously, 80 == 0 mod 4, hence 80 cannot be the answer.
The second answer is 76. Obviously, 76 == 0 mod 4, hence 76 cannot be the answer.
The first answer is 53. Obviously, 53 == 2 mod 3, 53 == 1 mod 4 and 53 == 4 mod 7. Hence 53 is the answer.
Third method:
Let the number be X. Given X == 2 mod 3, means X-2 = 3A, means X = 3A + 2
Now, the same number is divided by 4, leaves the remainder 1.
It means it should be in the form of 4(3A + 2) + 1 = 12A + 9
Now, the same number is divided by 7, leaves the remainder 4.
It means it should be in the form of 7{12A + 9} + 4 = 84A + 67. Hence remainder is 67. Of course, despite the fact that theoretically approved method has been applied, this is a wrong answer, as 67 is not equal to 2 mod 3. Again, we apply the same process but a little differently. Let the number be X. Given X == 4 mod 7, means X-4 = 7A, means X = 7A + 4. Now, the same number is divided by 4, leaves the remainder 1. It means it should be in the form of 4(7A + 4) + 1 = 28A + 17. Now, the same number is divided by 3, leaves the remainder 2. It means it should be in the form of 3{28A + 17} + 2 = 84A + 53. Hence remainder is 53. Don’t use this method.
Qu. 002: If a, a + 2 and a + 4 are prime numbers, then the number of possible solutions for a is: (1) one (2) two (3) three (4) more than three. (CAT 2003, Re-test)
Sol: The three numbers a, a + 2 and a + 4 are all either even numbers or odd numbers.
If they are even, not more than one of them is prime {i.e. if a = 2}.
If they are odd, we have to show that one of them is a composite number, and the easiest way is to prove that one of them is divisible by 3.
Since we are dividing the given numbers by 3, the remainders shall be only 0, 1 or 2.
If, suppose for number ‘a’, the remainder is 0, means the number is divisible by 3 and number is 3 and 3 is a prime number (we shall not consider 6 or 9 or 12 or ,….., because all these numbers are not primes).
Hence if a=3, a+2 = 5, a+4 = 7, all three are primes and this (a=3) is one solution.
Let a>3. Again on dividing by 3, the remainders are 0, 1 or 2.
If remainder is 0, the number is composite.
If remainder is 1, it can be written as, a == 1 mod 3. Adding 2 both the sides, we have, a+2 == 3 mod 3. Means a+2 is divisible by 3 and it is a composite number. Hence there is no solution.
Finally let the remainder by 2. Then a == 2 mod 3. Adding 4 both the sides, a+4 == 6 mod 3 => a+4 == 0 mod 3. Means a+4 is divisible by 3. Hence, once again there is no solution.
Hence the only solution for ‘a’ such that a, a + 2 and a + 4 are prime numbers, is a=3. Hence (1) is the answer.
