Use rules of divisibility to save time in CAT !

Before we start with the shortcut on how to solve using divisibility rules ,lets look at the basic divisibility rules ! Divisibility by 2 The last digit is even (0, 2, 4, 6, or 8). Divisibility by 3 The sum of the digits is divisible by 3.  Divisibility by 4 The last two digits divisible by 4.  Divisibility by 5 The last digit is 0 or 5. Divisibility by 6 The sum of the digits is divisible by 3 and the number itself is divisible by 2. Divisibility by 7 Subtract 2 times the last digit from the rest. Divisibility by 8 If the hundreds digit is even, examine the number formed by the last two digits. If the hundreds digit is odd, examine the number obtained by the last two digits plus 4. Divisibility by 9 The sum of the digits is divisible by 9. Divisibility by 10 The last digit is 0. Divisibility by 11 Add the digits in blocks of two from right to left. Example – 627: 6 + 27 = 33 Divisibility by 12 It is divisible by 3 and by 4. If you know the rules of divisibility, you can save considerable time on questions that would otherwise take you a long time to answer. It is worth learning how you tell if a number is divisible by 3, by 4, by 6 and so on. Let’s look at a Data Sufficiency question to know how useful these rules can be: If P and Q represent the hundreds and tens digits, respectively, in the four-digit number x=8PQ2, is x divisible by 8? (1) P=4 (2) Q=0 A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked C) BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked D) EACH statement ALONE is sufficient to answer the question asked E) Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed. Plug in numbers for statement 1 Consider statement 1 first. Plug in numbers. x could be 8402, which is not divisible by 8. x could also be 8432, which is divisible by 8. As such, Stat.(1) is insufficient, so our answer choices are now down to B, C, or E. Plug in numbers for statement 2 Consider statement 2. x ends in 02. You can check 8002, 8102, 8202 … all the way up to 8902 for divisibility by 8. If you do that, you will find eventually that all of them are not divisible by 8. As such, Stat.(2) is sufficient and you would get to the right answer of B. However, this would take some time. Is there a faster way? Efficient tricks for divisibility of 8 More importantly, do we even need to know the divisibility of 8? The rule of divisibility of 8 is if the last 3 digits form a 3-digit number that is divisible by 8, then the number is divisible by 8. But what if you had not memorized this rule? There is another way to go about answering this question. We know that 8 is equal to 2*4. Thus, to be divisible by 8, the number must pass the divisibility rules of both 2 and 4. The divisibility rule of 2 is easy: the number must be even. The divisibility rule of 4 is a bit trickier: the last two digits of the number must be divisible by 4. You know x ends in 02. These two digits are clearly not divisible by 4. Therefore, if a number is not divisible by 4, it is not divisible by 8. This is the faster way to determine that B is the answer. Learning the rules of divisibility is a worthwhile investment of your time. However, if you do not know one divisibility rule, you can lean on the divisibility rules of others. If you can get to an answer more quickly and be sure about it, it will leave you more time for the more difficult questions that will come up.     questions on divisibility rules cat questions on divisibility divisibility problems worksheet divisibility aptitude questions