Divisibility rules for CAT exam
Posted Date: 23-Jul-2009
A number is divisible by 2 if and only if the last digit is divisible by 2.
A number is divisible by 3 if and only if the sum of the digits is divisible by 3.
A number is divisible by 4 if and only if the last 2 digits is a number divisible by 4.
A number is divisible by 5 if and only if the last digit is divisible by 5.
A number is divisible by 6 if and only if it is divisible by 2 and 3.
A number is divisible by 8 if and only if the last 3 digits is a number divisible by 8.
A number is divisible by 9 if and only if the sum of the digits is divisible by 9.
A number is divisible by 10n if and only if the number ends in n zeros.
A number is divisible by 11 iff the sum of every other digit minus the sum of the rest of the digits is divisible by 11.
To find out if a number is divisible by seven, take the last digit, double
it, and subtract it from the rest of the number.
Example: If you had 203, you would double the last digit to get six, and
subtract that from 20 to get 14. If you get an answer divisible by 7
(including zero), then the original number is divisible by seven. If you
don't know the new number's divisibility, you can apply the rule again.
The greatest common divisor is found by looking at the prime factorizations or using the Euclidean algorithm.
The least common multiple of a and b is found by looking at the prime factorizations or (ab)/gcd(a,b).
Two numbers are said to be relatively prime in the greatest common factor is 1.
If gcd(a, b)=d, then there exist integers x and y so that ax+by=d.
If d divides both a and b, then d divides a+b and d divides a-b.
a=b mod m iff m divides a-b iff a and b both have the same remainder when divided by m.
ap-1 = 1 mod p (a is not a multiple of p)
af(m)=1 mod m ( gcd(a, m) =1)
If a probability experiment is repeated n times and the probability of success in one trial is p, then the
probability of exactly r successes in the n trials is nCr (p)r(1-p)(n-r).
Rules of Logarithms:
loga(M)=y if and only if M=ay
The number of zeros at the end of n! is determined by the number of 5’s. To find this you do the following
process: n/5 = n1 and some remainder. Drop the remainder and compute n1/5 = n2 plus some remainder.
Drop the remainder and compute n2/5 = n3 plus some remainder, etc. The number of zeros is n1+n2+n3+n4...
The sum of any consecutive integers k through n, with n being the larger,
simply use this equation:
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Divisibility rules for CAT exam
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