Natural Numbers - N (1,2,3,4,.........)
Whole Numbers - W(0,1,2,3,.....)
Prime Numbers - a number other than 1 , if it is divisible only by 1 and itself
prime numbers
o the lowest prime no. is 2
o 2 is the only even prime no.
o the lowest odd prime no. is 3.
o the remainder when a prime no. p>=5 is divided by 6 is 1 or 5. however, if a no. on being divided by 6 gives remainder of 1 or 5 the no. need not be prime.
o the remainder of the division of the square of a prime number p>= divided by 24 is 1.
o for prime no. p>3, p2-1 is divisible by 24.
divisibility rules
o divisibility by 2: a no. is divisible by 2 if its last digit is 0 or divisible by 2.
o divisibility by 3: a no. is divisible by 3 if the sum of its digits is divisible by three.
o divisibility by 4: a no. is divisible by 4 if its last two digits are '00 or divisible by 4.
o divisibility by 5: a no. is divisible by 5 if its last digit is 0 or 5.
o divisibility by 6: a no. is divisible by 6 if it is divisible by 2 and 3 both.
o divisibility by 7: a no. is divisible by 7 if the difference of the number of its thousands and the remainder of its division by 1000 is divisible by 7.
o divisibility by 8: a no. is divisible by 8 if its last three digits are '000 or divisible by 8.
o divisibility by 9: a no. is divisible by 9 if the sum of its digits is divisible by 9
o divisibility by 11: a no. is divisible by 11 if the difference of the sum of the digits at even places and sum of the digits at odd places is zero or divisible by 11.
o divisibility by 13: a no. is divisible by 13 if the difference of the number of its thousands and the remainder of its division by 1000 is divisible by 13.
certain rules pertaining to number system
o of n consecutive whole nos. a, a+1,.......a+n-1, one and only one is divisible by n.
o 3n will always have an even no. of tens.e.g. 27,81, 243,729 etc.
o a sum of five consecutive whole numbers will always be divisible by 5. e.g. 1+2+3+4+5=15/5=3
o xy-yx is divisible by 9.
o the product of three consecutive natural nos.is divisible by 6.
o odd no.x odd no.=odd no.
o odd no.x even no.=even no.
o even no.x even no.=even no.
o odd no.+ odd no.=even no.
o odd no.- odd no.=even no.
o odd no.+ even no.=odd no.
o even no.+ even no.=even no.
o even no.- even no.=even no.
o the product of 'r' consecutive numbers is divisible by r!
if m and n are two numbers then (m+n)! is divisible by m!n!
if a and b are any two odd prime then a2-b2 is composite. also, a2+b2 is composite
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Q1) IS 349 a prime number?
Ans- the square root of 349 is approx 19.The prime no. less than 19 are 2,3,5,7,11,13,17
clearly ,349 is not divisible by any of them. Therefore 349 is a prime number.
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'VBODMAS' where V ,B,O,D,M,A,S stand for viniculum,bracket, of ,division ,multiplication,addition and subtraction respectively.
Simplify: 3/7 of (6+8* 3-2) +[ 1/5 - {3/7 +8/14}
Sol : 13/7 of (6+8*1) + {1/5 / 7/25 - 14/14 }
1/ 3/7 of (6+8) + {1/5* 25/7 -1}
1 / 3/7 of 0f 14 +{5/7 - 1 }
1 / 6 +{ -2/7}
1/6 -2/7 = 7-12/42 = -5/42
Some formulaes:
ax *ay = ax+y
ax /ay = ax-y
ax = 1, then x is 0 for all values of a (except 0)
Ascending or descending orders in Rational Numbers:
Q: Which is greater 5/8 or 9/14 ?
Ans : the fraction whose numerator after cross multiplication gives the greater value is greator.
5/8 9/14 we have 5 * 14 = 70 & 8*9 = 72
as 72 is greater than 70 ,therefore 9/14 is greater of the two.
Some rules on counting numbers:
1) sum of all the first n natural numbers = n(n+1)/2
eg 1+2+3+......+105 =105(105+1)/2 = 5565
2) sum of 1st n odd numbers = n2
eg 1+3+5+7 = 42 = 16
3) sum of 1st n even numbers =n(n+1)
2+4+6+8+....+100= 50(50+1) = 2550
4) sum of squares of first n natural numbers = n(n+1)(2n +1)/6
5) sum of cubes of first n natural numbers = {n(n +1)/2 }2
Digit in unit place
1) for odd numbers
when there is an odd digit in the unit place (except 5) , multiply the no. by itself until you get 1 in the unit place.
3) for even numbers
when there is an even digit in the unit place ,multiply the no. by itself until you get 6 in the unit place
For 1,5,6 the unit digit will remain the same , be multiplied any no.times
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Q1: A watch ticks 90 times in 95 seconds and another watch ticks 315 times in 323 sec. If both the watches are started together ,how many times will they tick together in the first hour?
Ans- The first watch ticks every 95/90 sec and the sec watch ticks every 323/315 sec
they will tick together after (LCM of 95/90 and 323/315) sec
LCM of 95/90 &323/315 = LCM of 95,323/ HCF of 90, 315
= 19*5*17/45
The no. of times they will tick in the first 3600 sec = 3600 / 19*5*17/45= 3600*45/19*5*17 =100..
Once they have already ticked in the beginning ; so in 1 hour they will tick 100+1= 101 times.
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Q2) Find the least number which , when divided by 13, 15, 19 leaves the remainders 2, 4 and 8 respectively.
Ans - 13-2 =15-4= 19-8 = 11
now LCM of 13,15,19 = 3705
Therefore the required least no = 3705 - 11 = 3694
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FIND THE REMAINDER:
Q1) Find the remainder when 713 + 1 is divided by 6?
Ans- 713 = (6 +1)13 has each term except 113 exactly divisible by 6. Thus , when 713 is divided by 6 we have the remainder 113 = 1 and hence , when 713 +1 is divided by 6 the remainder is 1+1 = 2
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To find the number of different divisors of a composite number
find the prime factors of the no. and increase the index of each factor by 1. The continued product of increased indices will give the result including unity and the number itself.
Q1) 50 = 2*5*5 = 2 * 52
the no of total divisors = (1+1)(2+1) = 2*3 = 6
or the no. of divisors excluding 1 and 50 = 6-2 =4
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Q2) How many nos. up to 200 are divisible by 4 and 3 together?
Ans - LCM of 4& 3 is 12
Now, divide 200 by 12 and the quotient obtained is the required number of numbers .
200 = 16 * 12 + 8
Thus , there are 16 numbers
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Q3) If the places of last two digits of a three- digit number are interchanged , a new number greater than the original number by 54 is obtained.What is the difference between the last two digits of that number?
Ans - Let the three -digit no be 100x +10y +z
A/Q (100x +10z + y) - ( 100x +10y +z ) =54
or 9z - 9y = 54 or z-y = 6
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