APTITUDE-Problems on Numbers

Introduction:

Natural Numbers:

All positive integers are natural numbers.
Ex 1,2,3,4,8,……

There are infinite natural numbers and number 1 is the least natural number.Based on divisibility there would be two types of natural numbers. They are Prime and composite.

Prime Numbers:

A natural number larger than unity is a prime number if it does not have other divisors except for itself and unity.

Note:-Unity i e,1 is not a prime number.

Properties Of Prime Numbers:

->The lowest prime number is 2.
->2 is also the only even prime number.
->The lowest odd prime number is 3.
->The remainder when a prime number p>=5 s divided by 6 is 1 or 5.However, if a number on being divided by 6 gives a remainder 1 or 5 need not be prime.
->The remainder of division of the square of a prime number p>=5 divide by 24 is 1.
->For prime numbers p>3, p²-1 is divided by 24.
->If a and b are any 2 odd primes then a²-b² is composite. Also a²+b²
is composite.
->The remainder of the division of the square of a prime number p>=5
divided by 12 is 1.

Process to Check A Number s Prime or not:

Take the square root of the number.
Round of the square root to the next highest integer call this number as Z.
Check for divisibility of the number N by all prime numbers below Z. If
there is no numbers below the value of Z which divides N then the number will be prime.

Example 239 is prime or not?
√239 lies between 15 or 16.Hence take the value of Z=16.
Prime numbers less than 16 are 2,3,5,7,11 and 13.
239 is not divisible by any of these. Hence we can conclude that 239
is a prime number.

Composite Numbers:

The numbers which are not prime are known as composite numbers.

Co-Primes:

Two numbers a an b are said to be co-primes,if their H.C.F is 1.
Example (2,3),(4,5),(7,9),(8,11)…..
Place value or Local value of a digit in a Number:

place value:

Example 689745132
Place value of 2 is (2*1)=2
Place value of 3 is (3*10)=30 and so on.
Face value:-It is the value of the digit itself at whatever
place it may be.

Example 689745132
Face value of 2 is 2.
Face value of 3 is 3 and so on.

Tests of Divisibility:

Divisibility by 2:-A number is divisible by 2,if its unit’s digit is
any of 0,2,4,6,8.
Example 84932 is divisible by 2,while 65935 is not.

Divisibility by 3:-A number is divisible by 3,if the sum of its digits is divisible by 3.

Example 1.592482 is divisible by 3,since sum of its digits
5+9+2+4+8+2=30 which is divisible by 3.

Example 2.864329 is not divisible by 3,since sum of its digits
8+6+4+3+2+9=32 which is not divisible by 3.

Divisibility by 4:
-A number is divisible by 4,if the number formed by last two digits is divisible by 4.

Example 1.892648 is divisible by 4,since the number formed by the last
two digits is 48 divisible by 4.

Example 2.But 749282 is not divisible by 4,since the number formed by
the last two digits is 82 is not divisible by 4.

Divisibility by 5:-A number divisible by 5,if its unit’s digit is either 0 or 5.
Example 20820,50345

Divisibility by 6:-If the number is divisible by both 2 and 3.
example 35256 is clearly divisible by 2
sum of digits =3+5+2+5+21,which is divisible by 3
Thus the given number is divisible by 6.

Divisibility by 8:-A number is divisible by 8 if the last 3 digits of the number are divisible by 8.

Divisibility by 11:-If the difference of the sum of the digits in the odd places and the sum of the digitsin the even places is zero or divisible
by 11.

Example 4832718
(8+7+3+4) – (1+2+8)=11 which is divisible by 11.

Divisibility by 12:-All numbers divisible by 3 and 4 are divisible by 12.

Divisibility by 7,11,13:-The difference of the number of its thousands and the remainder of its division by 1000 is divisible by 7,11,13.

BASIC FORMULAE:

->(a+b)²=a²+b²+2ab
->(a-b)²=a²+b²-2ab
->(a+b)²-(a-b)²=4ab
->(a+b)²+(a-b)²=2(a²+b²)
->a²-b²=(a+b)(a-b)
->(a-+b+c)²=a²+b²+c²+2(ab+b c+ca)
->a³+b³=(a+b)(a²+b²-ab)
->a³-b³=(a-b)(a²+b²+ab)
->a³+b³+c³-3a b c=(a+b+c)(a²+b²+c²-ab-b c-ca)
->If a+b+c=0 then a³+b³+c³=3a b c

DIVISION ALGORITHM

If we divide a number by another number ,then

Dividend = (Divisor * quotient) + Remainder

MULTIPLICATION BY SHORT CUT METHODS

1.Multiplication by distributive law:

a)a*(b+c)=a*b+a*c
b)a*(b-c)=a*b-a*c

Example
a)567958*99999=567958*(100000-1)
567958*100000-567958*1
56795800000-567958
56795232042

b)978*184+978*816=978*(184+816)
978*1000=978000

2.Multiplication of a number by 5n:-Put n zeros to the right of the
multiplicand and divide the number so formed by 2n

Example 975436*625=975436*54=9754360000/16=609647500.

PROGRESSION:

A succession of numbers formed and arranged in a definite order according to certain definite rule is called a progression.

1.Arithmetic Progression:-If each term of a progression differs from its
preceding term by a constant.
This constant difference is called the common difference of the A.P.
The n th term of this A.P is Tn=a(n-1)+d.
The sum of n terms of A.P Sn=n/2[2a+(n-1)d].

Important Results:

a.1+2+3+4+5………………….=n(n+1)/2.
b.12+22+32+42+52………………….=n(n+1)(2n+1)/6.
c.13+23+33+43+53………………….=n2(n+1)2/4

2.Geometric Progression:-A progression of numbers in which every
term bears a constant ratio with ts preceding term.
i.e a,a r,a r2,a r3……………
In G.P Tn=a r n-1
Sum of n terms Sn=a(1-r n)/1-r

Problems

1.Simplify
a.8888+888+88+8
b.11992-7823-456

Solution: a.8888
888
88
8
9872
b.11992-7823-456=11992-(7823+456)
=11992-8279=3713

2.What could be the maximum value of Q in the following equation?
5PQ+3R7+2Q8=1114

Solution: 5 P Q
3 R 7
2 Q 8
11 1 4
2+P+Q+R=11
Maximum value of Q =11-2=9 (P=0,R=0)

3.Simplify: a.5793405*9999 b.839478*625

Solution:
a. 5793405*9999=5793405*(10000-1)
57934050000-5793405=57928256595
b. 839478*625=839478*54=8394780000/16=524673750.

4.Evaluate 313*313+287*287

Solution:
a²+b²=1/2((a+b)²+(a-b)²)
1/2(313+287)² +(313-287)²=1/2(600 ² +26 ² )
½(360000+676)=180338

5.Which of the following is a prime number?
a.241 b.337 c.391

Solution:
a.241
16>√241.Hence take the value of Z=16.
Prime numbers less than 16 are 2,3,5,7,11 and 13.
241 is not divisible by any of these. Hence we can
conclude that 241 is a prime number.
b. 337
19>√337.Hence take the value of Z=19.
Prime numbers less than 16 are 2,3,5,7,11,13 and 17.
337 is not divisible by any of these. Hence we can conclude
that 337 is a prime number.
c. 391
20>√391.Hence take the value of Z=20.
Prime numbers less than 16 are 2,3,5,7,11,13,17 and 19.
391 is divisible by 17. Hence we can conclude
that 391 is not a prime number.

6.Find the unit’s digit n the product 2467 153 * 34172?

Solution: Unit’s digit in the given product=Unit’s digit in 7 153 * 172
Now 7 4 gives unit digit 1
7 152 gives unit digit 1
7 153 gives 1*7=7.Also 172 gives 1
Hence unit’s digit in the product =7*1=7.

7.Find the total number of prime factors in 411 *7 5 *112 ?

Solution: 411 7 5 112= (2*2) 11 *7 5 *112
= 222 *7 5 *112
Total number of prime factors=22+5+2=29

8.Which of the following numbers s divisible by 3?
a.541326
b.5967013

Solution: a. Sum of digits in 541326=5+4+1+3+2+6=21 divisible by 3.
b. Sum of digits in 5967013=5+9+6+7+0+1+3=31 not divisible by 3.

9.What least value must be assigned to * so that th number 197*5462 is divisible by 9?
Solution: Let the missing digit be x
Sum of digits = (1+9+7+x+5+4+6+2)=34+x
For 34+x to be divisible by 9 , x must be replaced by 2
The digit in place of x must be 2.

10.What least number must be added to 3000 to obtain a number exactly divisible by 19?
Solution:On dividing 3000 by 19 we get 17 as remainder
Therefore number to be added = 19-17=2.

11.Find the smallest number of 6 digits which is exactly divisible by 111?
Solution:Smallest number of 6 digits is 100000
On dividing 10000 by 111 we get 100 as remainder
Hence,required number =10011.

12.On dividing 15968 by a certain number the quotient is 89 and the remainder is 37.Find the divisor?
Solution:Divisor = (Dividend-Remainder)/Quotient
=(15968-37) / 89
=179.

13.A number when divided by 342 gives a remainder 47.When the same number is divided by 19 what would be the remainder?

Solution:Number=342 K + 47 = 19 * 18 K + 19 * 2 + 9=19 ( 18K + 2) + 9.The given number when divided by 19 gives 18 K + 2 as quotient and 9 as remainder.

14.A number being successively divided by 3,5,8 leaves remainders 1,4,7 respectively. Find the respective remainders if the order of divisors are reversed?

Solution:Let the number be x.

3 x 5 y – 1 8 z – 4 1 – 7 z=8*1+7=15
y=5z+4 = 5*15+4 = 79
x=3y+1 = 3*79+1=238
Now
8 238
5 29 – 6
3 5 – 4
1 – 2
Respective remainders are 6,4,2.

15.Find the remainder when 231 is divided by 5?

Solution:210 =1024.unit digit of 210 * 210 * 210 is 4 as
4*4*4 gives unit digit 4
unit digit of 231 is 8.
Now 8 when divided by 5 gives 3 as remainder.
231 when divided by 5 gives 3 as remainder.

16.How many numbers between 11 and 90 are divisible by 7?

Solution:The required numbers are 14,21,28,………..,84
This is an A.P with a=14,d=7.
Let it contain n terms
then T =84=a+(n-1)d
=14+(n-1)7
=7+7n
7n=77 =>n=11.

17.Find the sum of all odd numbers up to 100?

Solution:The given numbers are 1,3,5………99.
This is an A.P with a=1,d=2.
Let it contain n terms 1+(n-1)2=99
=>n=50
Then required sum =n/2(first term +last term)
=50/2(1+99)=2500.

18.How many terms are there in 2,4,6,8……….,1024?

Solution:Clearly 2,4,6……..1024 form a G.P with a=2,r=2
Let the number of terms be n
then 2*2 n-1=1024
2n-1 =512=29
n-1=9
n=10.

19.2+22+23+24+25……….+28=?

Solution:Given series is a G.P with a=2,r=2 and n=8.
Sum Sn=a(1-r n)/1-r=Sn=2(1-28)/1-2.
=2*255=510.

20.A positive number which when added to 1000 gives a sum ,
which is greater than when it is multiplied by 1000.The positive integer is?

a.1 b.3 c.5 d.7

Solution:1000+N>1000N
clearly N=1.

21.The sum of all possible two digit numbers formed from three
different one digit natural numbers when divided by the sum of the
original three numbers is equal to?

a.18 b.22 c.36 d. none

Solution:Let the one digit numbers x,y,z
Sum of all possible two digit numbers=
=(10x+y)+(10x+z)+(10y+x)+(10y+z)+(10z+x)+(10z+y)
= 22(x+y+z)
Therefore sum of all possible two digit numbers when divided by sum of
one digit numbers gives 22.

22.The sum of three prime numbers is 100.If one of them exceeds another by 36 then one of the numbers is?
a.7 b.29 c.41 d67.

Solution:x+(x+36)+y=100
2x+y=64
Therefore y must be even prime which is 2
2x+2=64=>x=31.
Third prime number =x+36=31+36=67.

23.A number when divided by the sum of 555 and 445 gives two times their difference as quotient and 30 as remainder .The number is?
a.1220 b.1250 c.22030 d.220030.

Solution:Number=(555+445)*(555-445)*2+30
=(555+445)*2*110+30
=220000+30=220030.

24.The difference between two numbers s 1365.When the larger number is divided by the smaller one the quotient is 6 and the remainder is 15.
The smaller number is?

a.240 b.270 c.295 d.360

Solution:Let the smaller number be x, then larger number =1365+x
Therefore 1365+x=6x+15
5x=1350 => x=270
Required number is 270.

25.In doing a division of a question with zero remainder,a candidate
took 12 as divisor instead of 21.The quotient obtained by him was 35.
The correct quotient is?

a.0 b.12 c.13 d.20

Solution:Dividend=12*35=420.
Now dividend =420 and divisor =21.
Therefore correct quotient =420/21=20.