Number system
 Basic Formulae
 (a + b)(a – b) = (a^{2} – b^{2})
 (a+ b)^{2} = (a^{2} + b^{2} + 2ab)
 (a– b)^{2} = (a^{2} + b^{2} – 2ab)
 (a+ b + c)^{2} = a^{2} + b^{2} + c^{2} + 2(ab + bc + ca)
 (a^{3}+ b^{3}) = (a + b)(a^{2} – ab + b^{2})
 (a^{3}– b^{3}) = (a – b)(a^{2} + ab + b^{2})
 (a^{3}+ b^{3} + c^{3} – 3abc) = (a + b + c)(a^{2} + b^{2} + c^{2} – ab – bc – ac)
 When a+ b + c = 0, then a^{3} + b^{3} + c^{3} = 3abc
 Types of Numbers
 Natural Numbers
Counting numbers 1,2,3,4,5,…1,2,3,4,5,… are called natural numbers
 Whole Numbers
All counting numbers together with zero form the set of whole numbers.
Thus,
(i) 0 is the only whole number which is not a natural number.
(ii) Every natural number is a whole number.
III. Integers
All natural numbers, 0 and negatives of counting numbers i.e., …,?3,?2,?1,0,1,2,3,……..,?3,?2,?1,0,1,2,3,….. together form the set of integers.
(i) Positive Integers: 1,2,3,4,…..1,2,3,4,….. is the set of all positive integers.
(ii) Negative Integers: ?1,?2,?3,…..?1,?2,?3,….. is the set of all negative integers.
(iii) NonPositive and NonNegative Integers: 0 is neither positive nor negative.
So, 0,1,2,3,….0,1,2,3,…. represents the set of nonnegative integers,
while 0,?1,?2,?3,…..0,?1,?2,?3,….. represents the set of nonpositive integers.
 Even Numbers
A number divisible by 2 is called an even number, e.g.,2,4,6,82,4,6,8, etc.
 Odd Numbers
A number not divisible by 2 is called an odd number. e.g.,1,3,5,7,9,11,1,3,5,7,9,11, etc.
 Prime Numbers
A number greater than 1 is called a prime number, if it has exactly two factors, namely 1 and the number itself.
 Prime numbers up to 100 are :2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97.:2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97.
 Prime numbers Greater than 100: Let pp be a given number greater than 100. To find out whether it is prime or not, we use the following method:
Find a whole number nearly greater than the square root of pp. Let k>?jpk>?jp. Test whether pp is divisible by any prime number less than kk. If yes, then pp is not prime. Otherwise, pp is prime.
Example: We have to find whether 191 is a prime number or not. Now, 14>V19114>V191.
Prime numbers less than 14 are 2,3,5,7,11,13.2,3,5,7,11,13.
191 is not divisible by any of them. So, 191 is a prime number.
VII. Composite Numbers
Numbers greater than 1 which are not prime, are known as composite numbers, e.g., 4,6,8,9,10,12.4,6,8,9,10,12.
Note:
(i) 1 is neither prime nor composite.
(ii) 2 is the only even number which is prime.
(iii) There are 25 prime numbers between 1 and 100.
 3. Remainder and Quotient
“The remainder is rr when pp is divided by k” means p=kq+rp=kq+r the integer qq is called the quotient.
For instance, “The remainder is 1 when 7 is divided by 3” means 7=3?2+17=3?2+1. Dividing both sides of p=kq+rp=kq+r by k gives the following alternative form pk=q+rkpk=q+rk
Example:
The remainder is 57 when a number is divided by 10,000. What is the remainder when the same number is divided by 1,000?
(A) 5 (B) 7 (C) 43 (D) 57 (E) 570
Solution:
Since the remainder is 57 when the number is divided by 10,000, the number can be expressed as 10,000n+5710,000n+57, where nn is an integer.
Rewriting 10,000 as 1,000?101,000?10 yields 10,000n+57=1,000(10n)+5710,000n+57=1,000(10n)+57
Now, since nn is an integer, 10n10n is an integer. Letting 10n=q10n=q , we get
10,000n+57=1,000?q+5710,000n+57=1,000?q+57
Hence, the remainder is still 57 (by the p=kq+rp=kq+r form) when the number is divided by 1,000. The answer is (D).
Method II (Alternative form):
Since the remainder is 57 when the number is divided by 10,000, the number can be expressed as 10,000n+5710,000n+57. Dividing this number by 1,000 yields
10,000n+57100010,000n+571000 =10,000n1000+571000=10,000n1000+571000 =10n+571000=10n+571000
Hence, the remainder is 57 (by the alternative form pk=q+rkpk=q+rk ), and the answer is (D).
 Even, Odd Numbers
A number n is even if the remainder is zero when nn is divided by 2:n=2z+02:n=2z+0, or n=2zn=2z.
A number nn is odd if the remainder is one when nn is divided by 2:n=2z+12:n=2z+1.
The following properties for odd and even numbers are very useful – you should memorize them:
even * evenodd * oddeven * oddeven + evenodd + oddeven + odd=even=odd=even=even=even=oddeven * even=evenodd * odd=oddeven * odd=eveneven + even=evenodd + odd=eveneven + odd=odd
Example:
If nn is a positive integer and (n+1)(n+3)(n+1)(n+3) is odd, then (n+2)(n+4)(n+2)(n+4) must be a multiple of which one of the following?
(A) 3 (B) 5 (C) 6 (D) 8 (E) 16
Solution:
(n+1)(n+3)(n+1)(n+3) is odd only when both (n+1)(n+1) and (n+3)(n+3) are odd. This is possible only when nn is even.
Hence, n=2mn=2m, where mm is a positive integer. Then,
(n+2)(n+4)=(2m+2)(2m+4)=2(m+1)2(m+2)=4(m+1)(m+2)(n+2)(n+4)=(2m+2)(2m+4)=2(m+1)2(m+2)=4(m+1)(m+2)
=4 * (product of two consecutive positive integers, one which must be even)=4 * (product of two consecutive positive integers, one which must be even) =4 * (an even number), and this equals a number that is at least a multiple of 8=4 * (an even number), and this equals a number that is at least a multiple of 8
Hence, the answer is (D).
Questions
LevelI
1.  If onethird of onefourth of a number is 15, then threetenth of that number is:  

2.  Three times the first of three consecutive odd integers is 3 more than twice the third. The third integer is:  

3.  The difference between a twodigit number and the number obtained by interchanging the positions of its digits is 36. What is the difference between the two digits of that number?  

4.  The difference between a twodigit number and the number obtained by interchanging the digits is 36. What is the difference between the sum and the difference of the digits of the number if the ratio between the digits of the number is 1 : 2 ?  

5.  A twodigit number is such that the product of the digits is 8. When 18 is added to the number, then the digits are reversed. The number is:  

6.  The sum of the digits of a twodigit number is 15 and the difference between the digits is 3. What is the twodigit number?  

7.  The sum of the squares of three numbers is 138, while the sum of their products taken two at a time is 131. Their sum is:  

8.  A number consists of two digits. If the digits interchange places and the new number is added to the original number, then the resulting number will be divisible by:  

9.  In a twodigit, if it is known that its unit’s digit exceeds its ten’s digit by 2 and that the product of the given number and the sum of its digits is equal to 144, then the number is:  

10.  Find a positive number which when increased by 17 is equal to 60 times the reciprocal of the number.  

LevelII
11.  The product of two numbers is 9375 and the quotient, when the larger one is divided by the smaller, is 15. The sum of the numbers is:  

12.  The product of two numbers is 120 and the sum of their squares is 289. The sum of the number is:  

13.  A number consists of 3 digits whose sum is 10. The middle digit is equal to the sum of the other two and the number will be increased by 99 if its digits are reversed. The number is:  

14.  The sum of two number is 25 and their difference is 13. Find their product.  

15.  What is the sum of two consecutive even numbers, the difference of whose squares is 84?  

 16. If both 11^{2}and 3^{3}are factors of the number a * 4^{3} * 6^{2} * 13^{11}, what is the smallest possible value of ‘a’?
 121
 3267
 363
 33
 37
 17. Find the largest five digit number that is divisible by 7, 10, 15, 21 and 28.
 99840
 99900
 99990
 99960
 99970
 18. Anita had to multiply two positive integers. Instead of taking 35 as one of the multipliers, she incorrectly took 53. As a result, the product went up by 540. What is the new product?
 1050
 540
 1440
 1520
 1590
Answers:
LevelI
Answer:1 Option D
Explanation:
Let the number be x.
Then,  1  of  1  of x = 15 x = 15 x 12 = 180. 
3  4 
So, required number =  3  x 180  = 54.  
10 
Answer:2 Option D
Explanation:
Let the three integers be x, x + 2 and x + 4.
Then, 3x = 2(x + 4) + 3 x = 11.
Third integer = x + 4 = 15.
Answer:3 Option B
Explanation:
Let the ten’s digit be x and unit’s digit be y.
Then, (10x + y) – (10y + x) = 36
9(x – y) = 36
x – y = 4.
Answer:4 Option B
Explanation:
Since the number is greater than the number obtained on reversing the digits, so the ten’s digit is greater than the unit’s digit.
Let ten’s and unit’s digits be 2x and x respectively.
Then, (10 x 2x + x) – (10x + 2x) = 36
9x = 36
x = 4.
Required difference = (2x + x) – (2x – x) = 2x = 8.
Answer:5 Option B
Explanation:
Let the ten’s and unit digit be x and  
x 
Then,  10x +  8  + 18 = 10 x  8  + x  
x  x 
10x^{2} + 8 + 18x = 80 + x^{2}
9x^{2} + 18x – 72 = 0
x^{2} + 2x – 8 = 0
(x + 4)(x – 2) = 0
x = 2.
Answer:6 Option D
Explanation:
Let the ten’s digit be x and unit’s digit be y.
Then, x + y = 15 and x – y = 3 or y – x = 3.
Solving x + y = 15 and x – y = 3, we get: x = 9, y = 6.
Solving x + y = 15 and y – x = 3, we get: x = 6, y = 9.
So, the number is either 96 or 69.
Hence, the number cannot be determined.
Answer:7 Option A
Explanation:
Let the numbers be a, b and c.
Then, a^{2} + b^{2} + c^{2} = 138 and (ab + bc + ca) = 131.
(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2(ab + bc + ca) = 138 + 2 x 131 = 400.
(a + b + c) = 400 = 20.
Answer:8 Option D
Explanation:
Let the ten’s digit be x and unit’s digit be y.
Then, number = 10x + y.
Number obtained by interchanging the digits = 10y + x.
(10x + y) + (10y + x) = 11(x + y), which is divisible by 11.
Answer:9 Option A
Explanation:
Let the ten’s digit be x.
Then, unit’s digit = x + 2.
Number = 10x + (x + 2) = 11x + 2.
Sum of digits = x + (x + 2) = 2x + 2.
(11x + 2)(2x + 2) = 144
22x^{2} + 26x – 140 = 0
11x^{2} + 13x – 70 = 0
(x – 2)(11x + 35) = 0
x = 2.
Hence, required number = 11x + 2 = 24.
Answer:10 Option A
Explanation:
Let the number be x.
Then, x + 17 =  60 
x 
x^{2} + 17x – 60 = 0
(x + 20)(x – 3) = 0
x = 3.
Answer:11 Option C
Explanation:
Let the numbers be x and y.
Then, xy = 9375 and  x  = 15. 
y 
xy  =  9375 
(x/y)  15 
y^{2} = 625.
y = 25.
x = 15y = (15 x 25) = 375.
Sum of the numbers = x + y = 375 + 25 = 400.
Answer:12 Option B
Explanation:
Let the numbers be x and y.
Then, xy = 120 and x^{2} + y^{2} = 289.
(x + y)^{2} = x^{2} + y^{2} + 2xy = 289 + (2 x 120) = 529
x + y = 529 = 23.
Answer:13 Option B
Explanation:
Let the middle digit be x.
Then, 2x = 10 or x = 5. So, the number is either 253 or 352.
Since the number increases on reversing the digits, so the hundred’s digits is smaller than the unit’s digit.
Hence, required number = 253.
Answer:14 Option B
Explanation:
Let the numbers be x and y.
Then, x + y = 25 and x – y = 13.
4xy = (x + y)^{2} – (x– y)^{2}
= (25)^{2} – (13)^{2}
= (625 – 169)
= 456
xy = 114.
Answer:15 Option C
Explanation:
Let the numbers be x and x + 2.
Then, (x + 2)^{2} – x^{2} = 84
4x + 4 = 84
4x = 80
x = 20.
The required sum = x + (x + 2) = 2x + 2 = 42.
Answer:16 Option C
Explanation:
11^{2} is a factor of the given number.
In the given expression, a * 4^{3} * 6^{2} * 13^{11} none of the other factors, viz., 4, 6 or 13 is either a power or multiple of 11.
Hence, “a” should include 11^{2}
The question states that 3^{3} is a factor of the given number. 6^{2} is a part of the number.
6^{2} can be expressed as 3^{2} * 2^{2}.
Therefore, if 3^{3} has to be a factor of the given number a * 4^{3} * 6^{2} * 13^{11}, then we will need another 3 as part of the number.
Therefore, “a” should be at least 11^{2} * 3 = 363 if the given number has to have 11^{2} and 3^{3} as its factors.
Correct answer choice (C)
Answer:17 Option C
Explanation:
The number should be divisible by 10 (2, 5), 15 (3, 5), 21 (3, 7), and 28 (4, 7).
Hence, it is enough to check whether the number is divisibile by 3, 4, 5 and 7.
Test of divisibility by 3: Sum of the digits will be divisible by 3 if a number is divisible by 3.
Test of divisibility by 4: The righmost two digits viz., the units and tens digits will be divisible by 4 if a number is divisible by 4. For e.g, 1232 is divisible by 4 because 32 is divisible by 4.
Test of divisibility by 5: The units digit is either 5 or 0.
99960 is the only number which is divisible by 3, 4, 5 and 7.
Correct answer choice (C)
Answer:18 Option E
Explanation:
Let the number that Anita wanted to multiply be ‘X’.
She was expected to find the value of 35X.
Instead, she found the value of 53X.
The difference between the value that she got (53X) and what she was expected to get (35X) is 540.
i.e., 53X – 35X = 540
or (53 – 35) * X = 540
X = 30
Therefore, the correct product = 53 * 30 = 1590
Correct answer choice (E)