Summary Engineering mathematics and General Aptitude important formulas and concepts for Gate

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NUMBER SYSTEM
1. Method to multiply 2-digit number.
(i) AB × CD = AC / AD + BC / BD
35 × 47 = + = / 35 = 1645
(ii) AB × AC = A2 / A (B + C) / BC 74 × 76 = (4 + 6) / 4 × 6
= / 24 = / 24 = 5624
(iii) AB × CC = AC / (A + B)C / BC
= 35 × 44 = 3 × 4 / (3 + 5) × × 4
= / 20 = / 20 = 1540
2. Method to multiply 3-digit no.
ABC × DEF = AD / AE + BD / AF + BE + CD / BF + CE / CF
456 × 234 = 4 × × 3 + 5 × × 4 + 5 × 3 + 6 × × 4 + 6 × × 4
= + + 15 + + 18 /
24 = / / 24 = 106704
3. If in a series all number contains repeating 7. To find their sum, we start from the left multiply 7 by 1, 2, 3, 4, 5 & 6. Look at
the example below.
777777 + 77777 + 7777 + 777 + 77 + 7 = ?
=7×1/7×2/7×3/7×4/7×5/7×6
= / / = 864192
4. 0.5555 + 0.555 + 0.55 + 0.5 = ?
To find the sum of those number in which one number is repeated after decimal, then first write the number in either increasing
or decreasing order. Then -find the sum by using the below method.
0.5555 + 0.555 + 0.55 + 0.5
=5×4/5×3/5×2/5×1
= / = 2.1605
5 Those numbers whose all digits are 3.
(33)2 = 1089 Those number. in which all digits are number is 3 two or more than 2 times repeated, to find the square
of these number, we repeat 1 and 8 by (n – 1) time. Where n ® Number of times 3 repeated.
2
(333) = 110889
(3333)2 = 11108889
6. Those number whose all digits are 9.
(99)2 = 9801
(999)2 = 998001
(9999)2 = 99980001
(99999)2 = 9999800001
7. Those number whose all digits are 1.
A number whose one’s, ten’s, hundred’s digit is 1 i.e., 11, 111, 1111, ....
In this we count number of digits. We write 1, 2, 3, ..... in their square the digit in the number, then write in decreasing order
up to 1. 112 = 121
1112 = 12321




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S-2 101 Shortcuts in Quantitative Aptitude

11112 = 1234321
8. Some properties of square and square root:
(i) Complete square of a no. is possible if its last digit is 0, 1, 4, 5, 6 & 9. If last digit of a no. is 2, 3, 7, 8 then complete square
root of this no. is not possible.
(ii) If last digit of a no. is 1, then last digit of its complete square root is either 1 or 9. (iii) If last digit of a no. is 4, then last
digit of its complete square root is either 2 or 8.
(iv) If last digit of a no. is 5 or 0, then last digit of its complete square root is either 5 or 0.
(v) If last digit of a no. is 6, then last digit of its complete square root is either 4 or 6. (vi) If last digit of a
no. is 9, then last digit of its complete square root is either 3 or 7.
9. Prime Number :
(i) Find the approx square root of given no. Divide the given no. by the prime no. less than approx square root of no. If given
no. is not divisible by any of these prime no. then the no. is prime otherwise not. For example : To check 359 is a prime
number or not.
Sol. Approx sq. root = 19
Prime no. < 19 are 2, 3, 5, 7, 11, 13, 17
359 is not divisible by any of these prime nos. So 359 is a prime no.
For example: Is 25001 + 1 is prime or not?


Þ Reminder = 0,
\ 2 5001
+ 1 is not prime.
(ii) There are 15 prime no. from 1 to 50.
(iii) There are 25 prime no. from 1 to 100.
(iv) There are 168 prime no. from 1 to 1000.
10. If a no. is in the form of xn + an, then it is divisible by (x + a); if n is odd.
11. If xn ¸ (x – 1), then remainder is always 1.
12. If xn ¸ (x + 1)
(i) If n is even, then remainder is 1. (ii)If n is odd, then remainder is x.
4P +1 +1
13. (i) Value of P + P + P +..........¥ =
2

4P +1 - 1
(ii) Value of P- P- P - ..........¥ =
2

(iii)Value of P.P.P........... ¥ =P


(iv)Value of PPPPP =P
(212n - )¸ n



[Where n ® no. of times P repeated].


Note: If factors of P are n & (n + 1) type then value of P + P + P +....¥ = (n +1) and P - P - P -....¥ = n.
14. Number of divisors :
(i) If N is any no. and N = an × bm × cp × .... where a, b, c are prime no.
No. of divisors of N = (n + 1) (m + 1) (p + 1) ....
e.g. Find the no. of divisors of 90000.
N = 90000 = 22 × 32 × 52 × 102 = 22 × 32 × 52 × (2 × 5)2 = 24 × 32 × 54
So, the no. of divisors = (4 + 1) (2 + 1) (4 + 1) = 75
(ii) N = an × bm × cp, where a, b, c are prime
Then set of co-prime factors of N = [(n + 1) (m + 1) (p + 1) – 1 + nm + mp + pn + 3mnp]




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101 Shortcuts in Quantitative Aptitude S-3



(an 1+ -1)(bm 1+ -1)(cp 1+ -1)
(iii) If N = an × bm × cp..., where a, b & c are prime no. Then sum of the divisors = (a -1)(b-1)(c-1)
15. To find the last digit or digit at the unit’s place of a n.
(i) If the last digit or digit at the unit’s place of a is 1, 5 or 6, whatever be the value of n, it will have the same digit at unit’s
place, i.e.,

(.....1)n = (........1)
(.....5)n = (........5)
(.....6)n = (........6)

(ii) If the last digit or digit at the units place of a is 2, 3, 5, 7 or 8, then the last digit of a n depends upon the value of n and
follows a repeating pattern in terms of 4 as given below :
n last digit of (....2)n last digit of (....3)n last digit of (....7)n last digit of
(....8)n
4x+1 2 3 7 8
4x+2 4 9 9 4
4x+3 8 7 3 2
4x 6 1 1 6
n
(iii) If the last digit or digit at the unit’s place of a is either 4 or 9, then the last digit of a depends upon the value of n and
follows repeating pattern in terms of 2 as given below. n last digit of (....4) n last digit of (....9)n
2x 6 1
2x + 1 4 9


16. (i) Sum of n natural number =
(ii) Sum of n even number = (n) (n + 1) (iii)
Sum of n odd number = n2

n n( +1)(2n +1)
17. (i) Sum of sq. of first n natural no. = 6



n 4n ( 2 -1)
(ii) Sum of sq. of first n odd natural no. =
3


2n n( +1)(2n +1)
(iii) Sum of sq. of first n even natural no. =
3


n2 (n +1)2 én n( +1)ù2
18. (i) Sum of cube of first n natural no. = =ê ú
4 ë 2 û

(ii) Sum of cube of first n even natural no. = 2n 2 (n + 1)2




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