Number System - Types & Tricks

NUMBER SYSTEM

Speed Test-One Liner-Mentally Solve

Question 1: Simplify 3146 - 4217 + 8246 - 3185 + 2655 - 1562?

Solution:

Question 2: Multiply: a.) 3842796 x 25 ;  b.) 37428 x 999 ;  c.) 73 x 87 ;  d.) 213 x 143?

Solution: a.) 3842796 x 25 = 384279600/4 = 96069900
b.) 37428 x 999 = 37428000 - 37428 = 37390572

c.)  Step 1:  Multiply the rightmost digits (unit digits) of both the numbers.
            ⇒ 3 x 7 = 21
We can write 73 x 87 as: _ / _ /⃔21
Transfer '2' from unit digit to left place.

 Step 2:  Now do cross multiplication of all 4 digits and add the product.
         ⇒ 7 x 7 + 3 x 8 = 21
Add '2' of unit digit to this = 73 + 2 = 75
We can write 73 x 87 as:  _/⃔75/1

 Step 3:  Multiply the leftmost digits of both the numbers.
         ⇒ 7 x 8 = 56
Add '7' from tens digit to it = 56 + 7 = 63
We can write 73 x 87 as:63/5/1

Answer will be: 6351

d.)

Question 3: Find a.) (73)2 ;  b.) (437)2 ;  c.) (5555)2 ?

Solution: a.) (73)2 = 72/2 x 7 x 3 /32
                                  ⇒  49/⃔42/9
                                  ⇒ 5329

b.) (437)2 = 4(437 + 37)| (37)2
                 ⇒ 1896/⃔1369
                 ⇒190969
In this question we used two methods, one is squaring of two digit numbers which can be done in 4 sec plus squaring of three digits number in which we only need to add last two digits of numbers with itself and multiply by leftmost hundreds digit.

c.) (5555)2 = 52 x (1111)2 = 25 x 1234321 = 123432100/4 = 30858025
Here we used two formulas first one is squaring of any repeated digit number in which we count the digits and reverse the order and second is multiplication with 25 in which we wrote two zeros at the end of number and divided it by 4 so that we can do it mentally.

Question 4: Find the root of perfect square 53824?

Solution: Here unit place is 4, after squaring 2 & 8 we get 4 at unit place. So first write them.

After omitting last two digits which is “24” we get “538” only, now write those numbers whose squares are nearest to 538, but not greater, and that is “23”.
So answer can be ’232’ or ’238’. To find the correct one, check the middle term containing 5 at unit place which is ’235’.
Square of ‘235’ is ‘55225’ and number ‘53824’ is less than ‘55225’, it means correct answer is ‘232’.

U may think number is large but all calculation can be done mentally, try it.

Question 5: Find the cube root of 804357?

Solution: Here unit place is ‘7’, after cubing ‘3’ we get ‘7’ at the unit place. So write ‘3’.

After omitting last three digits which are “357” we get “804” left, now write number whose cube is nearest to “804”, but not greater, and that is “9”.
So answer is 93.

Question 6: Find the approximate value of root 160?

Solution: √160 = 12 + (160 - 144)/24
                       = 12 + 2/3 = 12.66 approx
It takes hardly 5 sec to find its value.

Question 7: Find the smallest fraction: 17/26, 9/13, 33/52, 28/29

Solution: 17 - 9 = 8 and 26 - 13 = 13
⇒ Ratio = 8/13 which is smaller than 9/13 so 17/26 is smaller.
Again 28 - 17 = 11 and 29 - 26 =3
⇒ Ratio = 11/3 which is greater than 17/26 so 17/26 is smaller
Again 33 - 17 = 16 and 52 - 26 = 26
⇒ Ratio = 16/26 which is less than 17/26 so 33/52 is smallest
No need to write a single line, just calculate ratio in mind.

Question 8: A number when divided by 899 gives a remainder of 63. If the same number is divided by 29 what will be remainder. If same no. is divided by 28 what will be the remainder?

Solution: As 29 is a factor of 899, divide 63 by 29, 5 will be remainder. So if same number is divided by 29, 5 will be remainder.
Now 28 is not a factor of 899, so its not possible to calculate.

Question 9: Find the last two digits of 22 x 31 x 4 x 27 x 43?

Solution: To find last two digits, divide this product by 100 and find the remainder.
⇒ (22 x 31 x 4 x 27 x 43)/100 = (22 x 31 x 27 x 43)/25
⇒ [(-3) x 6 x 2 x (-7)]/25 = [(-18) x (-14)]/25
⇒ (7 x 11)/25 = 2 remainder
As we took 4 common so multiply remainder with 4 we will get 08. So last two digits are 08

Question 10: Find the remainder when -24.8 is divided by 6?

Solution: -24.8 = -6 x 4 - 0.8
But remainder can't be negative, so remainder will be = 6 -0.8 = 5.2

Question 11: When a number is divided by 5 & 7 successively leaves remainder 2 & 4 respectively. Find the remainder when the same number is divided by 35? What will be the remainder when same number will be divided by 7 & 5 successively?

Solution: Number = D1 x r2 + r1 = 5 x 4 + 2 = 22
Remainder when divided by 35 = 22
Remainder when divided by 7 & 5 successively = 1 and 3

Question 12: Simplify: a.) 17.83 + 0.007 + 310.0202  ; b.) 0.006379 x 10000  ; c.) 36.343 /7?

Solution: a.) 17.83 + 0.007 + 310.0202 = 17.838383...+ 0.007777... + 310.020222

b.) 0.006379 x 10000 = 63.79379
Period of decimal always remains same if we multiply with any multiple of 10.

c.) 36.343 /7 = 36.34343.../7 = 5.19
No need to convert decimal fraction into vulgar fractions to solve such questions.

Question 13: Is 55277838 divisible by 7 ?

Solution: 838 - 277 + 55 = 616 = 88 x 7
             ⇒ It is divisible by 7

Question 14: N! Has 23 zeroes. What is the maximum possible value of n?

Solution: This can never happen. Number of zeroes = n!/5 + n!/52 + ...
And 99! = 22 zeroes and 100! = 24 zeroes.

Question 15: N! Has 13 zeroes. The highest and least values of n are?

Solution: 55! = 11 + 2 = 13 zeroes
Also 59! = 11 + 2 = 13 zeroes
So minimum and maximum values are 55 and 59.

Question 16: Find the number of zeroes in the product of 11 x 22 x 33 x....x 4848?

Solution: As fives are less than 2, we need to count the number of 5’s only.
55 x 1010 x 1515 x 2020 x 2525 x 3030 x 3535 x 4040 x 4545 = 5 + 10 + 15 + 20 + 25x2 + 30 + 35 + 40 + 45 = 250 zeroes..

Question 17: Find the number of zeroes in 11! X 22! X 33! x 44! x 55! X ... x 1010!?

Solution: Number of 5’s are = 55! x 1010! = 5! + 10!

Question 18: Find the maximum value of n such that 157! Is perfectly divisible by 10n and 12n?

Solution: 10n = (2 x 5)n
Number of 5’s will be less than number of 2’s in 157!
Therefore, Number of 5’s = 157/5 + 157/52 + 157/53 = 38
So maximum value of n in 10n will be = 38

No. Of 2’s in 157! = 152
No. Of 22s = 152/2 = 76
No. Of 3’s in 157! = 75
The answer would be given by lower of these values, because 1275 has 75 4’s and 75 3’s and we already know that 157! Has 76 4’s. So it will be divisible perfectly by 1275

In case we take higher of these two values which is 76, then there will be 76 3’s in denominator while there are only 75 3’s in numerator so it won’t divide the numerator perfectly.
So correct answer is 75.

Question 19: Find the number of zeros in 1! x 2! x 3! x .... x 50!?

Solution: From 1! To 4! there won’t be any zeros because there are no number of 5’s.

From 5! To 9! Every factorial has one 5, so total number of 5’s = 5

From 10! To 14! Every factorial has two 5, so total number of 5’s = 10

Total number of zeros from 5! To 49! = 5 + 10 + 15 + 20 + 30 + 40 + 45 + 50 = 250

Number of zeros for 50! = 12
⇒ Total number of zeros = 250 + 12 = 262

Question 20: How many 3-digit even numbers can you form such that if one of the digits is 5, the following digit must be 7??

Solution: If there is a ‘5’ then it must be followed by ‘7’. It means if there is a ‘57’ in number then for number to be even, possible numbers are 570, 572, 574, 576, 578. Total 5 numbers.

But if there is no ‘5’ then possible even numbers of three digits between 100 to 198 = 50
But it also include 150, 152, 154, 156, 158 which is not possible according to question because 5 must be followed by 7. So total even numbers between 100 to 198 = 50 – 5 = 45

⇒ Total three digit even numbers = 45 x 8 + 5 = 365
Note: We excluded numbers from 500 to  598 and included only 5 numbers from this period.

Question 21: A man sold 38 pieces of clothing [combined in the form of shirts, trousers and ties]. If he sold at least 11 pieces of each item and he sold more shirts than trousers and more trousers than ties, then the number of ties he must have sold is?

Solution: Shirts > trousers > ties. Minimum number of pieces of each is 11.
⇒ Total minimum clothing will be = 33

Now remaining 5 needs to be distributed among ties, trousers, and shirts..
Let total number of ties = 11
Remaining 27 pieces of clothing can be distributed in two ways: Either 12 trousers and 15 shirts or 13 trousers and 14 shirts.

If we take 12 ties, then it won’t be possible to distribute.

Question 22: Find the unit's digit of  (1273)122!?

Solution: Unit digit will be '1' because 122! is a number of the form 4n. And cyclicity of 3 is 4, it means 34n always gives '1'.

These are few questions on number system just to practice the methods which i explained in earlier posts.