Question #1

What is the minimum number of square marbles required to tile

a floor of length 5 metres 78 cm and width 3 metres 74 cm?

As we have to use a whole number of marbles, the side of the

square should a factor of both 5 m 78 cm and 3m 74. And

it should be the highest factor of 5 m 78 cm and 3m 74.

5 m 78 cm = 578 cm and 3 m 74 cm = 374 cm.

The HCF of 578 and 374 = 34.

Hence, the side of the square is 34.

The number of such square marbles required,

=578×37434×34=578×37434×34

=17×11==17×11= 187 marbles

Question #2

Find the number, when 15 is subtracted from 7 times the number,

the result is 10 more than twice of the number

Let the number be x.

7x -15 = 2x + 10 => 5x = 25 => x = 5

Question #3

What is the least value of x, so that 23x57 is divisible by 3?

Hint: SUM of all digits should be divisible by 3.

Question #4

A set M contains element all even number between 1 and 23 and

all odd numbers 24 and 100. if all the elements of the set

multiplied than how many trailing 0, resulting number will contain?

the result trailing 0 occurs when one 2 and one 5 multiplied.

here we see the set M:

( 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 25, 35, 45, 55, 65, 75, 85, 95 )

after 22 I have taken only the term which contains factor of 5

now we find prime factor

2 = 2

4 = 2 * 2

6 = 2 * 3

8 = 2 * 2 * 2

10 = 2 * 5

12 = 2 * 2 * 3

14 = 2 * 7

16 = 2 * 2 * 2 * 2

18 = 2 * 3 * 3

20 = 2 * 2 * 5

22 = 2 * 11

25 = 5 * 5

35 = 5 * 7

45 = 5 * 9

55 = 5 * 11

65 = 5 * 13

75 = 5 * 5 * 3

85 = 5 * 17

95 = 5 * 19

here we find

total number of 5 is 12

and

total number of 2 is 19

and

we get total number of 5 * 2 pair is 12

so number of trailing zero will be 12

Question #5

Find remainder of (9^1+9^2+.........+9^n)/6

n is multiple of 11.

9/6 remainder is 3

9^2/6 remainder is 3

9^3/6 remainder is 3

9 to the power of any number when divided by 6 ,the remainder will always be 3.

Now, ( 3+3+3 ......11 times)/6 =(3*11)/6; therefore

the remainder will be 3.

If we take the even multiple of 11, then remainder will be zero.

Therefore answer is cannot be determine.

Question #6

Which of the following numbers must be added to 5678 to give a

remainder of 35 when divided by 460?

(5678+x)=(460*y)+35

5678-35+x=460*y

5643+x=460*y

so in x last digit must be 7 so ans should be c

Question #7

What is the unit digit of(2^3)^123456

(2^3^)^123456=2^3*123456=2^370368

2 cycle size=4

so,370368/4=remainder(0)

2 unit values=2,4,8,6=so,ans is 6

Question #8

IF a=0,b=1,c=2.....................z=25

Then one+one=?

This problem is based on Base 26 rather than regular base 10

(decimal system) that we normally use. In base 10 there are

10 digits 0 to 9 exist. In base 26 there are 26 digits 0 to 25 exist.

To convert any number into base 26, we have to divide the

number with 26 and find the remainder.

(Study this Base system chapter).

Here, ONE + ONE =

E has value of 4. So E + E = 8 which is equal to I.

Now N + N = 13 + 13 = 26. But in base 26, there is no 26. So (26)10=(10)26(26)10=(10)26

So we put 0 and 1 carry over. But 0 in this system is A.

Now O + O + 1 = 14 + 14 + 1 = 29

Therefore, (29)10=(13)26(29)10=(13)26

But 1 = B and 3 = D in that system. So ONE + ONE = BDAI

Question #9

how many squares are there in series 2006,2013,2020,2027....2300?

as there r only 3 squares b/w 2006-2300 that are 45^2=2205,

46^2=2116, 47^2=2209

now this series is an ap with 7 differnce.

so the difference of these squares & 2006 must be a multiple of 7.

only 2209-2006=203 which is multiple of 7.

so correct ans is 1.

Question #10

99^n is such a number begin with 8, least value of n?

99(100 - 1) = 9900-99= 9801

9801(100 - 1) = 980100-9801= 971299

971299(100 - 1) = 97129900 - 971299 = 96157601

.....

.....

observe the pattern, 98, 97, 96, .... for power of 2, 3, 4, ...

So for 90 the power could be 10.

For 11,we get a number starts with 8.

or his is not rigorous, but this is what I will do if I were asked

such a question in an exam.

We want (100−1)n=100n−(n1)100n−1+…

(100−1)n=100n−(n1)100n−1+….

These are really the terms that contribute to the answer,

so definitely n<10n<10 is not possible. Even for n=10n=10,

we get a number starting with '9' here and the subsequent

term is +ve and everything together is going to have a

positive effect.

This leaves us with n=11n=11 which I have verified as the

correct value.

More Intuition

We know that 99×99=980199×99=9801. Now when we

further multiply, the essential digits come from 98×9998×99

which is 97029702 (ignore the 99 at the end). One more

multiplication with 9999 and the product will start with 96…96….

This goes on and at n=11n=11, we get the value 89…89….

(This was essentially explained using the binomial expansion above)

Question #11

What is the next number of the following sequence 7, 14, 55,

110, ..?

7 + 7 = 14

14 + 41 = 55

55 + 55 = 110

110 + 011 = 121

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