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

Let the number be x.
7x -15 = 2x + 10 => 5x = 25 => x = 5

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

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

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.

(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

(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

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

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.

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)

7 + 7 = 14
14 + 41 = 55
55 + 55 = 110
110 + 011 = 121