Number system Quiz 1

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

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)

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.

Dividend = divisor × quotient + remainder
5678 = 460 ×12 + 158
Here 158 is the remainder. we need the remainder to be 35 by adding a number.
Let the number to be added be x.

Since 158>35, the quotient we will get by dividing  (5678+x) by 460 should be greater than 12. It may be 13, 14, 15 etc. But i am taking the lowest one here i.e.13.

So (5678 +x) = 460 × 13 + 35  (here 35 is remainder)
⇒ 5678 + x - 35 = 460 × 13
⇒5643 + x = 460 × 13
⇒ x = 460 × 13 - 5643
⇒ x = 337

However, none of the options match right ???

• If rather than 13th number as quotient, we would've used 14th then?


• Thus, add 460 (Divisor) to 337, we get 797

(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

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.

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

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