TCS Divisibility Question Answer 3

Condition for divisibility by 3:
Sum of digits of number must be divisible by 3
i.e a+b+6+4+a+b = 0 mod 3
=> 10 + 2(a+b) = 0 mod 3
=> 1 + 2(a+b) = 0 mod 3 (10 = 1 mod 3)
=> 2(a+b) = -1 mod 3
=> 2(a+b) = 2 mod 3
=> a+b = 1 mod 3
So, a+b must be of the form 3n + 1 for some positive integer n.
Max value of (a+b) is 18 as a and b can take values upto 9 since they are digits.
The possible values are 1,4,7,10,13,16.

Condition for divisibility by 4:
For divisibility by 4, Last two digits must be divisible by 4.
i.e 10a+b = 0 mod 4
=> 2a + b = 0 mod 4 (Since, 8a = 0 mod 4)
So, 2a + b can be 4,8,12,16,20,24,28.
Combining these ,we get
a + b = 3n +1
2a + b = 4k
The smallest values which satisfies these
two equation is a = 4 ,b = 0.
So, the minimum sum of a + b is 4

lcm of 12,14,21,33,54=8316
now checking through options,
on adding \frac{7249+8136}{8316}=not perfetily divisible
similarly on checking through options,\frac{(9383+7249)}{8316}=2
answer=9383

Video solution:

let one number be x
the other number will be 50-x
\frac{1}{x} + \frac{1}{(50-x)}= \frac{1}{12}
50-x+x =\frac{x(50-x)}{12}
600 = 50x-x^{2}
x^{2}-50x+600=0
x^{2}-30x-20x+600=0
x(x-30)-20(x-30)=0
(x-30)(x-20)=0
x= 20or 30 the numbers are 20 and 30

Solution- 6^{17}+17^{6} = (7-1)^{17} + (21-4)^{6}

= (7-1)^{17} + (7 \times 3 -4)^{6}....[eqn1]
if the [eqn1] is expanded then every term of the expansion except [(-1)^{17} +(-4)^{6}] will have 7 as one of its factors.
Just think a little bit about the binomial expansion of both [(7-1)^{17}] and [(7 \times 3-4)^{6}] ,then u can readily point out that only the last term of both the expansions ,
i.e , [(-1)^{17}] & [(-4)^{6}] respectively,will don't have 7 as one of its factors.
So, we have to calculate the remainder when [(-1)^{17} + (-4)^{6}] is divided by 7.
Now, clearly (-1)^{17} = -1
and, (-4)^{6} = 4^{6} = 2^{12} = (2^{3})^{4} = (7+1)^{4}.
Now, a same reasoning related to binomial expansion mentioned previously
explains why, when (7+1)^{4} is divided by 7 will leave a remainder 1.
So, (7+1)^{4} will be of the for (7A + 1); where A is some +ve integer, to know
whose value isn't important in this case.
So, when
[(-1)^{17} + (-4)^{6}] will be divided by 7
or, [-1 + (7+1)^{4}] will be divided by 7
or, when [-1 + 7A +1] will be divided by 7
or, when [ 7A ] will be divided by 7 , clearly therefore the remainder will be zero,
i.e, 0.

no will be divisible by 45 when it is divided y both 9 and 5
a no is divisible by 9 if sum of digits is divisible by 9
now sum of digits of given no is 270
so the no is div by 9
and since last digit is 4 when divisible by 5 it leaves remainder 4
let us say numbers such as 9,54...
we see they are divisible by 9 but leaves remainder 4 when divisible by 5..
so when div by 45 they leave rem 9
so req ans=9

144= 2^{4} \times 3^{2}
Hence the Number of factors = (4+1) x (2+1) = 5 x 3 = 15.
We know that if a number represented in standard form (a^{m} \times b^{n}) , then the
number of factors Is given by (m+1)(n+1).
so 15 will be answer

let x is divided by 406 & gives quotient y & remainder 115 then
x = 406 \times y + 115
now, \frac{x}{29} = \frac{(406 \times  y + 115)}{29}

= \frac{406 y}{29} + \frac{115}{29}
=> rem = \frac{115}{29} = 28

We can solve this question by Fermat's little theorem. Let me give a prelude to the theorem.

Fermat's little theorem
It states that if p is a prime number, then a^{p-1} =1 (mod p), given that p & a are co-primes.
Here, 11 is a prime number. Also, 30 and 11 are co-primes.

30^{10} = 1 (mod 11).

It is required to simplify 30^{72^{87}} to the form 30^{10^{x}} to apply Fermat's theorem.

It is enough to find the unit's digit of expansion 72^{87} since any value of 30^{10^{x}} (mod 11) will be 1.

To find the units digit of 72^{87}, it's enough to find last digit of 2^{87}.

Taking cyclicity of 2, last digit of 2^{87} is 8.

Now, we have 30^{72^{87}} = 30^{10^{x}} . 30^{2^{87}} = 1 . 30^{8} = 30^{8} (mod 11)

Applying basic remainder theorem, 30^{8} (mod 11) = 8^{8} (mod 11) = 64 ^{4} (mod 11) = (-2)^{4} (mod 11) = 16 (mod 11) = 5.

For more detailed explanations regarding the Remainder theorem. Click here

LCM of 15, 20 and 36 is 180

Now 180 = 3 x 3 x 2 x 2 x 5

To make it perfect square, it must be multiplied from 5.

So required no. = 3² x 2² x 5² = 900

6^{3}=216 this gives remainder 1
6^{4}=1596 this gives remainder 6
6^{5}=7776 this gives remainder 36
6^{6}=46656 this gives remainder 1
similarly,
6^{7} gives remainder 6(remainder is repeating)
Hence,6^{50} gives remainder 36