TCS Divisibility Question Answer 4

use this simple technique.[ (1 x 3) + 2] = 5
[(5 x 5) + 0] = 25
Procedure:
Let the number = N
Now N = 5K
K = 3L + 2
L = 2M + 1
K = 3(2M + 1) + 2 = 6M + 5
N = 5(6M + 5) = 30 M + 25
For M = 0 we get the least number as 25. Now when 25 is divided by 2, we get 12
as quotient
and 1 as remainder. When 12 is divided by 3 we get 4 as quotient, and 0 as
remainder. When 4 is
divided by 5 we gt 4 as remainder.

Video solution:

2299 is divisible by 11.
2299 = 11×11×19×1×1=11× −11×19× −1×1=
Two of the terms in the given expression should equal to 1. As all the digits are
distinct, two of
the terms should be negative.
One possible solution = (75 - 64)(75 - 56)(75 - 86)(75 - 74)(75 - 76)
Then a + b + c + d + e = 64 + 56 + 86 + 74 + 76 = 356
But as the sum of only 4 terms was asked, we have to subtract one term.
So given answer can be one of 292, 306, 270, 282, 280

Video solution:

for divisible by 5 unit digit should be 0 or 5

Video solution:

_ _ _ _ _ _ _ _ _ (9 digit number)
Digits that can be used = 1, 2, 3, 4 and 5
For a number to divisible by 4, the last two digits have to be divisible by 4
So, the units digit cannot be 1, 3 or 5
If the units digit is 2, only 12, 32 adn 52 are acceptable as the last two digits
If the units digit is 4, only 24 and 44 are acceptable as the last two digits.
So, total 5 cases 12,24,32,44,52 are possible for the last two digits.
Now, the first 7 digits of the 9 digit number can be filled in 5^{7} ways as each place can be filled by 1, 2, 3, 4 or 5.
So, desired answer = 5^{7}\times 5 = 5^{8}

Video solution:

If a number has to be divisible by each number from 1 to 7, that number
should be L.C.M
of(1,2,3,4,5,6,7) = 420

Video solution:

n= 4p = 2^{2} \times p^{1}
so total no. of factors: (2+1) x (1+1)= 6
total no. of odd factors, since p is odd as it is a prime>2: p¹ and p⁰ = 2
Total no. of even factors: 6 - 2 = 4

Video solution:

We need to break down the question first.

6^{17}

=6⁽¹⁶⁺¹⁾
=(6)(6¹⁶) as [a^m x aⁿ = a^{m + n}]
=(6)((6²)⁸)

=(6)((36)⁸)

now, rem(\frac{6}{7})*rem(\frac{36}{7}\times  \frac{36}{7} \times  \frac{36}{7} \times... 8 times)

=(-1)(1\times 1\times 1\times ...8 times)= -1

2nd: rem \frac{117^{6}}{7})

= rem(\frac{117}{7}\times \frac{117}{7} \times \frac{117}{7} ... 6 times)

= rem(\frac{(-2)^{6}}{7})=rem(\frac{64}{7})=1
now just add up 1st and 2nd
-1 + 1 = 0
so, reminder is zero

Video solution:

There is a standard way of solving these questions quickly, which you should mug up/write in notes:

dividend = divisor × quotient + remainder

• The remainder when (m + n) is divided by 12 is 8


• What will be the minimum possible value of (m+n) when the quotient is 1

• (m + n) = 12 x 1 + 8


• (m + n) = 20

• (m - n) = 12 x 1 + 6


• (m - n) = 18

• 2m = 38


• m = 19


• replacing m in any of the above equation we will get n = 1

The divisibility rule for 9 is sum of the digits is to be divisible by 9. So
We calculate separately, sum of the digits in hundreds place, tenths place, and units place.
Sum of the digits in hundreds place: 1 x 50 = 50
Sum of the digits in tenths place : 0 x 9 + 1 x 10 + 2 x 10 + 3 x 10 + 4 x 10 + 5 x 1 = 105
Sum of the digits in units place : (1 + 2 + 3 + ...+ 9) x 5 = 225
So total = 380
So remainder = \frac{380}{9} = 2

Video solution:

Let both numbers are a, b
a-b= 2001
a=b×9+41
=9b+41-b=2001
8b=1960
b= 245
a=2001+245= 2246
Sum of the digits of the larger number= 2+2+4+6= 14

Video solution: