TCS Divisibility Question Answer 4

There is a standard way of solving these questions quickly, which you should mug up/write in notes:dividend = divisor × quotient + remainder

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
So,
(m + n) = 12 x 1 + 8
(m + n) = 20
Similarly, for (m - n)
(m - n) = 12 x 1 + 6
(m - n) = 18
adding the above two equations we get
2m = 38
m = 19
replacing m in any of the above equation we will get n = 1
Thus, remainder then mn is divided by 12, i.e. \frac{19}{6}, remainder: 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

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

2x+1
3 \times (2x+1)+2
5 \times [3 \times (2x+1)+2] =30x+25
\frac{30x+25}{2}=15x+12 (rem=1)
\frac{[15x+12]}{3}=5x+4 (rem=0)
\frac{(5x+4)}{5}=x (rem=4)
Answer: 1,0,4

For a number to be perfectly divisible by 72, it must be divisible by its co-prime factors, which are 8 and 9 ( 8 \times 9 = 72 ).

Check the divisibility rule for 8. A number is divisible by 8 if its last three digits are divisible by 8.
- The last three digits of the number are 73x.
- We know that 720 is divisible by 8 ( 8 \times 90 = 720 ). Therefore, the next multiples of 8 are 728 and 736.
- So, 736 is divisible by 8, which means x = 6.

Check the divisibility rule for 9 to verify the value of x. A number is divisible by 9 if the sum of its digits is divisible by 9.
- Sum of digits = 4 + 2 + 5 + 7 + 3 + x = 21 + x.
- Substituting x = 6, the sum becomes 21 + 6 = 27.
- Since 27 is completely divisible by 9, our confirmed value for x is correct.

Therefore, the value of x is 6.

For divisible by 5 unit digit should be 0 or 5

_ _ _ _ _ _ _ _ _ (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}

For a number to be exactly divisible by 88, it must be divisible by its co-prime factors, 8 and 11 ( 8 \times 11 = 88 ).

Check the divisibility rule for 8 to find y. A number is divisible by 8 if its last three digits are divisible by 8.
- The last three digits of the given number are 91y.
- When 91y is divided by 8, we know 8 \times 110 = 880 , leaving a remainder and a resulting digit 3y.
- The only number in the 30s divisible by 8 is 32 ( 8 \times 4 = 32 ). Therefore, y = 2.
- The number becomes 179x0912.

Check the divisibility rule for 11 to find x. A number is divisible by 11 if the difference between the sum of digits at odd places and the sum of digits at even places is either 0 or a multiple of 11.
- Sum of digits in odd places (from the left): 1 + 9 + 0 + 1 = 11.
- Sum of digits in even places (from the left): 7 + x + 9 + 2 = 18 + x.
- Difference: (18 + x) - 11 = 7 + x.
- For 7 + x to be a multiple of 11, the smallest single-digit value for x must be 4, since 7 + 4 = 11.
- Therefore, x = 4.

Calculate the final value of (5x - 8y).
- Substitute x = 4 and y = 2.
- (5 \times 4) - (8 \times 2) = 20 - 16 = 4.

Therefore, the value of (5x - 8y) is 4.

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

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