TCS Divisibility Question Answers 2

Solution:

When she bought 15 pencil packets it contains 180 pencils ( 15×12 =180 )

she have to bought 10 eraser packets(10×18 = 180) and 18 sharpener packets ( 18×10 = 180 )

Video solution:

Solution:

By going through every option we can say that:

1.  9910 is divisible by 5 and 10 but not divisible by 15 and 20


2.  9990 is divisible by 5, 10, and 15 but not divisible by 20


3.  9980 is divisible by 5, 10, and 20 but not divisible by 15


4.  9960 is divisible by 5, 10, 15, and 20.

44⁴⁴
= 2⁸⁸ x 11⁴⁴

Immediately, this has to be divded by 8ⁿ, which is same as 2³ⁿ

So, = \frac{2^{88} \times 11^{44}}{2^{3n}}
Thus, 2^88-3n should be greater than 0.

Now, the only way it is greater is for 88 - 3n be greater than 0. For n = 29 its at its max greater than 0.

1728=2^{6}\times 3^{3}
Hence the Number of factors = (6+1) x (3+1) = 7 x 4 = 28.
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).
Answer is 28

We recommend reading this page first to understand the solution (Has some cool tricks)

How to calculate remainders of large powers

Using Fermats Theorem (Given on page above)

If p is a prime, and HCF (a, p) = 1 (a and p are co-primes), then Rem[\frac{a^{p-1}}{p}] = 1

• Clearly, 16937 and 31 are co primes


• Power is 30 which is in format p - 1

- as remainder is 13 & 17 so dividend must be greater than 17
272738 = n x a+13 => 272725=n x a ---(1)
232342 = n x b+17 => 232325=n x b ---(2)
last two digit of (1)&(2) is 25 so n must be 25, no other two digit no. greater than
25 satisfies this
272725=25 x 10909
232325=25 x 9293
n=25, sum of digits of n = 2+5 = 7

\frac{32^{31^{301}}}{9}=\frac{5^{31^{301}}}{9}

\text{Euler\ totient\ theorem\ says\ that}

\left[\frac{a^{\phi(n)}}{n}\right]_{\text{Rem}}=1

\phi(n)=n\left(1-\frac{1}{a}\right)\left(1-\frac{1}{b}\right)\ldots\ \text{here}\ n=a^p b^q

\phi(9)=9\left(1-\frac{1}{3}\right)=6

\text{Therefore, }5^6\ \text{when divided by }9\ \text{leaves remainder }1

\frac{31^{301}}{6}=1^{301}=1

31^{301}=6k+1

5^{31^{301}}=(5^6)^k\cdot5^1

\frac{5^{31^{301}}}{9}=\frac{(5^6)^k\cdot5^1}{9}

=\frac{1^k\cdot5}{9}

=5

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

To get the original number add the 357 with remainder 5, you
will get 362. Then divide it by 17 you will get 5 as remainder.

If any number's last two digits are 25 or 75, then that should be divisible by 25.
b should be either 2 or 7.
The problem states that the two-digit number ab is a multiple of 13. The two-digit multiples of 13 are:
13, 26, 39, 52, 65, 78, ...
When ab=52, the number is 26525 and \frac{52}{13}=4,
So ab=52