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1

### JEE Main 2021 (Online) 27th August Morning Shift

Numerical
A number is called a palindrome if it reads the same backward as well as forward. For example 285582 is a six digit palindrome. The number of six digit palindromes, which are divisible by 55, is ____________.

## Explanation

 5 a b b a 5

For divisible by 55 it shall be divisible by 11 and 5 both, for divisibility by 5 unit digit shall be 0 or 5 but as the number is six digit palindrome unit digit is 5.

A number is divisible by 11 if the difference between sum of the digits in the odd places and the sum of the digits in the even places is a multiple of 11 or zero.

Sum of the digits in the even place = a + b + 5

Sum of the digits in the odd places = a + b + 5

Difference between the two sums = (a + b + 5 ) - (a + b + 5) = 0

0 is divisible by 11.

Hence, 5abba5 is divisible by 11.

So, required number = 10 $$\times$$ 10 = 100
2

### JEE Main 2021 (Online) 26th August Morning Shift

Numerical
The number of three-digit even numbers, formed by the digits 0, 1, 3, 4, 6, 7 if the repetition of digits is not allowed, is ______________.

## Explanation

(i) When '0' is at unit place Number of numbers = 20

(ii) When 4 or 6 are at unit place Number of numbers = 32

Total three digit even number = 20 + 32 = 52
3

### JEE Main 2021 (Online) 26th August Morning Shift

Numerical
If $${}^1{P_1} + 2.{}^2{P_2} + 3.{}^3{P_3} + .... + 15.{}^{15}{P_{15}} = {}^q{P_r} - s,0 \le s \le 1$$, then $${}^{q + s}{C_{r - s}}$$ is equal to ______________.

## Explanation

$${}^1{P_1} + 2.{}^2{P_2} + 3.{}^3{P_3} + .... + 15.{}^{15}{P_{15}}$$

= 1! + 2 . 2! + 3 . 3! + ..... 15 $$\times$$ 15!

$$= \sum\limits_{r = 1}^{15} {(r + 1)! - (r)!}$$

= 16! $$-$$ 1

= $${}^{16}{P_{16}}$$ $$-$$ 1

$$\Rightarrow$$ q = r = 16, s = 1

$${}^{q + s}{C_{r - s}} = {}^{17}{C_{15}}$$ = 136
4

### JEE Main 2021 (Online) 27th July Evening Shift

Numerical
Let n be a non-negative integer. Then the number of divisors of the form "4n + 1" of the number (10)10 . (11)11 . (13)13 is equal to ___________.

## Explanation

N = 210 $$\times$$ 510 $$\times$$ 1111 $$\times$$ 1313

Now, power of 2 must be zero,

power of 5 can be anything,

power of 13 can be anything

But, power of 11 should be even.

So, required number of divisors is

1 $$\times$$ 11 $$\times$$ 14 $$\times$$ 6 = 924

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