PE 2 - Numbers | Algebra - Number Theory

Answer: 97

Explanation :

The lowest such possible perfect square is 16 (4²).

The highest such possible perfect square is 10000 itself (100²).

∴ Number of perfect squares from 10 to 10,000 is (100 – 4 + 1) = 97.

Hence, 97.

Answer: Option D

Explanation :

Consider a 4-digit number – a b c d. 
The number formed by reversing the last two digits is – a b d c. 

Given,
1000a + 100b + 10c + d - (1000a + 100b + 10d + c) = 72
9c - 9d = 72
c -d = 8.

Alternately,
We could have safely ignored the first 2 digits, as they would have not mattered after subtraction, this gives us {for last 2 digits}.
(10c + d) – (10d + c ) = 72
⇒ 9(c – d) = 72
⇒ c – d = 8

Hence, option (d).

Answer: Option E

Explanation :

Factorizing 143 we get, 143 = 11 × 13

The pairs of x & y, which satisfy LCM(x, y) = 143 are
1.    1, 143
2.    11, 143
3.    13, 143
4.    143,143
5.    11, 13     

Out of these cases, (1) & (4) can be eliminated as they do not follow the criteria y > x; x > 1. 

Still there are three different cases possible.

Hence, option (e).

Answer: Option A

Explanation :

Here, XX and YY are multiples of 11, hence their sum ABA should also be a multiple of 11.

Also, sum of 2 2-digit numbers cannot exceed 198.
∴ A = 1 and ABA = 1B1.

Now, 1B1 should be a multiple of 11. This is only possible when B = 2.

Hence, A = 1 and B = 2

⇒ A + B + A = 1 + 2 + 1 = 4

Hence, option (a).

Answer: Option E

Explanation :

Let the middle number be ‘a’.

According to the question: (a – 1)(a + 1) = a² - 1
⇒ a² – 1 = a² - 1

Now, this is true for all values of ‘a’.

Hence, option (e).

Answer: 1300

Explanation :

A = 1¹ × 2² × 3³ × 4⁴ × … 100¹⁰⁰

Product of one 2 and one 5 gives one trailing zero.

Here number of 2’s will definitely be more than number of 5’s, hence number of trailing zeros will be same as number of 5’s.

Powers of 5 will occur at 5⁵, 10¹⁰, 15¹⁵ … 100¹⁰⁰

Multiples of 5 
power at least 1 = 5 + 10 + 15 + … + 100 = 1050
additional power of 5 (i.e., power at least 2) = 25 + 50 + 75 + 100 = 250

Total number of 5s = 1050 + 250 = 1300

i.e., there will be 1300 trailing zeros in A.

Hence, 1300.

Answer: Option B

Explanation :

One of x and x⁴ + 1, will definitely be even and the other odd.

The only even prime number is 2. Hence, either x = 2 or x⁴ + 1 = 2

If x⁴ + 1 = 2 ⇒ x = 1 (not a prime number, hence rejected)

∴ x = 2 and x⁴ + 1 = 17.

∴ Only one value of x is possible.

Hence, option (b).

Answer: 8

Explanation :

‘14’ as a product of 2 integers can be written as 1 × 14, 2 × 7, - 1 × -14, -2 × -7 i.e. 4 ways.

‘14’  as a product of 3 integers can be written as 1 × 2 × 7, -1 × -2 × 7, -1 × 2 × -7, 1 × -2 × -7 i.e. 4 ways.

∴ Total number of ways = 4 + 4 = 8.

Hence, 8.

Answer: 22

Explanation :

Let N = ‘ab’, where a and b are digits
N = 10a + b

Now, since a is a factor of ‘ab’, a should divide (10a + b) completely.
i.e., $R\left[\frac{10a+b}{a}\right]$ = 0

⇒ $R\left[\frac{10a}{a}\right]$ + $R\left[\frac{b}{a}\right]$ = 0

⇒ $R\left[\frac{b}{a}\right]$ = 0

∴ ‘b’ should be divisible by a.

If a = 1, b can be 0, 1, 2, 3, …, 9 ⇒ 10 possible values
If a = 2, b can be 0, 2, 4, 6, 8 ⇒ 5 possible values
If a = 3, b can be 0, 3, 6, 9 ⇒ 4 possible values
If a = 4, b can be 0, 4, 8 ⇒ 3 possible values

Number of possible values of ‘N’ = 10 + 5 + 4 + 3 = 22.

Hence, 22.

Answer: Option C

Explanation :

The required number should be of the form A × x (where x is any positive integer)

Also, B should be divisible by A × x

Now, 2020²⁰²² = 2020²⁰²⁰ × 2020²

Number of possible values of x will be same as the number of factors of 20202.

2020² = 2⁴ × 5² × 101²

Number of factors of 2020² = (4 + 1) × (2 + 1) × (2 + 1) = 45.

Hence, option (c).