# PE 2 - Numbers | Algebra - Number Theory

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**PE 2 - Numbers | Algebra - Number Theory**

How many even numbers from 10 to 10,000 are not perfect squares (including both)?

Answer: 4947

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**Explanation** :

The lowest such possible perfect square is 16 (4^{2}).

The highest such possible perfect square is 10,000 itself (100^{2}).

∴ The perfect squares are: 4^{2}, 6^{2}, ..., 100^{2 }= (2 × 2)^{2}, (2 × 3)^{2}, ..., (2 × 50)^{2}.

∴ Number of even perfect squares from 10 to 10,000 is (50 – 2 + 1) = 49.

Number of even numbers from 1 to 10,000 = 10,000/2 = 5000

Number of even numbers from 10 to 10,000 = 5000 - 4 = 4996 (excluding 2, 4, 6 and 8).

∴ Number of even numbers from 10 to 10,000 which are not perfect squares = 49996 - 49 = 4947.

Hence, 4947.

Workspace:

**PE 2 - Numbers | Algebra - Number Theory**

If the last 2 digits of a four-digit number are interchanged, the new number obtained is greater than the original number by 72. What is the difference between the last two digits of the number?

- (a)
9

- (b)
1

- (c)
6

- (d)
8

- (e)
Cannot be determined

Answer: Option D

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**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).

Workspace:

**PE 2 - Numbers | Algebra - Number Theory**

LCM of two numbers x and y is 143. Find the value of |x – y|, given that y > x, x > 1.

- (a)
30

- (b)
15

- (c)
41

- (d)
165

- (e)
Cannot be determined

Answer: Option E

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**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).

Workspace:

**PE 2 - Numbers | Algebra - Number Theory**

XX and YY are 2-digits numbers such that their sum is ABA. Find the value of A + B + A?

- (a)
4

- (b)
3

- (c)
2

- (d)
1

- (e)
5

Answer: Option A

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**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).

Workspace:

**PE 2 - Numbers | Algebra - Number Theory**

Three consecutive whole numbers are such that product of smallest and the largest number is one less than the square of the middle number. Find the middle number.

- (a)
6

- (b)
18

- (c)
12

- (d)
31

- (e)
All of these

Answer: Option E

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**Explanation** :

Let the middle number be ‘a’.

According to the question: (a – 1)(a + 1) = a^{2} - 1

⇒ a^{2} – 1 = a^{2} - 1

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

Hence, option (e).

Workspace:

**PE 2 - Numbers | Algebra - Number Theory**

A = 1^{1} × 2^{2} × 3^{3} × 4^{4} × … 100^{100}. How many zeroes will be there at the end of A?

Answer: 1300

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**Explanation** :

A = 1^{1} × 2^{2} × 3^{3} × 4^{4} × … 100^{100}

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 5^{5}, 10^{10}, 15^{15} … 100^{100}

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.

Workspace:

**PE 2 - Numbers | Algebra - Number Theory**

If x is a prime such that (x^{4} + 1) is also a prime, then how many values are possible for x.

- (a)
0

- (b)
1

- (c)
2

- (d)
more than 2 values

Answer: Option B

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**Explanation** :

One of x and x^{4} + 1, will definitely be even and the other odd.

The only even prime number is 2. Hence, either x = 2 or x^{4} + 1 = 2

If x^{4} + 1 = 2 ⇒ x = 1 (not a prime number, hence rejected)

∴ x = 2 and x^{4} + 1 = 17.

∴ Only one value of x is possible.

Hence, option (b).

Workspace:

**PE 2 - Numbers | Algebra - Number Theory**

In how many ways can 14 be written as a product of two or more distinct integers, without considering two integers with the same absolute value but opposite signs?

Answer: 8

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**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.

Workspace:

**PE 2 - Numbers | Algebra - Number Theory**

‘N’ is a two-digit number less than 50 such that when the unit’s digit of ‘N’ is erased, the resulting number is a factor of ‘N’. How many possible values of ‘N’ are there?

Answer: 22

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**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.

Workspace:

**PE 2 - Numbers | Algebra - Number Theory**

How many integers exist such that not only are they multiples of A = 2020^{2020} but also are factors of B = 2020^{2022}?

- (a)
36

- (b)
48

- (c)
45

- (d)
None of these

Answer: Option C

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**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, 2020^{2022} = 2020^{2020} × 2020^{2}

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

2020^{2} = 2^{4} × 5^{2} × 101^{2}

Number of factors of 2020^{2} = (4 + 1) × (2 + 1) × (2 + 1) = 45.

Hence, option (c).

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