PE 4 - Numbers | Algebra - Number Theory

Answer: 7

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Explanation :

Any number that when divided by 11 gives a remainder of 5 will be of the form 11k + 5.

Since we only need two-digit numbers, k will range from 1 to 8, as 11 × 1 + 5 = 16 and 11 × 9 + 5 = 93.

These numbers form an AP with first term = 16 and last term as 93.

Thus, sum of these numbers

S = 8/2 × (16 + 104) = 480

Dividing S by 11, the remainder comes out to be 7.

Hence, 7.

Answer: 5

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Explanation :

Let the number of rows be r and number of plants in each row be p. 

∴ r × p = 50   …(1)

Number of flowers on a plant in nth row = n²
∴ Total number of flowers in nth row is n² × p.

Total number of flowers in the garden = p × (1² + 2² + 3² + … + r²) = 550

⇒ pr(r + 1)(2r + 1)/6 = 550

⇒ 50 × (r + 1)(2r + 1)/6 = 550 [from (1): rp = 50]

⇒ (r + 1)(2r + 1) = 66 = 6 × 11

⇒ k = 5.

Hence, 5.

Answer: Option D

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Explanation :

(a + b) is greater than both a and b, hence (a + b)! will always be divisible by a! or b!.

Now, let us check a! × b!

$\frac{(a+b)!}{a!\times b!}$ = $\frac{1\times 2\times 3\times \dots \times a\times (a+1)\times (a+2)\times \dots \times b}{a!\times b!}$ = $\frac{(a+1)\times (a+2)\times \dots \times b}{b!}$

Now, (a + 1) × (a + 2) × … × b is product of b natural numbers, which will always be divisible by b!

∴ $\frac{(a+1)\times (a+2)\times \dots \times b}{b!}$ is an integer

⇒ $\frac{(a+b)!}{a!\times b!}$ is an integer i.e., (a + b)! is divisible by a! × b!

Hence, option (d).

Answer: 2

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Explanation :

To find remainder of any number when divided by 8, we need to find remainder of last 3 digits of the number when divided by 8.

Here we need to find the last 3 digits of 1223334444… (500 digits)

The given number is formed by writing consecutive natural numbers as many times.

1 is written once, 2 is written twice, 3 is written thrice and so on.

By the time number 1 is over, only 1 digit is written.
By the time number 2 is over, (1 + 2 =) 3 digits are written.
By the time number 3 is over, (1 + 2 + 3 =) 6 digits are written.
…
By the time all single digit numbers are written, (1 + 2 + 3 + … + 9 =) 45 digits are written.

By the time number 11 is over, 45 + 2 × 11 = 67 digits are written.
…
By the time number 23 is over, 45 + 2 × (11 + 12 + 13 + … + 23 =) 487 digits are written.

Now thirteen digits are remaining which will be formed by writing 24 repeatedly.

The last three digits of this number will be 242.

∴ The last 3 digits of the given number is 242.

Remainder when 242 is divided by 8 = 2.

∴ Remainder when the given number is divided by 8 = 2.

Hence, 2.

Answer: 4

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Explanation :

30! is divisible by all prime numbers less than 30. 

∴ The smallest prime number that is not its factor is 31. 

But N should not be a prime factor. 
∴ N must be kept as small as possible i.e., 31 × 2 = 62.

The factors of 62 are 1, 2, 31 and 62, i.e., 4.

Hence, 4.

Answer: Option B

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Explanation :

For a number to be divisible by 36, it should be divisible by 4 as well as 9.

Divisibility by 9: Sum of digits should be divisible by 9.
Sum of the digits = 6 + 4 + 5 + 4 + 3 + 8 + 9 + p + 6 = 45 + p
(45 + p) should be divisible by 9, hence p can be 0 or 9.

Divisibility by 4: Last two digits should be divisible by 4.
p6 should be divisible by 4.
Case 1: p = 0, 06 is not divisible by 4
Case 2: p = 9, 96 is divisible by 4

∴ For p = 9, the given number is divisible by both 9 and 4, hence divisible by 36.

Hence, option (b).

Answer: Option C

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Explanation :

Numbers of trailing zeros in n! is same as highest power of 5 in n!.

We will have to check options here. Let us first check for 941, since it is the middle number of all the options.

Highest power of 5 in 941! = 

$\frac{941}{5}$ + $\frac{941}{5^2}$ + $\frac{941}{5^3}$ + $\frac{941}{5^4}$ = 188 + 37 + 7 + 1 = 233

Now let us check the next higher option i.e., option (c)

$\frac{1003}{5}$ + $\frac{1003}{5^2}$ + $\frac{1003}{5^3}$ + $\frac{1003}{5^4}$ = 200 + 40 + 8 + 1 = 249 

∴ The number of trailing zeros in 1003! is 249. 

Hence, option (c).

Answer: Option C

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Explanation :

Using the cyclicity property we calculate the last digit of each term in the above series.

Last digit of 

1¹ = 1
2² = 4
3³ = 7
4⁴ = 6
5⁵ = 5
6⁶ = 6
7⁷ = 3
8⁸ = 6
9⁹ = 9

The sum of all the last digits = 47, which when divided by 5 leaves a remainder of 2.

∴ The remainder when 1¹ + 2² + 3³ + … + 9⁹ is divided by 5 is 2.

Hence, option (c).

Answer: Option C

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Explanation :

Let m = 7a and n = 7b (both are multiples of 7)

Option (a): m – n = 7(a - b) is divisible by 7.
Option (b): m² – n² = 49(a² - b²) is divisible by 49.
Option (c): m + n = 7(a + b) is divisible by 14 only when (a + b) is even.

∴ Option (c) is not necessarily true.

Hence, option (c).

Answer: Option C

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Explanation :

When a number N is divided by 5, 7 and 11, the remainder is 3.

N = k × LCM (5, 7, 11) + 3, where k = 1, 2, 3… and so on.

N = 385k + 3

When k = 5, N = 1928.

So, the number to be subtracted from 1995 is = 1995 – 1928 = 67.

Hence, option (c).