Number System, Problems on Number

Question: If n is a whole number greater then 1, then n² (n² - 1) is always divisible by -

Solution: 
Given that,
n²(n² - 1)
= 2²(2² - 1)     ; [put n = 2]
= 4(4 - 1)
= 4 × 3
= 12
Check the option it is divisible by 12.
Take n = 3
= 3³(3² - 1)
= 9(9 - 1)
= 9 × 8
= 72
It is divisible by 12

LCM of 5, 6, 7 and 8 = 840
Hence the number can be written in the form (840k + 3) which is divisible by 9.
If k = 1, number = (840 × 1) + 3 = 843 which is not divisible by 9.
If k = 2, number = (840 × 2) + 3 = 1683 which is divisible by 9.
Hence 1683 is the least number which when divided by 5, 6, 7 and 8 leaves a remainder 3,
but when divided by 9 leaves no remainder.

1470 = 2 x 3 x 5 x 7 x 7
So, the least number is = 2 x 3 x 5 = 30

Question: What is the greatest prime factor of 25¹⁵ − 5²⁸?

Solution:
25¹⁵ − 5²⁸
= (5²)¹⁵ - 5²⁸
= 5³⁰ - 5²⁸
= 5²⁸(5² - 1)
= 5²⁸(25 -1)
= 5²⁸ × 24
= 5²⁸ × 8 × 3
= 5²⁸ × 2³ × 3

সুতরাং, সম্পূর্ণ রাশিটির মৌলিক গুণনীয়কগুলো (prime factors) হলো 5, 2, এবং 3.
এর মধ্যে সবচেয়ে বড় হল 5. 
সুতরাং, বৃহত্তম মৌলিক গুণনীয়ক (prime factors) হলো 5.

Question: Let N be the smallest positive integer that is divisible by both 12 and 15. How many distinct prime factors does N have?

Solution:
এখানে, N হলো 12 এবং 15 দ্বারা বিভাজ্য ক্ষুদ্রতম সংখ্যা।

সুতরাং, N হবে 12 এবং 15 এর ল.সা.গু।

এখন, 12 = 2 × 2 × 3 = 2² ×3

এবং 15 = 3 × 5

LCM(12,15) = 2² × 3 × 5
= 60

অতএব, N = 60
60 এর মৌলিক উৎপাদক = 2² × 3 × 5

স্বতন্ত্র মৌলিক উৎপাদকগুলি হলো 2, 3 এবং 5।

∴ N এর স্বতন্ত্র মৌলিক উৎপাদকের সংখ্যা হলো 3টি।

Question: The sum of the squares of three numbers is 138, while the sum of their products taken two at a time is 131. Their sum is-

Solution:
Let the numbers are a, b and c.
Then,
a² + b² + c² = 138
and ab + bc + ca = 131.

Now,
(a + b + c)² = a² + b² + c² + 2(ab + bc + ca) = 138 + (2 × 131) = 138 + 262 = 400
⇒ (a + b + c)² = 400
∴ a + b + c = 20.

Let, first number be a and second number be b
Here, a+b = 28
ATQ, 6a+4b=152
⇒ 6a+4(28-a) = 152
⇒ 6a + 112 - 4a = 152
⇒ 2a = 152 - 112 = 40
⇒ a = 20

Let the three consecutive numbers are x, x + 1, x + 2
According to question
x² + (x + 1)² + (x + 2)² = 110
⇒ x² + x² + 1 + 2x + x² + 4 + 4x = 110
⇒ 3x² + 6x + 5 = 110
⇒ 3x² + 6x - 105 = 0
⇒ x² + 2x - 35 = 0
⇒ x_² + 7x - 5x - 35 = 0
⇒ x(x + 7) -5(x + 7) = 0
⇒ (x + 7)(x - 5) = 0
⇒ x = 5 & -7
[∴ (-) value can not considered]
∴ Smallest number is = 5

Question: Which of the following is a perfect square?

Solution:
বর্গসংখ্যা (perfect square): সাধারণভাবে একটি স্বাভাবিক সংখ্যা m কে যদি অন্য একটি স্বাভাবিক সংখ্যা n এর বর্গ (n²) আকারে প্রকাশ করা যায় তবে m বর্গসংখ্যা।
m সংখ্যাগুলোকে পূর্ণবর্গসংখ্যা বলা হয়।

পূর্ণবর্গ সংখ্যার ধর্ম:
• যে সংখ্যার সর্ব ডানদিকের অঙ্ক অর্থাৎ একক স্থানীয় অঙ্ক ২ বা ৩ বা ৭ বা ৮ তা পূর্ণবর্গ নয় ।
• যে সংখ্যার শেষে বিজোড় সংখ্যক শূন্য থাকে, ঐ সংখ্যা পূর্ণবর্গ নয়।
• একক স্থানীয় অঙ্ক ১ বা ৪ বা ৫ বা ৬ বা ৯ হলে, ঐ সংখ্যা পূর্ণবর্গ হতে পারে। যেমন : ৮১, ৬৪, ২৫, ৩৬, ৪৯ ইত্যাদি বর্গসংখ্যা ।
• আবার সংখ্যার ডানদিকে জোড়সংখ্যক শূন্য থাকলে ঐ সংখ্যা পূর্ণবর্গ হতে পারে। যেমন : ১০০, ৪৯০০ ইত্যাদি বর্গসংখ্যা ।

প্রদত্ত অপশগুলোর মধ্যে
গ) 49 = 7² [যা পূর্ণবর্গসংখ্যা]

Let the number of oranges in first basket be x,
Number of oranges in second basket = 640 - x
ATQ, x - x/5 = 640 - x + x/5
⇒ 4x/5 = 640 - 4x/5
⇒ 4x/5 + 4x/5 = 640
⇒ 8x/5 = 640
⇒ x = 640 × (5/8)
⇒ x = 400
∴ Number of oranges in first basket = 400.

Clearly, the numbers are (23 x 13) and (23 x 14).
Larger number = (23 x 14) = 322.

Question:  In a seminar, the number of participants in Bangla, English and Mathematics are 60, 84 and 108 respectively. Find the minimum number of rooms required, where in each room the same number of participants are to be seated; and all of them being in the same subject.


Solution: 
HCM of 60, 84 and 108 = 12

Minimum number of rooms required = (60/12) + (84/12) + (108/12)
= 5 + 7 + 9 
= 21

Let the numbers be x and x + 2.
Then, (x + 2)² - x² = 84
⇒ 4x + 4 = 84
⇒ 4x = 80
⇒ x = 20.
∴ The required sum
= x + (x + 2)
= 2x + 2
= 42

Let the ten’s digit be x.
Then, unit’s digit = (x+3).
Sum of the digits = x + (x +3) = 2x + 3.
Number = 10x + (x+3) = 11x + 3.
(11x + 3)/(2x + 3) = 4/1
11x + 3 = 8x + 12
11x - 8x = 12 - 3
3x = 9
x = 3
Hence, Required number = 11x + 3 = 36.

Let the numbers be x and (184-x).
Then,
x/3 - (184 -x)/7 = 8
⇒ 7x – 3(184-x) = 168
⇒ 10x = 720
⇒ x = 72.
So, the numbers are 72 and 112.
Hence, a smaller number = 72.

L.C.M. of 21, 36, 66 = 2772.
Now, 2772 = 2 x 2 x 3 x 3 x 7 x 11
To make it a perfect square, it must be multiplied by 7 x 11.
So, required number = 2² x 3² x 7² x 11² = 213444

6 = 2 × 3
10 = 3 × 5
35 = 5 × 7
27 = 3 × 9; Here 9 is not a prime number

(2 + 3 + 5 + x)/4 = 4
Or, 2 + 3 + 5 + x = 4×4
Or, x = 16 - 10 = 6

Question: If both 5² and 3² are factors of m where m = z × 2⁵ × 6² × 7³ , what is the smallest positive value of z? 

Solution:
Given, m = z × 2⁵ × 6² × 7³

We simplify this by expressing all terms in their prime factorizations except z,
∴ 6² = (2 × 3)² = 2² × 3²
∴ m = z × 2⁵ × 2² × 3² × 7³
= z × 2⁷ × 3² × 7³
 
If both 5² and 3² are factors, then they must be present in the number.
From the simplified expression of m, we can see that the factor 3² is already present but the required factor 5² is missing.
To make 5² a factor of m, it must come from z.

∴ The smallest positive value of z = 5² = 25.  

Given,
52416/312=168
⇔ 52416/ 168 =312
Now, 52.416/ 0.0168
= 524160/ 168
= (52416/ 168 )×10
=312×10
=3120

(2⁴)² − 1
= 2⁸ - 1
= 256 - 1
= 255
= 3 × 5 × 17

So, the smallest prime factor is 3

Let, the number be z,
Then, 7z – 15 = 2z + 10
⇒ 5z = 25
⇒ z = 5.
Hence, the required number is 5.

Let, the numbers be = x, x+2, x+4, x+6
ATQ, x + x + 2 + x + 4 + x + 6 = 68
Or, 4x = 68 - 12 = 56
Or, x = 14

So, the largest number is = 14 + 6 = 20

Let the numbers are 7x and 9x

According to the question,
⇒ 7x × 9x = 1575
⇒ 63x² = 1575
⇒ x² = 25
⇒ x = 5

Then greater number
= 9x
= 9 × 5
= 45

Question: A number is 5 more than its one-sixth part. The number is-

Solution:
Given that,
A number is 5 more than its one-sixth part.
Let the number be x.

According to the question,
x = (1/6)x + 5
⇒ x - (1/6)x = 5
⇒ (6x - x)/6 = 5
⇒ 5x/6 = 5
⇒ x = 5 × (6/5)
∴ x = 6

So the number is 6.

Given, 9 + 36 + 81 + 144
The series is = 3² + 6² + 9² + 12² + 15²
So, next term is 15² = 225

Numbers can be divided by 7 are: 7, 14, 21, 28, 35
Among them, 14, 28 are divided by 14 with no remainder
But when 7, 21, 35 these numbers are divided by 14, the remainder is 7

যেহেতু,
N ঋনাত্মক পূর্ণসংখ্যা।
ধরি, N = -7

∴ (N)² = (-7)² = 49
6 - N = 6 - (-7) = 13
- N = -(-7) = 7
6 + N = 6 + (-7) = -1

তবে অপশন D তে N এর মান -6 বা তার কম হলে এটার মানও ধণাত্মক হতো। এজন্যই প্রশ্নে 'Could have a negative value' টার্ম ব্যবহার করা হয়েছে

∴ 6 + N এর মান ঋনাত্মক হতে পারে।

Let the number be x and y, such that x > y.
then, 3x - 4y = 5 ..........(i)
And (x + y) - 6 (x - y) = 6 ⇔ -5x + 7y = 6 ........(ii)
Solving (i) and (ii), we get : x = 59 and y = 43
Answer : 59, 43.

If the remainder is same in each case and remainder is not given,
HCF of the differences of the numbers is the required largest number.

9997 - 7654 = 2343
9997 - 8506 = 1491
8506 - 7654 = 852
Hence, the greatest number which divides 7654, 8506 and 9997 and leaves same remainder
= HCF of 2343, 1491, 852
= 213
Now we need to find out the common remainder.
Take any of the given numbers from 7654, 8506 and 9997, say 7654
7654/213 = 35,
remainder = 199.

Let, first number be x and second number be y
Here,
x + y = 36 .... (i)
and, 5x + 3y = 142 ...(ii)

by multiplying the first equation by 3
3x + 3y = 108
5x + 3y = 142

by subtraction the second equation from the first
(3–5)x + (3–3)y = 108–142
Or, -2x = -34
Or, x = 17

by substituting x in the first equation
17 + y = 36
Or, y = 19