Number System, Problems on Number

Let the numbers be x and (x + 3)
Then,
⇔x²+(x+3)²=369
⇔x²+x²+9+6x=369
⇔2x²+6x−360=0
⇔x²+3x−180=0
⇔(x+15)(x−12)=0
⇔x=12 
So, the numbers are 12 and 15
∴ Required sum = (12 + 15) = 27

Question: M and N are two positive integers such that MN = 72. Which of the following cannot be the value of M + N?

Solution:
72-এর উৎপাদক জোড়াগুলো হল:
1 × 72 = 72 ⇒ M + N = 1 + 72 = 73
2 × 36 = 72 ⇒ M + N = 2 + 36 = 38
3 × 24 = 72 ⇒ M + N = 3 + 24 = 27
4 × 18 = 72 ⇒ M + N = 4 + 18 = 22
6 × 12 = 72 ⇒ M + N = 6 + 12 = 18
8 × 9 = 72 ⇒ M + N = 8 + 9 = 17

সুতরাং, M + N এর সম্ভাব্য মানগুলো: 17, 18, 22, 27, 38, 73

তাই, M + N = 25 হতে পারে না।

Question: If , the value of x is = ?

Solution: 
Given that, 
52/x = √(169/289)
⇒ 52/x = 13/17
⇒ x = (52 × 17)/13
⇒ x = 4 × 17
∴ x = 68

Let the numbers be 2x, 5x
ATQ,
(2x + 16) / (5x + 16) = 1/2
Or, 4x + 32 = 5x + 16
Or, x = 16
∴ The numbers: 2x = 2.16 = 32
5x = 5.16 = 80

Question: (x - 35) is divisible by 36, 48, 60. Find the value of x.

Solution:
Find the LCM of 36, 48, and 60.
Prime factorization
36 = 2² × 3²
48 = 2⁴ × 3¹
60 = 2² × 3¹ × 5¹

LCM formula
LCM = 2⁴ × 3² × 5 = 16 × 9 × 5 = 720
Let
x - 35 = 720
x = 720 + 35
x = 755

Question: Find the smallest positive integer that must be added to 11356 so that it becomes divisible by both 18 and 22.

Solution:
Since the number must be divisible by both 18 and 22, it must be divisible by their Least Common Multiple (LCM).

LCM of 18 and 22:
18 = 2 × 3²
22 = 2 × 11
∴ LCM = 2 × 3² × 11 = 198

Now, dividing 11356 by 198:
11356 = (198 × 57) + 70
Here, the remainder is 70.

Since we need to add a number to make it divisible by 198,
∴ Smallest integer to be added = Divisor - Remainder
= 198 - 70
= 128

Therefore, 128 must be added to 11356 to make it divisible by both 18 and 22.

We are given that the numbers m and n, when divided by 6, leave remainders of 2 and 3, respectively,
Hence, we can represent the numbers m and as 6p +2 and 6q + 3, respectively, where p and q are suitable integers.
Now, m - n = (6p + 2) - (6q + 3) = 6p - 6q - 1 = 6(p-q) - 1.
A remainder must be positive, so let's add 6 to this expression and compensate by subtracting 6 :
6(p - q) - 1 =
6 (p - q) - 6 + 6 - 1 =
6 (p - q) - 6 = 5 =
6 (p - q - 1) + 5
Thus, the remainder is 5,
and the answer is 5

এখানে,
3 দ্বারা বিভাজ্য সংখ্যা = 100/3
ভাগফল 33 এবং ভাগশেষ 1
সুতরাং 3 দ্বারা বিভাজ্য সংখ্যা = 33 টি
3 ও 8 এর ল, সা, গু = 24
এখন 100/24 =
ভাগফল 4 এবং ভাগশেষ 4
∴ 3 ও 8 দ্বারা বিভাজ্য সংখ্যা = 4 টি
সুতরাং, (33 - 4) = 29 টি সংখ্যা 3 দ্বারা বিভাজ্য কিন্তু 8 দ্বারা বিভাজ্য নয়।

Question: If 1 < p < 3, then which of the following could be true?
(I) p² < 2p
(II) p² = 2p
(III) p² > 2p

Solution:
Given: 1 < p < 3

Since p > 0,

(I) p² < 2p
⇒ p < 2
True for 1 < p < 2
Example: p = 1.5 gives 2.25 < 3.0
So, condition (I) could be true.

(II) p² = 2p
⇒ p = 2
Since 2 is within the given range (1< 2 < 3), condition (II) could be true.

(III) p² > 2p
⇒ p > 2
Example: p = 2.5, gives 6.25 > 5.0 
So, condition (III) could be true.

Conclusion: Since values of p in the range 1< p < 3 satisfy conditions (I), (II), and (III), the correct choice is: (E) I, II, and III.

Question: Which of the following is irrational?

Solution:
​√15 একটি অমূলদ সংখ্যা (irrational number)।
অমূলদ সংখ্যা (irrational number):
- যে সংখ্যাকে p/q আকারে প্রকাশ করা যায় না, যেখানে p ও q পূর্ণসংখ্যা এবং q ≠ 0, সে সংখ্যাকে অমূলদ সংখ্যা বলা হয়।
- পূর্ণবর্গ নয় এরূপ যে কোনো স্বাভাবিক সংখ্যার বর্গমূল কিংবা তার ভগ্নাংশ একটি অমূলদ সংখ্যা। যেমন, √2 = 1.414213..., √6 = 2.229489... ইত্যাদি অমূলদ সংখ্যা।
- কোনো অমূলদ সংখ্যাকে দুইটিপূর্ণ সংখ্যার অনুপাত হিসেবে প্রকাশ করা যায় না।
-অমূলদ সংখ্যাকে একটি মূলদ সংখ্যা দ্বারা গুণ করলে অমূলদ সংখ্যা পাওয়া যায়।
অর্থাৎ, non zero rational number × irrational number = irrational number.

Question: If the 11th number in a series of 11 consecutive integers has the value k + 15, what is the 1st number in the series expressed in terms of k?

Solution:
Let the 1st number in the series be x.
Since there are 11 consecutive integers, the series is: x, (x + 1), (x + 2), ..., (x + 10).

According to the question, the 11th number is k + 15.
So,
x + 10 = k + 15
⇒ x = k + 15 - 10
⇒ x = k + 5

∴ The 1st number in the series is k + 5.

ধরি,
বৃহত্তর সংখ্যাটি x এবং ক্ষুদ্রতর সংখ্যাটি y
প্রশ্নমতে,
x - y = 5 ............(i)
এবং x² - y² = 65...........(ii)
(ii) নং সমীকরণ হতে পাই,
x² - y² = 65
⇒ (x + y)(x - y) = 65
⇒ 5(x + y) = 65
⇒ x + y = 13 ................(iii)
এখন (i) ও (iii) নং সমীকরণ যোগ করে পাই,
(x - y) + (x + y) = 18
⇒ 2x = 18
⇒ x = 9
বৃহত্তর সংখ্যাটি 9.

Question: If k is a positive integer, what is the smallest possible value of k such that 1512 × k is the square of an integer?

Solution:
আমরা জানি, একটি সংখ্যা পূর্ণবর্গ হতে হলে এর মৌলিক গুণনীয়কের ঘাতসমূহ জোড় সংখ্যা হতে হবে।

1512 = 2 × 2 × 2 × 3 × 3 × 3 × 7
= 2³ × 3³ × 7

1512k = 2³ × 3³ × 7 × k
এখন k এর মান 2 × 3 × 7 = 42  হলে, 1512k একটি পূর্ণবর্গ সংখ্যা হবে।

1512 × 42 = (2³ × 3³ × 7¹) × (2 × 3 × 7)
=2⁴ × 3⁴ × 7² 
যেহেতু এই গুণফলের সব মৌলিক উৎপাদকের ঘাত জোড়, তাই এটি একটি পূর্ণবর্গ সংখ্যা।

সুতরাং, k = 42, হলে 1512 × k পূর্ণবর্গ সংখ্যা হয়।

The first number : The second Number = 3:2 = 3 × 3:2 × 3 = 9:6
The second Number : 3rd Number = 3:2 =3 × 2:2 × 2 = 6:4
1st:2nd:3rd = 9:6:4
Let,
The 1st, 2nd & 3rd are consecutively = 9x, 6x, 4x
Therefore,
(9x)² + (6x)² + (4x)² = 532
So, X = 2 and 1st number is 9 × 2 = 18

Question: A number when divided by 315 leaves a remainder of 47. If the same number is divided by 21, what will be the remainder?

Solution:
Let the number be x, and the quotient is q.

Then,
x = 315q + 47
= (21 × 15q) + (21 × 2) + 5
= 21(15q + 2) + 5

So, the given number when divided by 21 gives 5 as a remainder.

Question: Which of the following numbers can be removed from the set S = {0, 2, 4, 5, 9} without changing the average of set S?

Solution: 
The average of the elements in the original set S is (0 + 2 + 4 + 5 + 9)/5
= 20/5
= 4

If we remove an element that equals the average, then the average of the new set will remain unchanged.
The new set after removing 4 is {0, 2, 5, 9}.

∴ The average of the elements is (0 + 2 + 5 + 9)/4 
= 16/4 
= 4

Question: Among three numbers, the first number is twice the second and half of the third. If the average of the three numbers is 56, find the difference between the first and third numbers.

Solution:
Let, 
the second number be x.
Then first number = 2x, third number = 4x.

∴ 2x + x + 4x = 56 × 3
⇒ 7x = 168
⇒ x = 168/7
⇒ x = 24

Required difference:
= 4x - 2x
= 2x
= 2 × 24
= 48.

Question: If 2 tables and 3 chairs cost Tk. 3500 and 3 tables and 2 chairs cost Tk. 4000, then how much does a table cost?

Solution:
Let
The cost of a table and that of a chair be Tk. x and Tk. y respectively.

Then,
2x + 3y = 3500................(i)
and
3x + 2y = 4000................(ii)

(ii)× 3 - (i) × 2 ⇒
9x + 6y - 4x - 6y = 12000 - 7000
5x = 5000
x = 1000

The cost of a table Tk. 1000

91 - 5 =86 
86 - 10 = 76
76 - 15 = 61
So, 61 - 20 = 41

আমরা জানি,
n সংখ্যক স্বাভাবিক সংখ্যার যোগফল = {n(n + 1)/2}
∴ 100 টি স্বাভাবিক সংখ্যার যোগফল = {100(100 + 1)/2} = 5050
সুতরাং, এদের গড় = 5050/100 = 50.5

(n + 2)(n + 4)
= (2m + 2)(2m + 4)
= 2(m + 1)2(m + 2)
= 4(m +1)(m + 2)
= 4 × (product of two consecutive positive integers, one which must be even)
= 4 × (an even number), and this equals a number that is atleast a multiple of 8.
Hence, the answer is 8.

The sum of three consecutive integers can be written as n + (n + 1) + (n + 2) = 3n + 3
If the sum is 24, we need to solve the equation 3n + 3 = 24;
=> 3n = 21;
=> n = 7
The greatest of the three numbers is therefore 7 + 2 = 9

Question: How many pairs of natural numbers is there the difference of whose squares are 45.

Solution:
Let the two natural numbers be x and y where x > y.
x² - y² = 45
(x + y)(x - y) = 45

And Factor pairs of 45 Both (x + y) and (x - y)must be positive odd integers (since x and y are natural numbers), and x + y > x - y

Thus, the factors of 45 possibles are 1, 3, 5, 9, 15 and 45
Hence, numbers are 9 and 6 or 7 and 2 or 23 and 22
So, There are 3 pairs are (23, 22), (9, 6), (7, 2)

লাইভ পরীক্ষার প্রশ্নে 'z²' এর পরিবর্তে 'x²' এর মান জানতে চাওয়া হয়েছিল। 
প্রশ্ন অনুযায়ী সঠিক উত্তর - ঙ) None of these. 
---------------------------- 

Question: x, y and z are consecutive positive integers such that x < y < z . If the units digit of x² is 6 and the units digit of y² is 9, what is the units digit of x²?

Solution:
Given that x, y, and z are consecutive positive integers such that x < y < z, we are asked to find the units digit of z², based on the following information:
The units digit of x² is 6.
The units digit of y² is 9.

First, let's analyze the possible values of x and y by checking what numbers, when squared, have the given units digits:

For x², the units digit is 6. The numbers whose squares end in 6 are:
4² = 16(units digit is 6) and 6² = 36(units digit is 6).
Thus, x could be either 4 or 6.

For y², the units digit is 9. The numbers whose squares end in 9 are:
3² = 9 (units digit is 9) and 7² = 49(units digit is 9).
Thus, y could be either 3 or 7.

Since x < y, we will now check the possible pairs of x and y:
If x = 4, the next integer y would be 5, but the units digit of 5² is 5, not 9. So this case is not valid.
If x = 6, the next integer y would be 7, and indeed, the units digit of 7² is 9.
Thus, the valid pair is x = 6 and y = 7.

Since z is the next consecutive integer after y = 7, we have z = 8.
Now,
we compute the units digit of z²
8² = 64. The units digit of z² is 4.

The odds numbers are x - 4, x - 2, x, x + 2, x + 4 and x + 6
therefore, x + 6 = 15
so, x = 9, hence fourth number is 11

Let x = 13p + 11
and x = 17q + 9
Then,
13p + 11= 17q + 9
17q - 13p = 2
q = (2 + 13p)/17
The least value of p for which q = (2 + 13p)/17 is a whole number, is p = 26.
∴ x = (13 × 26 + 11)
= 338 + 11
= 349.

Since it is a non - terminating and non - repeating decimal,
So it is an irrational number

Question: x = 6, y = 4 and z = - 2, then x(y + z)/y(x + y + z) = ?

Solution:
Given that,
x =6, y = 4 and z = - 2

Then,
= x(y + z)/y(x + y + z)
= 6{4 + (- 2)}/{4(6 + 4 - 2)}
= 12/32
= 3/8

Let the numbers be a and b. Then, a + b = 12 and ab = 35.
(a + b)/ab = 12/35
arrow 1/b + 1/a = 12/35
Sum of reciprocals of given numbers = 12/35

21600=2⁵×3³×5²
To make it a perfect cube, it must be multiplied by (2 × 5), i.e., 10

Let the number of oranges in first basket be x,
Number of oranges in second basket = 672 - x
ATQ, x - x/5 = 672 - x + x/5
⇒ 4x/5 = 672 - 4x/5
⇒ 4x/5 + 4x/5 = 672
⇒ 8x/5 = 672
⇒ x = 672× (5/8)
⇒ x = 420
∴ Number of oranges in first basket = 420.

Question: Three numbers are in the ratio 3 : 4 : 6 and their products is 1944. The largest number is - 

Solution:
Let the number be 3x, 4x, 6x.

ATQ,
3x × 4x × 6x = 1944
⇒ 72x³ = 1944
⇒ x³ = 1944/72
⇒ x³ = 27
∴ x = 3

So largest number = 6x = 6 × 3 = 18

Question: Find the number of factors of 360.

Solution:
Factorize 360 into prime factors:
360 = 2³ × 3² × 5¹
The formula for the number of factors of a number n =p^aq^br^c…n is:
Number of factors = (a + 1) (b + 1)(c + 1)
Apply the formula:
(3 + 1) (2 + 1) (1 + 1) = 4 × 3 × 2 = 24

Let small number is x, so large number is (3/2)x
ATQ, x + 14 = (3/2)x
Or, 2x + 28 = 3x
So, x = 28

Question: The sum of 7 consecutive natural numbers is 77. Find how many of these are prime numbers?

Solution:
Given that,
The sum of seven consecutive natural numbers = 77

Now,
Let the numbers be n, n + 1, n + 2, n + 3, n + 4, n + 5, n + 6 respectively
∴ 7n + 21 = 77
⇒ 7n = 77 - 21
⇒ 7n = 56
⇒ n = 56/7
∴ n = 8

So the numbers is 8, 9, 10, 11, 12, 13, 14
Out of these 11, 13 are prime numbers

∴ Required prime numbers is 2

Question: The greatest number will divide 3026 and 5053 leaving remainders 11 and 13 respectively?

Solution: 
৩০২৬ - ১১ = ৩০১৫ ও ৫০৫৩ - ১৩ = ৫০৪০ এর গ.সা.গু 

৩০১৫ = ৩ × ৩ × ৫ × ৬৭ 
৫০৪০ = ২ × ২ × ২ × ২ × ৩ × ৩ × ৫ × ৭ 

৩০১৫ ও ৫০৪০ এর গ সা গু = ৩ × ৩ × ৫
= ৪৫