Number System, Problems on Number

Question: A and B are two positive integers such that A × B = 96. Which of the following cannot be the value of A + B?

Solution:
Factor pairs of 96:
(1, 96) → A + B = 97
(2, 48) → A + B = 50
(3, 32) → A + B = 35
(4, 24) → A + B = 28
(6, 16) → A + B = 22
(8, 12) → A + B = 20

So, possible values of A + B are:
97, 50, 35, 28, 22, 20
Among options, 49 is not possible.

Question: The difference between two numbers is 5 and the difference between their square is 65. What is the smaller number?

Solution:
Let the numbers be x and y
Then,
x - y = 5.........(i) 
x² - y² = 65
Or (x + y)(x - y) = 105
5(x + y) = 65
x + y = 13 ....... (ii)

by adding (i) and (ii) we get,
2x = 5 + 13
2x = 18
x = 9 

From (ii)  we get
x + y = 13
9 + y = 13
y = 13 - 9
y = 4

Hence, the smaller number is 4

ধরি, f(x) = 3x² + ax + a +3
যেহেতু, x + 2 দ্বারা বিভাজ্য সেহেতু,
f(-2) = 0 হবে
⇒ 3(-2)² + a(-2) + a + 3 = 0
⇒ 12 - 2a + a + 3 = 0
⇒ - a + 15 = 0
⇒ a = 15

Question: If p is a positive integer, what is the smallest possible value of p such that 2160 × p is a perfect square?

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

2160 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 5
= 2⁴ × 3³ × 5¹

2160p = 2⁴ × 3³ × 5¹ × p
এখানে, 2-এর ঘাত = 4 (জোড়), 3-এর ঘাত = 3 (বিজোড়), 5-এর ঘাত = 1 (বিজোড়)

পূর্ণবর্গ করতে হলে সব ঘাত জোড় হতে হবে।
তাই p = 3 × 5 = 15 হলে,

2160 × 15 = 2⁴ × 3³ × 5¹ × (3 × 5) = 2⁴ × 3⁴ × 5²
যেহেতু সব মৌলিক উৎপাদকের ঘাত জোড়, তাই এটি একটি পূর্ণবর্গ সংখ্যা।

সুতরাং, p = 15 হলে, 2160 × p পূর্ণবর্গ সংখ্যা হয়।

Question: A worker earns Tk. 250 on the first day and spends Tk. 200 on the second day, earns Tk. 250 on the third day and again spends Tk. 200 on the fourth day and so on. On which day would he have had Tk. 1000?

Solution:
১ম দিনে আয় করে ২৫০ টাকা।
২য় দিনে ব্যয় করে ২০০ টাকা।

∴ ২ দিনে তার জমা থাকে (২৫০ - ২০০) = ৫০ টাকা।

এখন, (১০০০ - ২৫০) = ৭৫০ টাকা।

৫০ টাকা জমা থাকে ২ দিনে
১ টাকা জমা থাকে ২/৫০ দিনে
৭৫০  টাকা জমা থাকে (২ × ৭৫০)/৫০ দিনে
= ৩০ দিন।

৩০ দিন পর তার হাতে থাকে ৭৫০ টাকা
এবং ৩১ তম দিনে সে আয় করে ২৫০ টাকা।
তাহলে মোট টাকা হয় (৭৫০ + ২৫০) = ১০০০ টাকা,

সুতরাং ৩১ দিনে তার কাছে ১০০০ টাকা ছিল।

Let the ten's digit be x.
Then, the unit's digit = x + 2.
Number = 10x + (x + 2) = 11x + 2
Sum of digits = x + (x + 2) = 2x + 2.
⇒ (11x + 2)(2x + 2) = 144
⇒ 22x² + 26x- 140 = 0
⇒ 11x² + 13x - 70 = 0
⇒ (x - 2)(11x + 35) = 0
⇒ x = 2
Hence, Required Number = 11x + 2 = 24

Question: What is the sum of all two-digit numbers that gives a remainder of 3 when they are divided by 7?

Solution: 
general formula for that number = 7n + 3  

n = 1, then the number is = 7 + 3 = 10 
n = 2, then the number is =14 + 3 = 17
.
.
.
n= 13,  then the number is = 94

sum  = 10 + 17 + ... + 94 
= 13 (10 + 94)/2 
= 676 

Question: The sum of three consecutive even integers is 30 more than the first of the numbers. What is the middle number?

Solution:
ধরি, তিনটি ক্রমিক জোড় সংখ্যা হলো x, x + 2, x + 4.

প্রশ্নমতে,
x + (x + 2) + (x + 4) = x + 30
⇒ 3x + 6 = x + 30
⇒ 3x − x = 30 − 6
⇒ 2x = 24
⇒ x = 24/2
⇒ x = 12

∴ প্রথম সংখ্যা = x = 12
দ্বিতীয় সংখ্যা (মাঝের সংখ্যা) = x + 2 = 14
তৃতীয় সংখ্যা = x + 4 = 16

∴ মাঝের সংখ্যাটি হলো = 14

Question: The difference between two numbers is 960. When the larger number is divided by the smaller, the quotient is 5 and the remainder is 12. What is the smaller number?

Solution:
Given that,
The difference of two numbers = 960
Quotient when the larger number is divided by the smaller number = 5
Remainder when the larger number is divided by the smaller number = 12

Now,
Let the smaller number be x.
Larger number = 5x + 12

ATQ,
⇒ 5x + 12 - x = 960
⇒ 4x + 12 = 960
⇒ 4x = 960 - 12
⇒ 4x = 948
⇒ x = 948/4
∴ x = 237

So the smaller number is 237.

Question: Which is the larger between two numbers if they are in the ratio of 6 : 13 and their least common multiple is 312?

Solution: 
Let, the numbers be 6x, 13x 
HCF is x and LCM is 312 

6x × 13x = 312 × x
⇒ x = 312/(6 × 13)
= 4 

Larger number = 13 × 4 = 52 

Let the number be P.
So,
P ÷ 44 = 432
⇒ P = 432 * 44
= 19008

P/31 = 19008 / 31
= 613, Remainder = 5

Question: - 5x - [4y - {9x - (3y - 7x)}] simplifies to 

Solution:
- 5x - [4y - {9x - (3y - 7x)}]
= - 5x - [4y - {9x - 3y + 7x}]
= - 5x - [4y - 9x + 3y - 7x]
=  - 5x - [7y - 16x]
= - 5x - 7y + 16x
= 11x - 7y

Multiples of 7 between 11 and 200 are 14, 21, 28, 35, 42, ..........., 189, 196.
T_m = 196
14 + (m - 1) × 7 = 196
⇒ (m - 1) × 7 = 196 - 14
⇒ (m - 1) × 7 = 182
⇒ (m - 1) = 182/7
⇒ (m - 1) = 26
⇒ m = 27.
Multiples of 7 and 3 both, i.e that of 21 are 21, 42, 63, ........, 189
T_n = 189
21 + (n - 1) × 21 = 189
⇒ (n - 1) × 21 = 189 - 21
⇒ (n - 1) × 21 = 168
⇒ (n - 1) = 168/21
⇒ (n - 1) = 8
⇒ n = 9
∴ Required number of terms = 27 - 9
= 18.

Question: If a is an even integer and b is an odd integer, which of the following must be an odd integer?

Solution: 
Let, a = 2 and b = 3

Then, 
ক) a/b
= 2/3
So a/b is not an integer, hence cannot be odd integer.

খ) 2(a + b)
= 2 × (2 + 3)
= 10
So 2(a + b) is always even.

গ) ab
= 2 × 3
= 6    ; [even × odd = even]
So ab is always even.

ঘ) 2a + b 
= (2 × 2) + 3
= 4 + 3
= 7
So 2a + b is always odd.

∴ 2a + b This must be an odd integer.

Question: In a class, 25 students play cricket, 25 students play football, and 10 students play both. 10 students play neither cricket nor football. What is the total number of students in the class?

Solution:
Number of students who play cricket, n(C) = 25
Number of students who play football, n(F) = 25
Number of students who play both cricket and football, n(C ∩ F) = 10
Number of students who play neither = 10

n(C ∪ F) = n(C) + n(F) - n(C ∩ F)
= 25 + 25 − 10 = 40

Total students in the class = students who play cricket or football + students who play neither
n(U) = n(C ∪ F) + neither = 40 + 10 = 50

∴ Total 50 students in the class.