Number System, Problems on Number

Difference between divisors and remainders = (8 - 1) / ( 12 - 5) / (16 - 9) / (20 - 13) = 7
LCM of 8,12,16,20 is 240
Least 5 digit number is = 10000
240)10000(41
         9840
____________
           160
So, the number is = 10000 + 240 - 160 - 7 = 10073

The pages of the book may be divided into 10 groups;
(1 -100), (101 - 200), (201 -300),......, (901 - 1000).
Clearly, for the first group, one needs 11 zeros,
For second to ninth groups, one needs 20 zeros each.
So, total number of zeros required = 11 + 8 × 20 + 21 = 192
Answer : 192

Question: 


Solution:

Question: The value of √ 0.01 + √0.81 + √1.21 + √0.0009 is = ?

Solution:
Given that,
√ 0.01 + √0.81 + √1.21 + √0.0009
= √ (1/100) + √(81/100) + √(121/100) + √(9/10000)
= (1/10) + (9/10) + (11/10)+ (3/100)
= 0.1 + 0.9 + 1.1 + 0.03
= 2.13

Let the number be x
Then,
⇒ 3/4x−1/3x = 60
⇒ 5x/12 = 60
⇒ x = (60×12)/5
∴ x = 144

Question: The sum of all the factors of 100 is-

Solution:
Factors of 100: 1, 2, 4, 5, 10, 20, 25, 50, 100

∴ the sum of these number is,
1 + 2 + 4 + 5 + 10 + 20 + 25 + 50 + 100
= 217

So the sum of all factors of 100 is 217.

Question: The number 2272 and 875 are divided by a 3 digit number N, giving the same remainders. The sum of the digit is-

Solution: 
Let the remainder in each case be x
Then, (2272 - x) and (875 - x) are exactly divisible by three digit number
Difference :
= (2272 - x) - (875 - x)
= 1397
Factor of 1397 = 11 × 127
Since, both 11 and 127 are prime number
Three digit number is 127

∴ Sum of digits = 1 + 2 + 7 = 10

6   7    9   13    21    37
   1   2    4     8      16

Question: If x² is an odd number, determine the nature of x² - x. 

Solution:
যেহেতু x² বিজোড় তাই x ও বিজোড় হবে।
এখন,
x² - x
= (x - 1)x
= x(x - 1)
∴ (x - 1) এবং x দুইটি ক্রমিক সংখ্যা।

x বিজোড় সংখ্যা হলে (x - 1) অবশ্যই জোড় সংখ্যা হবে।
কারণ দুইটি ক্রমিক সংখ্যার মধ্যে একটি বিজোড় হলে অন্যটি জোড় হবে।

সুতরাং, x ও (x - 1) এর গুনফল = x(x - 1) = x² - x একটি জোড় সংখ্যা।
[জোড় × বিজোড় = জোড়]

এখানে, আমরা প্রশ্নের option গুলো বিবেচনা করিঃ
a) m যদি বিজোড় হয়, তখন p শুধুমাত্র বিজোড় হলেই (m + p)m জোড় হবে।
b) যদি m বিজোড় হয়, তখন p জোড় হলে কখনই (m + p)m জোড় হবে না।
c + d) যদি m জোড় হয়, তাহলে p জোড় বা বিজোড় যাই হোক না কেন (m + p)m জোড় হবে।
অর্থাৎ, m যদি even হয়, তাহলে প্রশ্নোক্ত সমীকরণ থেকে p জোড় বা বিজোড় যেকোনোটাই হতে পারে যা 'must' শর্তকে মানে না। 
তাই উত্তর হবেঃ If m is odd, then p is odd.

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

34 - 24 = 10
34 - 28 = 6
28 - 24 = 4

Hence, the greatest number which divides 24, 28, and 34 and gives the same remainder
= HCF of 10, 6, 4
= 2.

When Divided by 7,
A = 7x + 4
So, numbers can be: 4, 11, 18, 25, 32, 39, 46, 53…….

Again,
when divided by 11,
A = 11y + 9
So, numbers can be: 9, 20, 31, 42, 53…….

Here, 53 is common.

Now, LCM of 7 & 11 is 77.
So, the numbers pattern is : 77x + 53

Then,
77x ≤ 1000
Or, x ≤ 12.2
x can be 0, 1, 2 ...... 12 = total 13 integers
It can't be x = 13, then the number will be bigger than 1000

Question: The ratio of two numbers is 3 : 4 and their H.C.F is 4. Their L.C.M is:
 
Solution:
ধরি,
সংখ্যা দুইটি যথাক্রমে 3x, 4x
 3x, 4x এর লসাগু = 12x
3x, 4x এর গসাগু = x

প্রশ্নমতে,
x = 4

∴  3x, 4x এর লসাগু = 12x
= 12 × 4
= 48

Question: The product of two numbers is 300, and the sum of their squares is 625. What is the sum of two numbers?

Solution:
Let the numbers be a and b.
As per the question:
ab = 300
a² + b² = 625

So,
(a + b)² = a² + b² + 2ab
= 625 + 2 × 300
= 625 + 600
= 1225

∴ a + b = √1225 = 35

The answer is divisible by 987.
So we can use the hit and trial method to find out the number divisible by 987 from the given choices.
553681/987 gives a remainder not equal to 0
555181/987 gives a remainder not equal to 0
556581/987 gives a remainder not equal to 0
But 555681/987 gives 0 as a remainder. Hence this is the answer

Question: The product of two consecutive even positive number is 528. Find the numbers.

Solution:
Let the integers be x and x + 2
So, The equation:
x(x + 2) = 528
⇒ x² + 2x - 528 = 0
⇒ x² + 24x - 22x - 528 = 0 [528 = (2 × 2 × 2 × 3) × (2 × 11) = 24 × 22]
⇒ x(x + 24) - 22(x + 24) = 0
⇒ (x - 22)(x + 24) = 0
⇒ x = 22 or x = -24
x = 22 [Ignoring the negative value]

so, x + 2 = 24

∴ Numbers = 22, 24

4 6 9 6 14 6 .......
এখানে, ২য়, ৪র্থ, ৬ষ্ঠ, ৮ম ......... ইত্যাদি অবস্থানে নির্দিষ্ট সংখ্যা 6 বিদ্যমান
অবশিষ্ঠ সংখ্যাগুলোর ধারাঃ 4     9          14        19
পার্থক্যঃ                           5         5          5
সুতরাং নির্ণেয় সংখ্যাটি হচ্ছে 14 + 5 = 19.

Question: A two-digit number has 5 in its ten's digit. The sum of its digits is one-sixth of the number itself. What is the number?

Solution:
ধরি,
একক স্থানীয় অঙ্ক = x
দশক স্থানীয় অঙ্ক = 5

∴ সংখ্যাটি = 50 + x

প্রশ্নমতে,
5 + x = (50 + x)/6
⇒ 6(5 + x) = 50 + x
⇒ 30 + 6x = 50 + x
⇒ 6x - x = 50 - 30 
⇒ 5x = 20
∴ x = 4

∴ সংখ্যাটি = 50 + 4 = 54

Between 100 and 200 (and including 100) there are 21 numbers evenly divisible by 5. 
201 to 900 only net 20 numbers per century range that qualify. So, now there are 20×7 = 140 numbers evenly divisible by 5.

In last 100 (901 to 999) we only have 19 (the 20th one is 1,000 which is not a 3 digit number).

So, in total, we have 21 + 140 + 19 = 180

Question: If y > 1 and y < 4, then which of the following expressions is positive?
I. (y - 1)(y - 4)
II. (1 - y)(y - 4)
III. (1 - y)(4 - y)

Solution:
Given,
y > 1 and y < 4

For expression I: (y - 1)(y - 4)
Since y > 1, (y - 1) will be positive.
Since y < 4, (y - 4) will be negative.
∴ (y - 1)(y - 4) = positive × negative = negative

For expression II: (1 - y)(y - 4)
Since y > 1, (1 - y) will be negative. 
Since y < 4, (y - 4) will be negative.
∴ (1 - y)(y - 4) = negative × negative = positive

For expression III: (1 - y)(4 - y)
Since y > 1, so (1 - y) will be negative. 
Since y < 4, so (4 - y) will be positive.
∴ (1 - y)(4 - y) = negative × positive = negative

∴ Only expression II is positive.

Sum of decimal places = 7.
Since the last digit to the extreme right will be zero (since 5 x 4 = 20),
so there will be 6 significant digits to the right of the decimal point.

84 × 59 × 13 × 76 = 4896528
Without the use of calculator, to count the unit digit = 4 × 9 × 3 × 6 = 36 × 18 = 648

So, 8 is the unit digit

Let the large number be x.
Then,
x - 20 is a 20% of x = 20x/100 = x/5
Or, x - x/5 = 20
Or, 4x = 100
∴ x = 25

Question: If one-third of one-fourth of a number is 15, then three-tenth of that number is - 

Solution:
Let the number be x.
The equation:
⇒ 1/3 × 1/4 × x = 15
⇒ 1/12 × x = 15
⇒ x = 12 × 15
⇒ x = 180

three-tenth of the number:
= 3/10 × 180
= 54

∴ That number is 54

Question: The square root of (6 + 5√2)(6 - 5√2) is:

Solution:
√{(6 + 5√2)(6 - 5√2)}
= √{6² - (5√2)²}
= √{36 - (25 × 2)}
= √(36 - 50)
= √(- 14)
= √{14 × (- 1)}
= √(14 × i²) [∵ i² = - 1]
= i√14

There are total 7 digits given - 2, 3, 4, 5, 6, 8 and 0
Between 500 and 1000 means from 501 to 999.

All of them are 3 digit numbers.
First digit - 5, 6 or 8 i.e. 3 possibilities (1 digit gets used here)
Second digit - Any one from 7-1 = 6 remaining digits i.e. 6 possibilities
Third digit - Any one from 6-1 = 5 remaining digits i.e. 5 possibilities

∴ Total numbers possible = 3 x 6 x 5 = 90.

Question: If the difference between the squares of two consecutive natural numbers is 21, what is the sum of the squares of these two numbers?

Solution:
Bigger number = (difference of squares + 1)/2 = (21 + 1)/2 = 11
Smaller number = (difference of squares - 1)/2 = (21 - 1)/2 = 10

So, sum of the squares of these numbers = 11² + 10²
= 121 + 100
= 221 

Question: 40 is subtracted from 60% of a number, the result is 50. Find the number.

Solution:
Let the number be x.

According to the question,
60% of x - 40 = 50

⇒ (60/100)x - 40 = 50
⇒ (3/5)x = 90
⇒ x = (5 × 90)/3
⇒ x = 150

∴ The required number is 150.