Number system and number property

To find the next sequence
=> 124/2 = 62 & 62 - 2 = 60
=> 60/2=30 & 30 - 2 =28
=> 28/2=14 & 14 - 2 =12

Sum of the infinite G.P = a/1 - r
(Here a = First term, r = ratio)
=1 / (1 - 1/2)
=2

If the last two digits of the number is divisible by 4, then we can say that the number is divisible by 4

To find maximum number of bows that can be made
1 yard = 36 inches
4 yards = 4*36=144 inches
No.of maximum bows that can be made = 144/15 = 9.6
So maximum 9 bows can be made

Let the digits of the No. be x & y & z
Hence , x + y + z = 10………(1)
x + z = y………(2)
As per the condition
(100z + 10y + x) - (100x + 10y + z) = 99
Solving, the value of (z - x) =1
From equation(1) and (2)
2y = 10, y = 5
So 253 satisfies all the condition

Unit digit in 34 = 1 Unit digit in (34)16 = 1
Unit digit in 365 = Unit digit in [ (34)16 x 3 ] = (1 x 3) = 3
Unit digit in 659 = 6
Unit digit in 74 Unit digit in (74)17 is 1.
Unit digit in 771 = Unit digit in [(74)17 x 73] = (1 x 3) = 3
Required digit = Unit digit in (3 x 6 x 3) = 4.

Unit digit in 7¹⁰⁵ = Unit digit in [ (7⁴)²⁶ x 7 ]

But, unit digit in (7⁴)²⁶ = 1

Unit digit in 7¹⁰⁵ = (1 x 7) = 7

3251
+ 587

369
----
4207
----
Let 4207 - x = 3007
Then, x = 4207 - 3007 = 1200

Let 7589 - x = 3434
Then, x = 7589 - 3434 = 4155

(217)² +(183)² = (200 + 17)² + (200 - 17)²

= 2 x [(200)² + (17)²] [Ref: (a + b)² + (a - b)² = 2(a² + b²)]

= 2[40000 + 289]

= 2 x 40289

= 80578

Unit digit in the given product = Unit digit in (4 x 8 x 7 x 3) = (672) = 2

3897 x 999 = 3897 x (1000 - 1)
= 3897 x 1000 - 3897 x 1
= 3897000 - 3897
= 3893103

(461 + 462 + 463 + 464) = 461 x (1 + 4 + 42 + 43) = 461 x 85
= 460 x (4 x 85)
= (460 x 340), which is divisible by 10

106 x 106 - 94 x 94= (106)² - (94)²

= (106 + 94)(106 - 94)   [Ref: (a² - b²) = (a + b)(a - b)]

= (200 x 12)

= 2400

8988 x 1/8 x 1/4 = 2247/8 = 280.875

3
+ 33
+ 333
+ 3.33
------
372.33
------

(1000)⁹ / 10²⁴ = (10³ )⁹ / 10²⁴ = (10)²⁷ / 10²⁴ = 10^(27-24) = 10³ = 1000

From the given expression [(a + b)² - 4ab]
Consider a = 476 and b = 424
= [(476 + 424)² - 4 x 476 x 424]
= [(900)² - 807296]
= 810000 - 807296
= 2704

4 | x

--------
5 | y -1
--------

| 1 -4
y = (5 x 1 + 4) = 9
x = (4 x y + 1) = (4 x 9 + 1) = 37
Now, 37 when divided successively by 5 and 4, we get

5 | 37
---------
4 | 7 - 2
---------
| 1 - 3

Respective remainders are 2 and 3

8796 x 223 + 8796 x 77= 8796 x (223 + 77)   [Ref: By Distributive Law ]
= (8796 x 300)
= 2638800

From the given expression[a² + b² - 2ab] Consider a = 287 and b = 269
= (a - b)² = (287 - 269)²
= (18²)
= 324

The square of a natural number never ends in 2
143642 is not the square of natural number.

Let S_n = (2 + 4 + 6 + ... + 30). This is an A.P. in which a = 2, d = 2 and l = 30

Let the number of terms be n. Then,

a + (n - 1)d = 30

2 + (n - 1) x 2 = 30

n = 15

Therefore, S_{n =} n/2(a + l) = 15/2 x (2 + 30) = (15 x 16) = 240

Let the number be x. Then
60% of 3/5 of x = 36
=> 60/100 * 3/5 * x = 36
=> x = (36 * 25/9) = 100
Required number = 100

80 = 2 x 5 x 8
Since 653xy is divisible by 2 and 5 both, so y = 0
Now, 653x is divisible by 8, so 13x should be divisible by 8
This happens when x = 6
x + y = (6 + 0) = 6

Let the two consecutive odd integers be (2n + 1) and (2n + 3). Then,

(2n + 3)² - (2n + 1)² = (2n + 3 + 2n + 1) (2n + 3 - 2n - 1)

= (4n + 4) x 2

= 8(n + 1), which is divisible by 8

Unit digit in (4137)⁷⁵⁴ = Unit digit in {[(4137)⁴]¹⁸⁸ x (4137)²}

=Unit digit in { 292915317923361 x 17114769 }

= (1 x 9) = 9

4 | x

-----------
5 | y -2

-----------
6 | z - 3

-----------
| 1 - 4
z = 6 x 1 + 4 = 10
y = 5 x z + 3 = 5 x 10 + 3 = 53
x = 4 x y + 2 = 4 x 53 + 2 = 214

Hence, required number = 214.

The smallest 6-digit number 100000

111) 100000 (900
999
-----
100
-------

Required number = 100000 + (111 - 100)
= 100011

Largest 5-digit number = 99999

91) 99999 (1098
91
----
899
819
----

809
728
-----
81
---

Required number = (99999 - 81)
= 99918

Here a = 3 and r = 6/3 =2 Let the number of terms be n

Then, t_n = 384      ar^n-1 = 384

=>3 x 2^{n - 1}= 384

=>2^n-1 = 128 = 2⁷

=>n - 1 = 7

=> n = 8

Number of terms = 8

By hit and trial, we put x = 5 and y = 1 so that (3x + 7y) = (3 x 5 + 7 x 1) = 22, which is divisible by 11

(4x + 6y) = ( 4 x 5 + 6 x 1) = 26, which is not divisible by 11

(x + y + 4 ) = (5 + 1 + 4) = 10, which is not divisible by 11

(9x + 4y) = (9 x 5 + 4 x 1) = 49, which is not divisible by 11

(4x - 9y) = (4 x 5 - 9 x 1) = 11, which is divisible by 11

9548

+ 7314

-----

16862
-----
16862 = 8362 + x
x = 16862 - 8362
x = 8500

Number = (12 x 35)

Correct Quotient = 420 ÷ 21 = 20

This is a G.P. in which a = 2, r = 22 = 2 and n = 9

S_n= a(rⁿ-1) / (r-1) = 2 *(2⁹ -1) / (2-1) = 2 *(512 - 1) = 2*511 = 1022

Let the two consecutive even integers be 2n and (2n + 2). Then,
(2n + 2)2 = (2n + 2 + 2n)(2n + 2 - 2n)
= 2(4n + 2)
= 4(2n + 1), which is divisible by 4

3²⁵ + 3²⁶ + 3²⁷ + 3²⁸) = 3²⁵ x (1 + 3 + 3² + 3³) = 3²⁵ x 40

= 3²⁴ x 3 x 4 x 10

= (3²⁴ x 4 x 30), which is divisible by 30

Let x = 6q + 3

Then, x² = (6q + 3)²

= 36q² + 36q + 9

= 6(6q² + 6q + 1) + 3

Thus, when x² is divided by 6, then remainder = 3

When n is even. (xⁿ - aⁿ) is completely divisibly by (x + a)
(17²⁰⁰ - 1²⁰⁰) is completely divisible by (17 + 1), i.e., 18
(17²⁰⁰ - 1) is completely divisible by 18
On dividing 17²⁰⁰ by 18, we get 1 as remainder

Let x = 6q + 3
Then, x² = (6q + 3)²
= 36q² + 36q + 9
= 6(6q² + 6q + 1) + 3
Thus, when x² is divided by 6, then remainder = 3

x = 13p + 11 and x = 17q + 9
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
Therefore,x = (13 x 26 + 11)
= (338 + 11)
= 349

S_n = (1 + 2 + 3 + ... + 50 + 51 + 52 + ... + 100) - (1 + 2 + 3 + ... + 50)
= 100/2 * (1 + 100) – 50/2 * (1 + 50)
= (50 x 101) - (25 x 51)
= (5050 - 1275)
= 3775

Let n = 4q + 3. Then 2n = 8q + 6 = 4(2q + 1 ) + 2
Thus, when 2n is divided by 4, the remainder is 2

From the given equation = [(a+b)²-(a-b)²] / ab = 4ab /ab = 4

Let the given number be 476xy0
Then (4 + 7 + 6 + x + y + 0) = (17 + x + y) must be divisible by 3
And, (0 + x + 7) - (y + 6 + 4) = (x - y -3) must be either 0 or 11
x - y - 3 = 0 y = x - 3
(17 + x + y) = (17 + x + x - 3) = (2x + 14)
x= 2 or x = 8
x = 8 and y = 5

Clearly, (2272 - 875) = 1397, is exactly divisible by N
Now, 1397 = 11 x 127
The required 3-digit number is 127, the sum of whose digits is 10

987 = 3 x 7 x 47
So, the required number must be divisible by each one of 3, 7, 47
553681 (Sum of digits = 28, not divisible by 3)
555181 (Sum of digits = 25, not divisible by 3)
555681 is divisible by 3, 7, 47

Let x be the number and y be the quotient. Then,
x = 357 x y + 39
= (17 * 21 x y) + (17 * 2) + 5
= 17 * (21y + 2) + 5)
Required remainder = 5

Clearly, 4864 is divisible by 4
So, 9P2 must be divisible by 3. So, (9 + P + 2) must be divisible by 3
P = 1.

When n is odd, (xⁿ + aⁿ) is always divisible by (x + a)

Each one of (47⁴³ + 43⁴³) and (47⁴⁷ + 43⁴⁷) is divisible by (47 + 43).

Let the number be x and on dividing x by 5, we get k as quotient and 3 as remainder
x = 5k + 3
x² = (5k + 3)²
= (25k² + 30k + 9)
= 5(5k² + 6k + 1) + 4
On dividing x² by 5, we get 4 as remainder

(xⁿ + 1) will be divisible by (x + 1) only when n is odd
(67⁶⁷ + 1) will be divisible by (67 + 1)
(67⁶⁷ + 1) + 66, when divided by 68 will give 66 as remainder

Let S_n =(1 + 2 + 3 + ... + 45). This is an A.P. in which a =1, d =1, n = 45
S_n= n/2[2a+(n-1)d] = 45/2 * [2*1+(45 - 1)*1)] = (45/2*46) =(45 x 23)
= 45 x (20 + 3)
= 45 x 20 + 45 x 3
= 900 + 135
= 1035

91 is divisible by 7. So, it is not a prime number

Let the three integers be x, x + 2 and x + 4
Then, 3x = 2(x + 4) + 3 x = 11
Third integer = x + 4 = 15

Let the ten's digit be x and unit's digit be y
Then, x + y = 15 and x - y = 3 or y - x = 3
Solving x + y = 15 and x - y = 3, we get: x = 9, y = 6
Solving x + y = 15 and y - x = 3, we get: x = 6, y = 9
So, the number is either 96 or 69
Hence, the number cannot be determined

Let the numbers be a, b and c
Then, a² + b² + c² = 138 and (ab + bc + ca) = 131
(a + b + c)² = a² + b² + c² + 2(ab + bc + ca) = 138 + 2 x 131 = 400
(a + b + c) = 400 = 20

Let the numbers be x and y
Then, xy = 120 and x² + y² = 289
(x + y)² = x² + y² + 2xy = 289 + (2 x 120) = 529
x + y = 529 = 23

Let the middle digit be x
Then, 2x = 10 or x = 5. So, the number is either 253 or 352
Since the number increases on reversing the digits,so the hundred's digits is smaller than the unit's digit.
Hence, required number = 253