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Divisibility rule for 6: The prime factor of 6 are 2 and 3 for a number to be divisible by 6, it must ALSO be divisible by 2 and 3. Therefore, we need to check if a number is EVEN and then check if the sum of the digits is divisible by 3.

All number are even so all are divisible by 2.

For 3,

23562 = 2 + 3 + 5 + 6 + 2 = 18 (Divisible by 3)

43742 = 4 + 3 + 7 + 4 + 2 = 20 (Not divisible by 3)

53422 = 5 + 3 + 4 + 2 + 2 = 16 (Not divisible by 3)

44444 = 4 + 4 + 4 + 4 + 4 = 20 (Not divisible by 3)

Let us assume (3²⁰⁰² - 1) be a MULTIPLE of 4

⇒ 3²⁰⁰² - 1 = 4q

⇒ 3²⁰⁰² + 1 = 4q + 2 is not DIVISIBLE by 4

⇒ They cannot have a common factor of 4

⇒ Statement I is false

Statement II:

(4⁸⁴ - 1) = (4⁴² + 1) (4⁴² - 1) [? (a² - B²) = (a + b) (a - b)]

⇒ (4⁴² + 1) (4²¹ - 1) (4²¹ + 1)

⇒ (4²¹ + 1) will be divisible by 5 [? (aⁿ + bⁿ) will be divisible by (a + b), when n is odd]

⇒ (4⁸⁴ - 1) will be divisible by 5

⇒ Statement II is true

∴ Only statement 2 is true.

Let the number be X

According to the QUESTION,

x - 39 = 79 - x

⇒ 2x = 118

⇒ x = 118/2

⇒ x = 59

∴ the number = 59

Short trick:

Difference between both no. = 79 - 39 = 40

Let the no. be 21

So, on dividing by 114 it leaves a REMAINDER of 21

The numbers are 1, 3, 5, 7…., 49

Sum of N terms of an AP = n/2 × {2a + (n - 1)d}

Where a is the first term, n is the number of terms and d is the common DIFFERENCE

First term of the series (a) is 1

Common difference (d) = 3 - 1 = 2 and n is 25

Sum = 25/2 × (2 + 24 × 2) = 625

The given expression:

$(\begin{ARRAY}{l}\frac{{\sqrt {80}- \sqrt {112} }}{{\sqrt {45}- \sqrt {63} }}\\ = \frac{{\sqrt {4 \TIMES 4 \times 5}- \sqrt {4 \times 4 \times 7} }}{{\sqrt {3 \times 3 \times 5}- \sqrt {3 \times 3 \times 7} }}\\ = \frac{{4\sqrt 5- 4\sqrt 7 }}{{3\sqrt 5- 3\sqrt 7 }}\\ = \frac{{4\left( {\sqrt 5- \sqrt 7 } \right)}}{{3(\sqrt 5- \sqrt {7)} }}\\ = \frac{4}{3} = 1\frac{1}{3}\end{array})$

Let ages of Man, SISTER and Brother be M, S, B years respectively.

Given,

⇒ M = 50 years.

⇒ B = 50 + 7 = 57 years

⇒ S = 57 – 12 = 45 years

Difference in his age and his sister age =

= 50 – 45 = 5 years

Given,

We know that LCM is always a MULTIPLE of HCF.

Let LCM = K(HCF)

⇒ HCF + k(HCF) = 37

⇒ (k + 1)(HCF) = 37

Possible values of k are 0 and 36.

If k = 0, HCF = 37, which is not possible as LCM will BECOME 0.

If k = 36, HCF = 1, and LCM = 36.

So, P and Q can be 1 and 36, or 4 and 9.

If they are 1 and 36, their sum is 37 and product is 36.

If they are 4 and 9, their sum is 13 and product is 36.

So, sum can be lesser or greater than product.

GIVEN, HCF of the TWO numbers is 37.

Thus the numbers can be assumed to be 37a and 37B.

Given, product of the two numbers is 4107

∴ 37a × 37b = 4107

⇒ 1369ab = 4107

⇒ ab = 3

Thus the ordered pair of (a, B) is (1, 3)

Numbers are 37a and 37b.

⇒ Numbers are 37 and 111

To find the new REMAINDER just DIVIDE the QUOTIENT with new number

⇒ 267 ÷ 57

Divisor × Quotient + Remainder = Dividend

⇒ 57 × 4 + 39 = 267

∴ Remainder = 39

From given data-

QUOTIENT is 16 and the divisor is 25 times the quotient

⇒ Divisor = 25 × 16 = 400

Also, divisor is 5 times the remainder

⇒ Remainder = divisor/5

⇒ Remainder = 400/5 = 80

We know that, DIVIDEND = quotient × divisor+ remainder

⇒ Dividend = 16 × 400+ 80

For this to be divisible by 9, SUM of numbers at DIFFERENT places MUST be divisible by 9

⇒ Sum = 2 + 1 + 3 = 6

instead of 6 there must be 9, so that it will be divisible by 9

We know that, any FACTOR of the (2⁵ × 3³ × 5²) can be expressed as 2^a × 3^b × 5^c, where a RANGES from 0 to 5, b ranges from 0 to 3 and c ranges from 0 to 2.

Here, we need to find even factors of 2⁵ × 3³ × 5², so there has to be at least one factor of 2 in the given NUMBER. So, a will range from 1 to 5.

As a result, a can take 5 values, b can take 4 values and c can take 3 values.

Given,

Let two numbers be a and b respectively.

Given,

⇒ a × b = 14(a – b)

⇒ ab = 14a – 14B

⇒ a(14 – b) = 14b

⇒ a = 14b/(14 – b)

Then,

⇒ a + b = 45

⇒ {14b/(14 – b)} + b = 45

⇒ 14b + 14b – b² = 630 – 45b

⇒ b² – 73b + 630 = 0

⇒ (b – 63)/(b – 10) = 0

b = 63 or b = 10

then,

If b = 63 then a = -18

If b = 10 then a = 35

LCM of 7, 28, 21 = 84

 25/7 = (25 × 12)/(7 × 12) = 300/84

15/28 = (15 × 3)/(28 × 3) = 45/84

20/21 = (20 × 4)/(21 × 4) = 80/84

GIVEN, 1 × 2 × 3 × 4 × 5 × …. × n) = n!

17! – 16! – 15!

= 17 × 16 × 15! – 16 × 15! – 15!

= 15! × (17 × 16 – 16 – 1)

= 15! × (16 × 16 – 1)

We can write,

165375 = 3³ × 5³ × 7²

Numbers MULTIPLE of 3 or 4 = numbers multiple of 3 + numbers multiple of 4 - numbers multiple of both 3 and 4

For numbers multiple of 3,

On dividing 100 by 3 we get quotient 33.

∴ there are 33 numbers from 1 to 100 which are multiple of 3.

For numbers multiple of 4,

On dividing 100 by 4 we get quotient 25

∴ there are 25 numbers from 1 to 100 which are multiple of 4.

For numbers multiple of both 3 and 4,

LCM of 3 and 4 is 12.

Numbers which are multiple of 12 from number 1 to 100 are the numbers which are multiple of both 3 and 4.

On dividing 100 by 12 we get quotient 8.

∴ there are 8 numbers from number 1 to 100 which are multiples of both 3 and 4.

Numbers multiple of 3 or 4 = numbers multiple of 3 + numbers multiple of 4 - numbers multiple of both 3 and 4

The largest 4 digit number is 9999.

When 9999 is divided by 14, a REMAINDER of 3 is obtained.

⇒ The NUMBER given is 680621.

Let’s FIND the SQUARE root of 680621

⇒ 824.9…

8) 680621

64

162) 406

324

1644) 8221

6576

164500

16489)148401

1609900……

⇒ √680621 = 824.9 which is slightly less than 825

∴ the NEXT perfect square is that of 825.

⇒ (825)² = 680625

∴ the smallest no. to be added to 680621 is 680625 – 680621 = 4

3⁵¹ + 3⁵² + 3⁵³ + 3⁵⁴

= 3⁵¹ (3⁰ + 3¹ + 3² + 3³)

= 3⁵¹ (1 + 3 + 9 + 27)

= 3⁵¹ × 40

⇒ N = 36Q + 12 (where Q is quotient and EQUAL to any whole number)

The same number (N) is divided by 9,

$( \Rightarrow N = \FRAC{{36Q + 12}}{9})$

? When 36 is divided by 9, the remainder will be zero.

Now, divide 12 by 9, we GET 3 as a remainder.

Hence, when N = 36Q + 12 is divided by 9, we get 3 as a remainder.

The SUM of ‘n’ terms of an A.P. is given as,

Sum = (n/2) × (first term + last term)

Given, n = 12

First term = 3

Last term = 47

∴ Sum of 12 terms of A.P. = (12/2) × (3 + 47) = 6 × 50 = 300

Solving the EXPRESSIONS,

⇒ [(3.75 – 2.25) × 2] = 3

⇒ [(1.42 + 2.08) × 3] = 10.5

⇒ [(2.68 – 0.18) × 4] = 10

⇒ [(1.42 + 3.38) × 5] = 24

∴ 10.5 is not an integer, as DECIMALS are not INTEGERS

Firstly, we have to find the L.C.M. of all the POWERS so that we can equate the bases.

L.C.M. of 3, 4, 6, 12 = 12

⇒ (5)^1/3 = (5⁴)^1/12 = (625)^1/12

⇒ (6)^1/4 = (6³)^1/12 = (216)^1/12

⇒ (12)^1/6 = (12²)^1/12 = (144)^1/12

⇒ (276)^1/12

Since, the powers are equal, we can equate the bases

Hence, (5)^1/3 is the largest

Putting the above options ONE by one and CHECKING its divisibility we FIND that P must be 9.

NOTE 342 = 7³ - 1. On further simplification we get,

= (7³)²⁸/342 = 343²⁸/342 = (342 + 1)²⁸/ 342 = 342N + 1/342 = 1/342

Hence, REMAINDER = 1.

A number P when divided by 24 LEAVES a remainder that is perfect SQUARE of an even number.

We can write, P = 24k + square of an even number = An even number

When this number is divided by 14, it leaves a remainder that is perfect square of an odd number.

We can write, P = 14n + square of an odd number = An odd number

P cannot be even and odd at the same time.

LET the NUMBER be x.

We know that,

Dividend = Divisor × QUOTIENT + Remainder

⇒ x = 16 × 1 + 8

⇒ x = 24

1261 × 3474 × 1269 = 5559126066

We can observe that, 6 is at the unit’s place of the product

Shortcut method:

Unit digit of 1261 × 3474 × 1269 will be same as unit digit of 1 × 4 × 9

1 × 4 × 9 = 36

Unit digit of 36 is 6

∴Unit digit of 1261 ×3474 ×1269 is 6.

$(\FRAC{{4 - \sqrt 6 }}{{2\sqrt 2 - \sqrt 3 }} = \;?)$

$(\Rightarrow \frac{{\sqrt 2 \left( {2\sqrt 2- \sqrt 3 } \right)}}{{2\sqrt 2- \sqrt 3 }} = \;?)$

⇒ ? = √2 = 1.414

? The number should be divisible by 9, 10 and 15 leaving a remainder of 7 in each case.

Thus a remainder of 7 will be ALSO obtained when it is divided by the LCM of 9, 10 and 15.

LCM of 9, 10 and 15 = 90

Dividing 1936 by 90 we GET,

Quotient = 21

Remainder = 46

The remainder should be 7

To SOLVE this TYPE of problems we have to follow the below steps.

Step I: Convert each of the decimals to like decimals.

Step II: Remove the decimal point and find the least common multiple (LCM) as usual.

Step III: In the LCM, PUT the decimal point as there are a number of decimal places in the like decimals.

LCM of 3.2, 2.72, 1.28 and 1.44

Step I: Like decimals are 3.2, 2.72, 1.28 and 1.44

Step II: LCM of 32/10, 272/100, 128/100 and 144/100 = LCM(32, 272, 128, 144)/HCF (10, 100) = 19584/10

Step III: LCM of 3.2, 2.72, 1.28, 1.44 = 1958.4

∴ LCM of 3.2, 2.72, 1.28, 1.44 = 1958.4

684 × 686 = 684 × (684 + 2) = 684² + 2 × 684

To make it PERFECT square we should add 1

⇒ 684² + 2 × 684 × 1 + 1² = (684 + 1)²

According to the given information,

HCF of the two NUMBERS =21

∴let the two numbers be 21x and 21Y where x and y are coprime.

21x + 21 y = 231

⇒21 (x + y) = 231

⇒x + y = 11

∴possible pairs of x and y are (9, 2), (8, 3), (7, 4), (6, 5), (1, 10)

LET the number be ‘x’

x = 6Y + 3

x² = (6y + 3)²

x² = 36y² + 36y + 9

x² = 6 × (6y² + 6y + 1) + 3

∴ When the square of the same number is divided by 6, the remainder will be 3.

Given,

√2 = 1.4142

Given EXPRESSION is,

$(\BEGIN{array}{l}2\sqrt 2+ \sqrt 2+ \frac{1}{{2 + \surd 2}} + \frac{1}{{\sqrt 2- 2}}\\ = 2\sqrt 2+ \sqrt 2+ \frac{{\left( {2 - \sqrt 2 } \right)}}{{\left( {2 + \sqrt 2 } \right)\left( {2 - \sqrt 2 } \right)}} + \frac{{\left( {\sqrt 2+ 2} \right)}}{{\left( {\sqrt 2- 2} \right)\left( {\sqrt 2+ 2} \right)}}\end{array})$

USING the identity: a² – b² = (a + b)(a – b)

$(\begin{array}{l} = 2\sqrt 2+ \sqrt 2+ \frac{{\left( {2 - \sqrt 2 } \right)}}{{{2^2} - {{\left( {\sqrt 2 } \right)}^2}}} + \frac{{\left( {\sqrt 2+ 2} \right)}}{{{{\left( {\sqrt 2 } \right)}^2} - {2^2}}}\\ = 2\sqrt 2+ \sqrt 2+ \frac{{\left( {2 - \sqrt 2 } \right)}}{{4 - 2}} + \frac{{\left( {\sqrt 2+ 2} \right)}}{{2 - 4}}\\ = 2\sqrt 2+ \sqrt 2+ \frac{{\left( {2 - \sqrt 2 } \right)}}{2} - \frac{{\left( {\sqrt 2+ 2} \right)}}{2}\\ = 2\sqrt 2+ \sqrt 2+ \frac{{2 - \sqrt 2- \sqrt 2- 2}}{2}\\ = 2\sqrt 2+ \sqrt 2- \frac{{2\sqrt 2 }}{2}\end{array})$

= 2√2+√2- √2

= 2√2

Putting value of √2 = 1.4142

The sum of first 50 odd numbers i.e. (1, 3, 5, …., 99)

Here, first number (a) = 1

And, LAST number (l) = 99

For the given A.P.,

First term, a = 3 and common difference, d = 2

The nth term = a + (N – 1) d

5^th term = 3 + 4 × 2 = 11

16^th term = 3 + 15 × 2 = 33

49^th term = 3 + 48 × 2 = 99

Hence, the G.P. is 11, 33, 99

Since, G.P. is of the form a, AR, ar²

Let the number be N.

Dividend = DIVISOR × Quotient + REMAINDER

⇒ N = 208 × Q + 43

⇒ N = (16 × 13 × Q) + (16 × 2) + 11

⇒ N = 16(13Q + 2) + 11

∴ Remainder = 11

Alternate Solution:

Let the number be 208 + 43 = 251

When 251 is divided by 16, the remainder will be 11.

For a NUMBER to divisible by 30, the number should be divisible by both 10 and 3.

For a number to divisible by 10, the last DIGIT should be zero

For a number to divisible by 3, the sum of digits should be divisible by 3

All the numbers are divisible by 10.

The sum of digits of 2530 is 10 which is not divisible by 3

The sum of digits of 1570 is 13 which is not divisible by 3

The sum of digits of 2370 is 12 which is divisible by 3. 2370/30 = 79

The sum of digits of 1520 is 8 which is not divisible by 3

LCM of 20, 32 and 34 = 2720

LCM of FRACTIONS = LCM of NUMERATORS/HCF of denominators

LCM of Numerators = LCM of 7 and 9 = 63

HCF of Denominators = HCF of 5 and 25 = 5

In the unit’s PLACE, STARTING with 2, 12 ….92, we have 10 2s used.

Also in 20 - 29, we have 2, 10 TIMES in the ten’s place.

Hence, total no. of 2s used = 10 + 10 = 20

Finding L.C.M of 2, 4, 5 and 6;

FACTORS of 2 = 2

Factors of 4 = 2²

Factors of 5 = 5

Factors of 6 = 2 × 3

L.C.M(2, 4, 5, 6) = 2² × 3 × 5 = 60

Divide 3475 by 60,

⇒ 3475/60 = $(57\frac{{55}}{{60}})$ [Here 55 represent the remainder which need to make 0 for that,60 - 55 need to do. Answer of subtraction is REQUIRED answer which need to add in 3475]

A number will be DIVISIBLE by 9 if the SUM of the digits of the number is divisible by 9

Sum of digits of 234561 = 2 + 3 + 4 + 5 + 6 + 1 = 21, Not Divisible by 9

∴ 234561 is not divisible by 9

Sum of digits of 444123 = 4 + 4 + 4 + 1 + 2 + 3 = 18, Divisible by 9

∴ 444123 is divisible by 9

Sum of digits of 555231 = 5 + 5 + 5 + 2 + 3 + 1 = 21, Not Divisible by 9

∴ 555231 is not divisible by 9

Sum of digits of 65422 = 6 + 5 + 4 + 2 + 2 = 19, Not Divisible by 9

The lowest 3-digit no. is 100 and the highest is 999.

The lowest 3-digit multiple of 9 is 108, and the highest is 999.

We know that,

LCM × HCF = product of TWO numbers

⇒ 1680 × HCF = 80640

⇒ HCF = 80640/1680 = 48

∴ the REQUIRED HCF is 48

LCM of 10, 17 = 170

We have to FIND the number of multiples of 170 < 1000

Number of multiples = integer value of (1000/170) = integer part of 5.88 = 5

∴ there are 5 multiples of both 10 and 17 LESS than 1000 are 5

We KNOW that (70)2$ = 4900 and (75)2$ = 5625 and the no. 547N is < (75)2$

ALSO,

⇒ (74)2$ = 5476

∴ N = 6