Number system questions

• LCM(6, 8, 9)

  6 = 2 × 2

  8 = 2 × 2 × 2 and

  9 = 3 × 3

• LCM(6, 8, 9) = 2 × 2 × 2 × 3 × 3 = 72

• Multiple of 72 → 72, 144, 216, 288, 360, ......

• There are only two numbers lies between 200 and 300.

• ∴ Only two number can divisible by 6, 8 and 9 between 200 and 300.

\(\frac{208}{9}=23.11\)

\(\frac{351}{9}=39\)

Numbers that will be divisible are from \(9 \times 24=216\) to \(9 \times 38=342\)

Numbers \(=38-24+1=15\)

Hence, Option A is correct.

Let the numbers be \(x,(x+2),(x+4)\)

\(3x+6=\frac{3x+6}{3}+38\)

\(3x+6=x+2+38\)

\(2x+4=38\)

\(2x=34\)

\(x=17\)

Least number divisible by 8, 15, 16, 21 and 5 =LCM of 8, 15, 16, 21 and 5

LCM of 8, 15, 16, 21 and 5 = \(=2 \times 2 \times 2 \times 2 \times 3 \times 5 \times 7=1680\)

Greatest five-digit number = 99999

When we divide 99999 by 1680 we obtain 879 as a remainder.

Hence, Required number = 99999 – 879 = 99120

The two digits in which the sum of the digits is 8 are \(17,26,35,44\).

Here, \(53-35=18\)

So the original number is 35 .

\(55 \%\) of the original number \(=\frac{55 \times 35}{100}=19.25\)

Hence, option \(A\) is correct.

Largest 5-digit number = 99999

\( 99999 \div 89 =\) we get remainder 52 \)

So the answer is \( 99999- 52 = 99947 \)

Required numbers are 24, 30, 36, 42, ..., 96

This is an A.P. in which a = 24, d = 6 and l = 96

Let the number of terms in it be n.

Then \(t_{n}\) = 96

24 + (n - 1)6 = 96

(n - 1)6 = 96- 24 = 72

(n - 1)\( = \frac{72}{6} \)

\( n = 13 \)

In mathematical form \(x-y=3\) and \( xy = 504\)

By solving this the numbers are 21, 24 or -21, -24

Largest 4-digit number = 9999

92) 9999 (108

     92
 ----
 799
  736
 ----
  63

Required number = (9999 - 63) = 9936.

5678 x 43267 y
divisibility8\( \rightarrow \) last 3 - digit by 8
when y=2 the last three digit is divisible
At x=4 the sum of digit is divisible by 9.
\( \sqrt{5 x+8 y} \)=\( \sqrt{36} \)
=6

If the number is divided by 16, the remainder is 5. When 5 is divided by 4, the remainder =1

hence the remainder will be 1.

To be divisible by 99, the number has to be divisible by 11 and 9 both.

For divisibility by 11,(7 + 8 + 6 + 8 + 9) -( 0 + x + y + z)= 11 or 0

(38 - x + y + z)= 5, 16, 27 or 38

For divisibility by 9,

(38 + x + y + z) has to be divisible by 9.

By option C),

x + y + z = 16

(38 - x + y + z) = 38 - 16 = 22 is divisible by 11.

(38 + x + y + z) = 38 + 16 = 54 is divisible by 9.

We know that to divide by 5 last digit of number should be 0 or 5.

And to divide by 8 last three digits of number should be divisible by 8.

Therefore, minimum possible value for dividing 40 is 57968320

Required, value of P = 2 and q = 0

let the numbers be \(x\) and \(x\) - \(5\)

\(x\) (\(x\)-\(5\)) = 336 , ie. \(x²\)-\(5\)\(x\) -\(336\)=\(0\)

by quadratic formula, \(x\) = \(\frac{5+√(25+1334)}{2}\) , ie. \(x\) =\(21\)

sum of two numbers = \(21\)+\(21\)-\(5\) = \(37\)

According to the question,

\(\frac{314}{8}=39.25\)

\(\frac{408}{8}=51\)

So between 51 and \(39=40\) to 50 (total 11 values)

So the answer \(=11\)

Hence, Option \(C\) is correct.

Factors of 72 is 8 and 9.

For a number divisible by 8, its last three digit must be divisible by 8. Therefore 78y should be divisible by 8.

y should be 4 to complete the divisibility of the number . y =4

For a number divisible by 9, sum all the digit should be divisible by 9

Therefore 9+8+5+x+3+6+7+8+4 = 50+x

Nearest digit is 54, which is divisible by 9 therefore x is 4.

Now (4x+5y) = 4× 4+5× 4 = 16+20 = 36

Here, the first divisor (36) is exactly divisible by the second divisor (12).

Required remainder

= Remainder obtained after 19 is divided by 12 = 7.

Rules of divisibility of 72 - number should be divisible by 9 and 8

Rules of divisibility of 9 – sum of digits of number should be divisible by 9

Rules of divisibility of 8 – last three digits of number should be divisible by 8

Therefore, possible number is = 65784432672

Value of x = 4, y = 2

Required value of \(\sqrt{x+6 y}=\sqrt{4+6 \times 2}=\sqrt{4+12}=\sqrt{16}=4\)

The smallest three digit number divisible by 6 and 7 is = 42 × 3 = 126

⇒ A = 126

And to find the largest four digit number digit number divisible by 6 and 7, we have to divide 9999 by 42

⇒ 9999 = 42 × 238 + 3

⇒ The largest four digit number divisible by 42 is 9999 – 3 = 9996

⇒ B = 9996

So, the required difference = 9996 – 126

∴ The difference between B and A is 9870

Largest four digit number = 9999.
\( \frac{9999}{88}\)leaves a remainder of 55
SO the required number is \(9999-55 = 9944\).

We know that 72 = 9 × 8

Therefore, the given number must be divisible by 9 and 8.

First, divisibility rule of 8: Last three digits of the number must be divisible by 8.

Therefore, the last three digits are 78y.

In addition, y = 4. (since we have to calculate least value)

Now, divisibility rule of 9: Digital sum of the number must be divisible by 9.

Now, digital sum of 785x36784 = 7 + 8 + 5 + x + 3 + 6 + 7 + 8 + 4

= 48 + x

The next multiple of 9 after 48 is 54.

Therefore, x = 54 – 48 = 6

Now, required x – y

= 6 – 4 = 2

Divisibility rules of 8 – last three digits of number should be divisible by 8

Divisibility rules of 5 – last digits of number should be 0 and 5

Possible value of such numbers are = 126348720, 126348760

Therefore, greatest possible of x and y = 6 and 0

number = 6 \(\times\)q + 3.

when divided by 3, remainder will be zero.

let numbers are \(x\) and \(y\).

According to question, \(x+y=19 \quad \ldots \ldots \ldots\) i

\(x-y=5 \ldots \ldots . .\) ii

From adding eq \(I\) and ii, we get \(x=12\)

\(y=19-12=7\)

Product of numbers \(=12 \times 7=84\).

The largest 5 digit number = 99999

\( 99999 \div 84 \) we get a remainder 39

\( So the answer is 99999- 39 = 99960 \)

Divisibility rules of 9 – sum of all digits of number should be divisible by 9

Divisibility rules of 11 – the difference between the sum of the digits at the odd and even place is equals 0 or divisible by 11

So, possible number is 847563948

Therefore, possible value of A = 4 and possible value of B = 8

Let the digit in the unit's place be \(2x\)

Digit in the ten 's place be \(3x\)

original number

=\(32x\)

New number obtained by interchanging the digits,

=\(23x\)

\(32x -23x =27\) [Given]

\(9x =27\)

\(x =\frac{27}{9}\)

=\(3\)

Original number =\(32\times 3 =96\)

• 6 digit number 4x573y is divisible by 72, so we can say number also divisible by both 8 and 9.

• Divisibility law of 8 ⇒ A number divisible by 8 if its last three digit is divisible by 8.

• Divisibility law of 9 ⇒ A number is divisible by 9 if sum of its digit is divisible by 9.

  ⇒ 4x573y is divisible by 8 if y = 6

  ⇒ 4x5736 is divisible by 9 if x = 2

  ⇒ (x + y)

  ⇒ 6 + 2

  ⇒ 8

The sum of the two digits of the number = 13

The number = 94

\( 94- 49 = 45 \)

• LCM \((3,7,11)=231\)

• By taking the largest 5 -digit number 67699 and divide it by 231 .

  \(\because 67699=(231 \times 293) +16\)

  \(\Rightarrow 67699=67683+16\)

  \(\Rightarrow 67699-16=67683\) (Completely divisible by 231 )

  \(\therefore 67683=676 xy\) (Where \(x=8, y=3\) )

  \((3 x-5 y)=3 \times 8-5 \times 3\)

  \(\Rightarrow 24-15=9\)

• \(\therefore\) The required result \(=9\).

908^6#8!7 is divisible by 33

It means it is divisible by both 11 and 3 because 33 = 11 × 3

We know, the divisibility rule of 3 is sum of the digits must be divisible by 3.

and the divisibility rule of 11 is difference of sum of alternate digits is equal to 0 or divisible of 11.

908^6#8!7/11 = 7 + 8 + 6 + 8 + 9 – (^ + # + !)

38 – (^ + # + !)

If we put ^ + # + ! = 16

then it is divisible by both 3 and 11

Largest six- digit number = 999999

When we divide 999999 by 243 ,54 is obtained as a remainder.

Hence, largest six-digit number exactly divisible by 243 = 999999 – 54 = 999945

First number \(= 16,\)
Last number \(= 200\)
Last number = First number \(+ (n – 1) \times 8\)
\(200 = 16 + (n – 1) \times 8\)
\(n = 24\)

\( \frac{19}{7}=5 \) remainder

• ⇒ 1st number = 1125

• ⇒ Last number = 4950

• ⇒ Number of terms = (4950 – 1125)/225 + 1

  ⇒ 3825/225 + 1

  ⇒ 17 + 1 = 18.

• Required numbers = 22 - 4 = 18

• If a number is divisible by 3, 7 and 11, then it also be divisible by 231
• Possible value of 5 digits number 688xy which will be divisible by 3, 7 and
• 11 = 68838
• Value of x = 3
• Value of y = 8

• Therefore, the required value of (5x + 3y)

  = 5 × 3 + 3 × 8

  = 15 + 24

  = 39.

The numbers lying between 300 and 1000, which are exactly divisible by 13, are

312,325,338, ...............................................,998 are in A.P

a=312,d=13,l=998

By using the formula, n=[(l-a)/d]+1

we get n= 53

Let the numbers be x and y.

Then, xy = 120 and \(x^2+ y^2 = 289. \)

∴ \((x + y)^2\)

= \(x^2+y^2+ 2xy = 289 + (2 \times 120) = 529\)

∴\( x + y = √529 = 23\)

Let the numbers be \(1x\) and \(5x\)

\(5x^{2} =320\)

\(x^{2} =\frac{320}{5}\)

=\(64\)

\(x =\sqrt{64}\)

=\(8\)

Sum =\(1x +5x =6x\)

=\(6\times 8 =48\)

Number \( (7 + 5) \times 3 + 2 = 38 \)

\(\Rightarrow \frac{38}{21}=17 \)

\( \therefore \)remainder = 17

\( 2\ 5 \ 9 \ 8 \ 7\ 6 \ p \ 0 \ 5 \)

is divisible by 11

Sum of digits at even places – Sum of digits at odd places = 0 or multiple of 11

\( (23+P) – 19 = 0 \) or 11

\( P = 7 \)satisfy

\( P^{2}+5 \)

\( 7^{2}+5 = 54 \)

Divisibility rules of 8 – last three digits of the number should be divisible by 8

Divisibility rules of 9 = sum of all digits of the number should be divisible by 9

Now, possible numbers is 42688480536

Value of x = 3 and y = 2

Therefore, required value of = \(\sqrt{x y}=\sqrt{2 \times 3}=\sqrt{6}\)

• First we need to take out the LCM of 4,5 and 7 = 140
• Numbers from 250 to 1000 which are divisible by 4, 5 and 7 are divisible by 140 also.
• The smallest number and highest number divisible by 140 are 280 and 980 respectively.
• So \(\frac{980-280}{140}\)+1 =6

Sum of the digits of the number is 11,

so number must be= \( 92, 83, 74, 65, 56, 47, 38, 29 \)

\( 92- 29 = 63 \)

Largest 5 digit number is 99999 if it is divided by 91 remainder is 81.so subtract \(99999-81 = 99918\)

As the number is divisible by 11

i.e. (ρ +6 +σ +5)-(3+9+0)=( ρ + σ -1) is divisible by 11

ρ +6 +σ +5-3-9 = 12 or 1

so, the probable value of ρ + σ-7 from the given option=5