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Q:

What smallest number of 6 digit is divisible by 111?

Answer

Smallest number of 6 digits is 100000

on dividing 100000 by 111 we get 100 as remainder

$\inline \fn_jvn \therefore$ Number to be added = (111 -100) = 11

$\inline \fn_jvn \therefore$ Required Number = 100011

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Subject: Numbers

Q:

The value of

$\left [ \frac{2^{n}+2^{n-1}}{2^{n+1}-2^{n}} \right ]$ is:

Answer

$\frac{2^{n} + \frac{2^{n}}{2}}{2 x 2^{n} - 2^{n}} = \frac{2^{n} (1 + \frac{1}{2})}{2^{n} (2 - 1)} = \frac{3}{2}$

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Subject: Numbers Exam Prep: AIEEE , Bank Exams , CAT
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Q:

A number when divided by 779 gives a remainder 47. By dividing the same number by 19, what would be the remainder?

Answer

Number = ( 779 x a) + 47, where "a" is the quotient

= (19 x 41 x a) + (19 x 2) + 9

= 19 x (41a + 2) + 9

= 19 x (New quotient) + 9

$\inline \fn_jvn \therefore$ Required remainder = 9

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Subject: Numbers

Q:

(51 + 52 + 53 + .........+100) is equal to

Answer

51 + 52 + 53 + ...........+ 100

= (1 + 2 + 3 + .... + 100) - (1 + 2 + 3 + ...... + 50)

= It is in the form of $\frac{n (n + 1)}{2} s e r i e s s u m m a t i o n$

= n1 = 100 , n2 = 50

= $[\frac{100 (100 + 1)}{2}] - [\frac{50 (50 + 1)}{2}]$

= (5050 - 1275) = 3775

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Subject: Numbers Exam Prep: AIEEE , Bank Exams , CAT
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Q:

The total number of prime factors in the expansion

$4^{11}\times 7^{5}\times11^{2}$

Answer

$2^{2^{11}} x 7^{5} x 11^{2} = > 2^{22} x 7^{5} x 11^{2}$

Total number of prime factors is (22+5+2) = 29

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Subject: Numbers Exam Prep: AIEEE , Bank Exams , CAT
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Q:

The sum of first 45 natural numbers is:

Answer

The sum of n natural numbers $\frac{n(n+1)}{2}=\frac{45\times 46}{2}=1035$

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Subject: Numbers

Q:

The difference between two numbers is 1365. When larger number is divided by the smaller one, the quotient is 6 and the remainder is 15. The smaller number is ?

Answer

Let the numbers be x and 1365+x

Then 1365+x = 6x+15

=> x = 270

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Subject: Problems on Numbers

Q:

If sum of three consecutive numbers is 42 . What will the sum of the digits of middle number .

Answer

Let the numbers be x, x+1, x+2

Then x + x+1 + x+2 = 42

=> 3x + 3 = 42

=> 3x = 39

=> x = 13

The numbers would be 13, 14, 15

Therefore, Sum of the digits of middle number = 1+4 = 5

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Subject: Problems on Numbers

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