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Explore the Back-Storyare natural numbers each greater than 1 . If , and there are terms on the left hand side, then the number of ordered pairs is

Value of will be greater than . So first we can find out the factors of .

So,

When the value of is , value of is and when the value of is 3 then the value of n is and vice versa.

So the ordered pairs will be,

So the number of ordered pairs is .

The number of times the digit occurs in the result of

111 (100digits) is

sum is

there are brackets

so comes times

Let us call a sum of integers a cool sum if the first and last terms are 1 and each term differs from its neighbours by at most. For example, and are cool sums. The minimum number of terms required to write 2019 as a cool sum is ……

Sequence is

So minimum number of terms is .

For each positive integer let . Then the sum of all which are prime is

is prime

sum of values

The product of four positive integers and is 9 ! The number satisfy 1224, and . The

given

Therefore

and

So

The number of 6 digit numbers of the form "ABCABC", which are divisible by 13 , where and are distinct digits, and being even digits is

where Now and are even digits and are different digits .

Case-I: When is zero

Case-II : When is not zero

Total number of digits

Number possible

In the sequence and 43 , the number of blocks of consecutive terms whose sums are divisible by 11 is

Exactly four

In the subtraction below, what is the sum of the digits in the result ? (100 digits) (50 digits)

---------------------------------------------

49 times 1,49 times 8 and 1 times 0 and 9

Sum

Therefore

The least odd prime factor of is

Let be an odd prime which divides

So

Now by Euler's theorem

So should be divisible by

Where is a prime

First two prime numbers which gives remainder when divided by is and

Case-1

(mode )

While

(mod )

So the answer is .

Let be positive integers each less than 50 , such that . The number of such triples is

As is a multiple of

So it means the last digit of and is same.

So can be .

So there are such pairs .

One more pair for is .

So total pairs are possible.

If , then the sum of the digits of is

4-digit no. (4921) is multiplied by a single digit no. (D) \& result is five digit no., so definitely

So by hit \& trial we put the values of D from 3 to 9 .

at

So

Now ABBBD

Sum of digits

What is the digit to the right of the decimal point, in the decimal representation of ?

[2 for pairs of 6 repeating number]

digit from right side to decimal is first digit in repetition

So correct answer is .

If is a 1000 digit number, is the sum of its digits, the sum of the digits of and the sum of the digits of , then the maximum possible value of is

digit no. If all digit are ' 9 ' so that maximum sum of digit of ' ' is 9000 So maximum value of is 9000

But for maximum sum of digit of is 35 for number (8999) So is maximum 35 .

Now for maximum sum of digit of is 11 for number So . Practical example: if Sum of digit of Sum of digit of Sum of digit of

Let be the number . 001 which has 2019 zeroes after the decimal point. Then which of the following numbers is the greatest?

(A) , (B) , (C) , (D)

From option

From option

From option

From option

So from options is greatest.

then the number of possible values for satisfying this equation where

If then the number of and are distinct digits is

_________

must be

Means or 11

Sum of is even is

So possible values of B is 0 or 5 .

But if we take as ' 0 ' so there is no carry forward \& Sum of , did not get different digit from D.

So B must be 5 .

Sum is convert into

_____

Now possible pairs of are

So total 4 possible solutions are there.

If and each alphabet represents a different digit, what is the maximum possible value

of FLAT?

T should be 0 or 5

But if we take as ' 5 ' sum is carry is forward and sum of 3 'A' and '1' never give unit digit 'A' So T must be '0'

Now again possible values of are but alphabet represents different digits so is 5 . For maximum value of FLAT, we take maximum value of as , but sum is is repeat.

So by taking as . We get maximum value of FLAT.

2029-2003=26

1+1+1+1+3+3+3+9=22

So numbers less than 4 is 7

The number of natural numbers such that is an integer is

So should be multiple of for integer value.

integer integer integer.

So should be multiple of for intege So, possible 'n' are

(reject)

So only three values of are possible

The number of perfect cubes that lie between and is

and and

Numbers lies between Total perfect cubes number .

are natural numbers each greater than 1 . If , and there are terms on the left hand side, then the number of ordered pairs is

Value of will be greater than . So first we can find out the factors of .

So,

When the value of is , value of is and when the value of is 3 then the value of n is and vice versa.

So the ordered pairs will be,

So the number of ordered pairs is .

The number of times the digit occurs in the result of

111 (100digits) is

sum is

there are brackets

so comes times

Let us call a sum of integers a cool sum if the first and last terms are 1 and each term differs from its neighbours by at most. For example, and are cool sums. The minimum number of terms required to write 2019 as a cool sum is ……

Sequence is

So minimum number of terms is .

For each positive integer let . Then the sum of all which are prime is

is prime

sum of values

The product of four positive integers and is 9 ! The number satisfy 1224, and . The

given

Therefore

and

So

The number of 6 digit numbers of the form "ABCABC", which are divisible by 13 , where and are distinct digits, and being even digits is

where Now and are even digits and are different digits .

Case-I: When is zero

Case-II : When is not zero

Total number of digits

Number possible

In the sequence and 43 , the number of blocks of consecutive terms whose sums are divisible by 11 is

Exactly four

In the subtraction below, what is the sum of the digits in the result ? (100 digits) (50 digits)

---------------------------------------------

49 times 1,49 times 8 and 1 times 0 and 9

Sum

Therefore

The least odd prime factor of is

Let be an odd prime which divides

So

Now by Euler's theorem

So should be divisible by

Where is a prime

First two prime numbers which gives remainder when divided by is and

Case-1

(mode )

While

(mod )

So the answer is .

Let be positive integers each less than 50 , such that . The number of such triples is

As is a multiple of

So it means the last digit of and is same.

So can be .

So there are such pairs .

One more pair for is .

So total pairs are possible.

If , then the sum of the digits of is

4-digit no. (4921) is multiplied by a single digit no. (D) \& result is five digit no., so definitely

So by hit \& trial we put the values of D from 3 to 9 .

at

So

Now ABBBD

Sum of digits

What is the digit to the right of the decimal point, in the decimal representation of ?

[2 for pairs of 6 repeating number]

digit from right side to decimal is first digit in repetition

So correct answer is .

If is a 1000 digit number, is the sum of its digits, the sum of the digits of and the sum of the digits of , then the maximum possible value of is

digit no. If all digit are ' 9 ' so that maximum sum of digit of ' ' is 9000 So maximum value of is 9000

But for maximum sum of digit of is 35 for number (8999) So is maximum 35 .

Now for maximum sum of digit of is 11 for number So . Practical example: if Sum of digit of Sum of digit of Sum of digit of

Let be the number . 001 which has 2019 zeroes after the decimal point. Then which of the following numbers is the greatest?

(A) , (B) , (C) , (D)

From option

From option

From option

From option

So from options is greatest.

then the number of possible values for satisfying this equation where

If then the number of and are distinct digits is

_________

must be

Means or 11

Sum of is even is

So possible values of B is 0 or 5 .

But if we take as ' 0 ' so there is no carry forward \& Sum of , did not get different digit from D.

So B must be 5 .

Sum is convert into

_____

Now possible pairs of are

So total 4 possible solutions are there.

If and each alphabet represents a different digit, what is the maximum possible value

of FLAT?

T should be 0 or 5

But if we take as ' 5 ' sum is carry is forward and sum of 3 'A' and '1' never give unit digit 'A' So T must be '0'

Now again possible values of are but alphabet represents different digits so is 5 . For maximum value of FLAT, we take maximum value of as , but sum is is repeat.

So by taking as . We get maximum value of FLAT.

2029-2003=26

1+1+1+1+3+3+3+9=22

So numbers less than 4 is 7

The number of natural numbers such that is an integer is

So should be multiple of for integer value.

integer integer integer.

So should be multiple of for intege So, possible 'n' are

(reject)

So only three values of are possible

The number of perfect cubes that lie between and is

and and

Numbers lies between Total perfect cubes number .

Cheenta is a knowledge partner of Aditya Birla Education Academy

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.

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