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
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
?
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
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.
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
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
?
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
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.
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
.