Number System - Questions & Discussions

MBA CAT 2022
2896

M is a two digit no. which has the property that the products of factorials of its digits > sum of the factorials of its digit.
How many values of M exist ?
a. 56
b. 64
c. 63
d. none of these

2897

Find the 28383rd term of the series 123456789101112..............

  1. 3
  2. 4
  3. 9
  4. 7
2898

How many integer values of x and y are there such that 4x+7y=3, while x

  1. 144
  2. 141
  3. 143
  4. 142
2899

x and y are two positive integers. Then what will be the sum of the coefficients of the expansion of the expression (x+y)^44?

  1. 2^43
  2. 2^43 + 1
  3. 2^44
  4. 2^44 - 1
2900

The series of no (1, 1/2, 1/3, 1/4.....................1/1972) is taken. Now two no.s are taken from the series(the first two) say x and y. Then the operation x+y+x.y is formed to get a consolidated no. The process is repeated. What will be the value of set after all the no.s are consolidated into one no.

  1. 1970
  2. 1971
  3. 1972
  4. none of these
2901
What is the remainder when 128^1000 is divide by 153 ?
a. 103
b. 145
c. 118
d. 52


My take is 52.

128^1000 mod 153.
153 = 9*17.
128^1000 mod 9.
E(9) = 6.
=> 1000 mod 6 = 4.
=> 128^4 mod 9 = 2^4 mod 9 = 7.
128^1000 mod 17.
E(17) = 16.
=> 1000 mod 16 = 0.
=> 128^8 mod 17 = 9^8 mod 17 = 1.
So, remainder would be of the form 9k+7 = 17p+1. Hence, 52.

M is a two digit no. which has the property that the products of factorials of its digits > sum of the factorials of its digit.
How many values of M exist ?
a. 56
b. 64
c. 63
d. none of these


My take is 63.

Let the number be ab.
So we need a!*b! > a!+b!.
This is only possible when a,b>1.
The given conditions will not be satisfied for following cases.
1)a = 1 , b = any value -> 10 values.
2) a > 1, b = 0 or 1 -> 16 values.
3) a =2, b = 2 -> 1 value.

We have a total of 90 two digit numbers.
Hence, answer = 90 - 27 = 63.

Find the 28383rd term of the series 123456789101112..............

  1. 3
  2. 4
  3. 9
  4. 7


My take is 3.

1-9 -> 9.
10-99 -> 180.
100-999 -> 2700.
1000-6999 -> 24000.
7000-7369 -> 1480.
total = 28369.
We need another 14 digits.
7370,7371,7372 = 12 digits.
So 3 is the 28383rd digit.

How many integer values of x and y are there such that 4x+7y=3, while x
  1. 144
  2. 141
  3. 143
  4. 142


My take is 144.

4x + 7y = 3.
=> x = (3 - 7y)/4.
Now, when y = 1, x = -1.
when y = 5, x = -8.
when y = 9, x = -15.
and so on, till.
when y = 285, x = -498.
=> 72 terms. But we need to find integer solutions. Hence 72*2 = 144.

x and y are two positive integers. Then what will be the sum of the coefficients of the expansion of the expression (x+y)^44?

  1. 2^43
  2. 2^43 + 1
  3. 2^44
  4. 2^44 - 1


My take is 2^44.

Put x=y=1.


The series of no (1, 1/2, 1/3, 1/4.....................1/1972) is taken. Now two no.s are taken from the series(the first two) say x and y. Then the operation x+y+x.y is formed to get a consolidated no. The process is repeated. What will be the value of set after all the no.s are consolidated into one no.

  1. 1970
  2. 1971
  3. 1972
  4. none of these


My take is 1972.

Just take -> {1,1/2}
=> 1+1/2+1/2 = 2.
Now take -> {1,1/2,1/3}
=> 2+1/3+2/3 = 2+1 = 3.
Hence when 1/1972 is taken then x+y+xy = 1972.
2902
x and y are two positive integers. Then what will be the sum of the coefficients of the expansion of the expression (x+y)^44?

  1. 2^43
  2. 2^43 + 1
  3. 2^44
  4. 2^44 - 1

nC0 + nC1 + nC2 + ......... nCn = 2^n.
2903
x and y are two positive integers. Then what will be the sum of the coefficients of the expansion of the expression (x+y)^44?

  1. 2^43
  2. 2^43 + 1
  3. 2^44
  4. 2^44 - 1


3.,the correct answer is 3.2^44
2904

is answer 6 ?

2905


4x + 7y = 3.
=> x = (3 - 7y)/4.
Now, when y = 1, x = -1.
when y = 5, x = -8.
when y = 9, x = -15.
and so on, till.
when y = 285, x = -498.
=> 72 terms. But we need to find integer solutions. Hence 72*2 = 144.


Acc to me solution is ccoming as 143. 72 + 71(this is for negative values of y. one term less is coming for -ve value of y's)
2906

Hello everyone , one of my friend a TIME student told me to solve this type of question
Q)Find the remainder when 123456789.....40 is divided by 36

He(my friend) told me that 36=9x4 therefore first find remainder when 123456...40 is divided by 9 which is found by diving sum of digits 244/9 => remainder = 1
Then find remainder when 123456...40 is divided by 4 which is remainder of dividing last 2 digits by 4 ; 12/4=> remainder= 0
Now to find the final remainder we find the lowest multiple of 4 which will give remainder 1 with 9 and such a no. can be 28 ; Hence final remainder answer is 28

But now if this approach is applied to one of the ARUN SHARMA question then the answer is not coming correct
The question is as follows
Q) Find the remainder when 123456789...4950 is divided by 16

Now my approach is 16=2x8 , Dividing 1234567...4950 by 2 will yield remainder=0
Dividing 123456...4950 by 8 which is found by dividing last 3 digits i.e 950/8 => remainder = 6
Now to find the final remainder we need to find lowest multiple of 2 that gives remainder 6 with 8 which will be 2x7 =14 as 14/8 => remainder = 6 ; the final answer by TIME's approach should be 14 BUT THE ANSWER IS GIVEN AS 6 IN THE ARUN SHARMA BOOK CAN SOMEBODY HELP??????????????????

2907
My take is option D.

Lets take examples.
1*2*3/3! = 1 (which is a perfect square).
4*5*6/3! = 20 (which is a natural number).



That what i think. but OA was given as 'a natural number' dunno if the author has missed '1'. It is from Ariahnt Publication by Sarvesh Kumar Verma.
2908
2)what vll b d last digit of 2^3^4^5 - 2^3^5^4 ???

a)0 b)2 c)4 d)6 e)none of dese

cat 2011 Says
hmmm 2nd is d correct ...can u pls tell me d approach...


Lets take the first term: 2^3^4^5

4^5 can be written as 4*256 , therefore first term becomes 2^3^(4*256) = 2^81^256.

Now we can find out the remainder of 81^256 = 81 and hence 2^81 has a unit digit as 2.

Following the same approach in second terms leads us to unit digit as 8.

Now situation is *****2 - *****8. Therefore, unit digit as 4.

Hope this helps. Let me know if i am missing something or need anything else.
2909
Hello everyone , one of my friend a TIME student told me to solve this type of question
Q)Find the remainder when 123456789.....40 is divided by 36

He(my friend) told me that 36=9x4 therefore first find remainder when 123456...40 is divided by 9 which is found by diving sum of digits 244/9 => remainder = 1
Then find remainder when 123456...40 is divided by 4 which is remainder of dividing last 2 digits by 4 ; 12/4=> remainder= 0
Now to find the final remainder we find the lowest multiple of 4 which will give remainder 1 with 9 and such a no. can be 28 ; Hence final remainder answer is 28

But now if this approach is applied to one of the ARUN SHARMA question then the answer is not coming correct
The question is as follows
Q) Find the remainder when 123456789...4950 is divided by 16

Now my approach is 16=2x8 , Dividing 1234567...4950 by 2 will yield remainder=0
Dividing 123456...4950 by 8 which is found by dividing last 3 digits i.e 950/8 => remainder = 6
Now to find the final remainder we need to find lowest multiple of 2 that gives remainder 6 with 8 which will be 2x7 =14 as 14/8 => remainder = 6 ; the final answer by TIME's approach should be 14 BUT THE ANSWER IS GIVEN AS 6 IN THE ARUN SHARMA BOOK CAN SOMEBODY HELP??????????????????


Your friend gave you half information.
See, when he told you that you would need to break 36 into 9*4, then something that needs to be pondered upon is why not break it into some other factors of 36? There are 2 reasons behind it, first one is that we know the divisibility rules of these numbers (9 and 4). And the second and most important thing to be noted here is that, the factors you need to break this into should be co-prime to each other (Chinese remainder theorem).
Now, we have 16 at hand here. 2^4 = 16. So we cannot break it into two co-prime factors. So this is what we do here.
We know that 10^4 mod 16 = 0.
Hence we break this number in sets of 4 and take remainder individually and then obtain the sum of all the 4 digits number and then find the remainder with 16. This makes it less arduous.

Hope it helps. 😃
2910

1 and 8 are the first two natural numbers for which 1+2+3+..... + n is a perfect square. Which number is the 4th such number?

2911
We need to find values of n for which n(n + 1) = 2k2
Here are a few points:
--there would be no factor in common between n and n + 1, i.e. two consecutive natural numbers do not have any factor in common except 1.
--this means that one of the numbers from n and n + 1 would be a perfect square and the other would be twice of a perfect square.

--out of n and n + 1, one would be an odd number and the other would be an even number. The odd number cannot be twice of a perfect square. Therefore, it will only be a perfect square.
Now we start examining all the odd perfect squares (9, 25, 49, 81, 121, ..) and see if the numbers lying above or below it are twice of a perfect square or not.
For example, numbers lying on both sides of 25 are 24 and 26 and none of them is twice of a perfect square. But for 49, 50 is twice of a perfect square.
As such, the third one will be 25 and the fourth one 289.
2912

hooo really tough , can u explain in more detailed manner?

2913
If a number A has 24 factors and A x B has 105 factors then, least possible number of factors that B can have?
a.12 b.9 c.16 d.7 e. none of these
2914
.Which of the following numbers gives the largest remainder when divided by 101?
OPTIONS
1) 123456789123
2) 231456789231
3) 312456789312
4) 213456789213
5) All of the above give the same remainder
2915

please solve this question?
how to find number of factors of 15! ?