Number System - Questions & Discussions

MBA CAT 2022
2065
_ _ _ _ _ _ _

we'll convert binary to decimal to find the sum-

7 digit no -> first digit will be always 1...remaining 6 places wil be filled by 3 1's and 3 0's...
so 2^6 will be repeated 6!/(3!*3!) ways
sum=2^6 *(6!/3!*3!)

now 6th digit- no of times 1 will be there

1 1 _ _ _ _ _

remaining blanks will be filled by 2 1's and 3 0's -> 5!(2!*3!) ways
sum = 2^5 * 5!/(2!*3!)

similarly for rest of terms....

so total sum =
2^6 (6!/3!*3!) + 5!/(2!*3!) (2^5+2^4+2^3+2^2+2^1+2^0)

= 1910

my take is 1910
2066

2^3^4^5 - 2^3^5^4 ?

a) 0
b) 2
c) 4
d) 6
e) None of these

pls mention approach for soloving this type of question

2067
2^3^4^5 - 2^3^5^4 ?

A) 0
b) 2
c) 4
d) 6
e) none of these

pls mention approach for soloving this type of question


2^3^20 - 2^3^20 =0
2068

Is the answer 0 by any chance to the above question?

2069
2^3^4^5 - 2^3^5^4 ?

a) 0
b) 2
c) 4
d) 6
e) None of these

pls mention approach for soloving this type of question


It depends what exactly is your expression:-

If its 2^(3^(4^5)) - 2^(3^(5^4)) = 2^(3^1024) - 2^(3^625) , then none of these

If its ((2^3)^4)^5 - ((2^3)^5)^4 = (2^3)^20 - (2^3)^20 = 0

So, I don't know which one you are refering??
2070

If n is a natural number then n^13 - n is always divisible by

a) 13
b) None
c) 15
d) 14

2071
If n is a natural number then n^13 - n is always divisible by

a) 13
b) None
c) 15
d) 14


(b)none of the above
2072

can someone help me in solving this

find the 28383rd term of series: 1234567891011121314....
Ans: 3

2073
can someone help me in solving this

find the 28383rd term of series: 1234567891011121314....
Ans: 3


by counting the number of digits:
1-9 = 9
10-99 = 90*2 = 180
100-999 = 900*3 = 2700
1000-9999 = 9000*4 = 36000


No. of digits till 999 is 2889.
We need the 28383rd digit,
So, we need to find the 28383-2889 = 25494th digit starting from 1000.

Since, we know it will be a 4 digit number, we can find the last number by dividing 25492/4

25492/4 = 6373rd number starting from 1000.Till 7372( 1000+6373-1) we have got 25492+2889 = 28381 digits.
So,Next number will be 7373,
and 28383rd digit will be 7373

so 3 is the answer..
2074

i think this question is more based on what will be the unit digit...

1) 2^3^4^5 is greater than 2^3^5^4, because number with same base ( 2^3 ) and greater power is greater. (leaving 2,3 as an exception 2^3 is smaller than 3^2)
2) for 2^3^4^5
4^5 will have unit digit 4
3^4^5 will have unit digit 1
2^3^4^5 will have unit digit 2

3) for 2^3^5^4
5^4 will have unit digit 5
3^5^4 will have unit digit 3
2^3^4^5 will have unit digit 8

from point 2 and point 3 above ------- unit digit 2-8 = 4

so my guess would be 4

2075

chilli bhai pls help me with these question

1) If the last 2 digit of a four-digit number are interchanged, the new number obtained is greater than the original number by 54. What is the difference between the last two digits of the number?

a) 9
b) 12
c) 6
d) 3
e) Data Inadequate

2) A = 1^1 x 2^2 x 3^3 x 4^4 x ...100^100 . How many zeroes will be there at the end of A?

a) 1300
b) 1320
c) 1325
d) 1050
e) None of these

3) If f(x)=sum of all digits of x, where x is a natural number, then what is the value of f(101) + f(102) + f(103) + ...+f(200)?

a) 1000
b) 901
c) 999
d) 1001
e) 1111

2076

i think the answar is none of this

2077
theairbender Says
(b)none of the above


none of this
2078

1) differnce between last two digits = 6
10c + d= 10d + c - 54
d-c=6

3) f(101) = 1 + 0 + 1

Now from 1 to 99 every digit (0 to 9) comes 20 times, so for unit and tens place digit sum from 1-99 = 20 * 45= 900
Count of hundred digit 1= 99
Hundred digit 2= 2

so, 900+99+2=1001

2079
chilli bhai pls help me with these question

1) If the last 2 digit of a four-digit number are interchanged, the new number obtained is greater than the original number by 54. What is the difference between the last two digits of the number?

a) 9
b) 12
c) 6
d) 3
e) Data Inadequate


abcd be the 4 digit number.
given that abdc - abcd = 54
implies 9(d-c) = 54
difference = 6
2) A = 1^1 x 2^2 x 3^3 x 4^4 x ...100^100 . How many zeroes will be there at the end of A?

a) 1300
b) 1320
c) 1325
d) 1050
e) None of these


Calculating the power of 5 in the whole expression would give the answer -

its 5 (1+2+3+.....+20) in which 5, 10, 15 and 20 gets repeated twice.(Because of 25,50,75,100 each of which contains 5^2)
Hence result-

5 *
= 1300
3) If f(x)=sum of all digits of x, where x is a natural number, then what is the value of f(101) + f(102) + f(103) + ...+f(200)?

a) 1000
b) 901
c) 999
d) 1001
e) 1111


For the range sum goes from 2 to 10
Similarly goes from 2 to 11
goes from 3 to 12
...
goes from 10 to 19

Summing up we get
= 54 + 65 + 75 + 85 + ..... + 145 + 2
= 56 + 5*(13 + 15 + 17 + .... + 29)
= 56 + 945
= 1001
2080
hi friends... plzz help ...

find the last digit of 1^1! + 2^2! + 3^3!.... 88^88!
2081
hi friends... plzz help ...

find the last digit of 1^1! + 2^2! + 3^3!.... 88^88!


Is the answer is 5.
2082
hi friends... plzz help ...

find the last digit of 1^1! + 2^2! + 3^3!.... 88^88!

is it 2???
(1)^1+(2)^2+(3)^6+.....+(8^88!
now apply the cyclicity of the powers to the bases
for all the single digit no's it is easy to find out,for the 2 digit no's take into count only the units digit of bases and raise the power to the units digit
hence soln. will be
(1)^1+(2)^2+(3)^3!+(4)^4!+(5)^5!+(6)^6!+(7)^7!+(^8!+(9)^9!+(0)^10!+(1)11!+....

if wrong thn pls lemme knw d ans...thx
2083
kdn1512 Says
Is the answer is 5.

process please..
2084
hi friends... plzz help ...

find the last digit of 1^1! + 2^2! + 3^3!.... 88^88!


Well, all digits from 1-9 have a cyclicity of 4 or lesser than 4.

Beyond 3!, all other factorials are multiple of 4.
So we can use the cyclicity of all the no.s which have powers >= 4!


For 1^1! + 2^2! + ... + 10^10!
Sum = 1 + 4+ 9 + 6 + 5 + 6 + 1 + 6 + 1 + 0 => Last digit = 9

For 11^11! + 12^12! + ... + 80^80!
Sum = 7*(1+6+1+6 + 5 + 6 + 1 + 6 + 1 + 0 ) => Last digit = 7*33 = 1

For 81!^81! + 82^82! + ... + 88^88!
Sum = (1+6+1+6 + 5 + 6 + 1 + 6 ) => Last digit = 2

So cumulative last digit = 9+1+2 = 2
Ans = 2