Number System - Questions & Discussions

MBA CAT 2022
2085
N is the smallest number such that N is the perfect square..N/3 is perfect cube..then number of divisors of N would be ???
A.12
B.16
C.20
D.24

I have the answer key with me..But i am unable to find the approach..
Please help...
Will post answer after some time as i don't want to misdirect you
2086
N is the smallest number such that N is the perfect square..N/3 is perfect cube..then number of divisors of N would be ???
A.12
B.16
C.20
D.24

I have the answer key with me..But i am unable to find the approach..
Please help...
Will post answer after some time as i don't want to misdirect you

One small mistake N/2 is perfect Square..sorry
2087
(c)20 N = (2^3)*(3^4)
2088
(c)20 N = (2^3)*(3^4)

Please give answer with approach ...
2089
hi5blast Says
Please give answer with approach ...


Answer : 2^3X3^4
Reason : Start with Number 2. Since N/2 is a perfect square. it means that the power of 2 would be odd. Also as the Number N/3 is a perfect cube. Therefore, The power of 2 should be an odd multiple of 3. The least value is 3.
Now as N/3 is a perfect cube. It means the number should have a factor of 3 also. And power of 3 should be of the form 3n+1 as N/3 is a perfect cube. The smallest power is 4.

So, the answer is 2^3X3^4.

Hope its clear..
2090
kdn1512 Says
Is the answer is 5.



no the answer is 2.. but how.. tat is my question
2091
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


yes it is thanks.. bt can u plz xplain hw to calculate the last digits for 7^7! onwards..
2092

it will b 6

2093
satakshigupta Says
it will b 6



no the answer is 2.. but can u plz explain the cyclicity process frm 7 ^7! onwards??
2094
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



can u plz xplain hw this cyclicity works after 7^7!..i mean the process
2095
Answer : 2^3X3^4
Reason : Start with Number 2. Since N/2 is a perfect square. it means that the power of 2 would be odd. Also as the Number N/3 is a perfect cube. Therefore, The power of 2 should be an odd multiple of 3. The least value is 3.
Now as N/3 is a perfect cube. It means the number should have a factor of 3 also. And power of 3 should be of the form 3n+1 as N/3 is a perfect cube. The smallest power is 4.

So, the answer is 2^3X3^4.

Hope its clear..

Thank You very much..Everything is now crystal clear
2096
kriti.v123 Says
yes it is thanks.. bt can u plz xplain hw to calculate the last digits for 7^7! onwards..

kriti.v123 Says
can u plz xplain hw this cyclicity works after 7^7!..i mean the process


cyclicity of numbers are as follows :
(4n+1) (4n+2) (4n+3) (4n)
2: 2 4 8 6
3: 3 9 7 1
4: 4 6 4 6
5: 5 5 5 5
6: 6 6 6 6
7: 7 9 3 1
8: 8 4 2 6
9: 9 1 9 1
where (4n+k) represents the power in mod 4 form(e.g. if the power is 39, then it is in 4n+3 form or 4*9+3 form)

another thing : 4! onwards, every power in factorial is in 4n form
therefore the last digit of 7^7! is 1

hope this clears...
2097
kriti.v123 Says
can u plz xplain hw this cyclicity works after 7^7!..i mean the process


Last digit(simply multiply the last digit with 7..no need to find actual value)

7^1=7 ..(multiply 7*7=49..last digit is 9..so 2nd one will be..
7^2=9 (9*7=63..last digit is 3 ...)
7^3=3 (3*7=21..last digit is 1..)
7^4=1 (1*7=21..last digit is 7..)
7^5=7 (it will again repeat)
-----------cycle of 4----------
hence 7^4k+1=7
7^4k+2=9
7^4k+3=3
7^4k= 1
Hope it is clear..
2098
cyclicity of numbers are as follows :
4n+1 4n+2 4n+3 4n
2: 2 4 8 6
3: 3 9 7 1
4: 4 6 4 6
5: 5 5 5 5
6: 6 6 6 6
7: 7 9 3 1
8: 8 4 2 6
9: 9 1 9 1
where (4n+k) represents the power in mod 4 form(e.g. if the power is 39, then it is in 4n+3 form or 4*9+3 form)

another thing : 4! onwards, every power in factorial is in 4n form
therefore the last digit of 7^7! is 1

hope this clears...



ohk!! tat means suppose taking 7^24 it also b written as 7^4 * 7^4... n eventually the last digit of each term will b 1 .. thank you so much
2099
kriti.v123 Says
can u plz xplain hw this cyclicity works after 7^7!..i mean the process

Same rule applies for other ones too..you need to just see the value of last digit..ignore rest higher digits ..
now you need to mug up the cyclity of digits from 0-9..
2100

Thankuu so much 😃

2101
N is the smallest number such that N/2 is the perfect square..N/3 is perfect cube..then number of divisors of N would be ???
A.12
B.16
C.20
D.24

I have the answer key with me..But i am unable to find the approach..
Please help...
Will post answer after some time as i don't want to misdirect you

Correct Answer is C..
Answer : 2^3X3^4

Reason : Start with Number 2. Since N/2 is a perfect square. it means that the power of 2 would be odd. Also as the Number N/3 is a perfect cube. Therefore, The power of 2 should be an odd multiple of 3. The least value is 3.
Now as N/3 is a perfect cube. It means the number should have a factor of 3 also. And power of 3 should be of the form 3n+1 as N/3 is a perfect cube. The smallest power is 4.

So, the answer is 2^3X3^4.
thanks krti
2102

If n is a natural number such that 1^1! + 2^2! + 3^3! + ...+ n^n! ends with the same unit digit as 1^1! + 2^2! + 3^3! + ...+(n-1)^(n-1)! and n
1) 3
2) 4
3) 6
4) 8
5) 9

2103
If n is a natural number such that 1^1! + 2^2! + 3^3! + ...+ n^n! ends with the same unit digit as 1^1! + 2^2! + 3^3! + ...+(n-1)^(n-1)! and n
1) 3
2) 4
3) 6
4) 8
5) 9

n =1,10,11,20,21,30,31,40,41
so 9 values
correct me if I am wrong...
2104
n =10,11,20,21,30,31,40,41
so 8 values
correct me if I am wrong...


I don't have OA. Its last year aimcat qustion