Number System - Questions & Discussions

MBA CAT 2022
2271
This one is from testfunda.. Please explain
The integer N is given by N = 2^M 5^386, where M is an integer. If N is known to have 388 digits, how many values can M take?
OPTIONS

1) 1
2) 2
3) 3
4) 4
5) None of these

spectramind07 Says
Option 3???

Yes... explaination????
Thanku

Since N has 388 digits and has 2 and 5 as factors,the 2's and 5's would combine to give zeroes.So N would have 386 zeroes at the end.
Now,remaining digits= 388-386=2
Now,power of 2 which give only 2 digits are 4,5,6(2^4=16,2^5=32,2^6=64)

Hence M takes 3 values.
2272
Since N has 388 digits and has 2 and 5 as factors,the 2's and 5's would combine to give zeroes.So N would have 386 zeroes at the end.
Now,remaining digits= 388-386=2
Now,power of 2 which give only 2 digits are 4,5,6(2^4=16,2^5=32,2^6=64)

Hence M takes 3 values.


Thanku
can you tell me for this
if 2^M*5^5 has 7 digits, then how many values does M can have??
In this case also remaining digits = 2. i find if m=4 den i get 50000
confused
2273
Thanku
can you tell me for this
if 2^M*5^5 has 7 digits, then how many values does M can have??
In this case also remaining digits = 2. i find if m=4 den i get 50000
confused


In this case, there are five 5's. So, we can have five zeroes.
Now, we need two extra digits, which could be contributed by 2.
So, similarly, as in last case, 2^4, 2^5, 2^6 which would each contribute to 2 extra digits.

As per your question, m = 4. But, m cannot be 4.
See, here there are five 5's. So, we need minimum five 2's to get five zeroes.
Apart from that, we need either 2^4, 2^5 or 2^6 zeroes.
Hence, total number of 2's would be 2^9, 2^10 or 2^11, i.e., m = 9,10 or 11

So, answer would be 3 ways.
2274
In this case, there are five 5's. So, we can have five zeroes.
Now, we need two extra digits, which could be contributed by 2.
So, similarly, as in last case, 2^4, 2^5, 2^6 which would each contribute to 2 extra digits.

As per your question, m = 4. But, m cannot be 4.
See, here there are five 5's. So, we need minimum five 2's to get five zeroes.
Apart from that, we need either 2^4, 2^5 or 2^6 zeroes.
Hence, total number of 2's would be 2^9, 2^10 or 2^11, i.e., m = 9,10 or 11

So, answer would be 3 ways.


thanku pkman.............. a litle misunderstanding in understanding the concept.......... Thanks a ton
Spectra bhai thanku............ U jst rock in QA as well
2275
thanku pkman.............. a litle misunderstanding in understanding the concept.......... Thanks a ton
Spectra bhai thanku............ U jst rock in QA as well

I was typing the solution to the second question when power failure occured
2276

A Five digit number is such that it is a Perfect Square and the Units place and Tens place have the same digit. How many such numbers are possible?????

how to get the answer????

2277
A Five digit number is such that it is a Perfect Square and the Units place and Tens place have the same digit. How many such numbers are possible?????

how to get the answer????


One concept.....suppose the sqaure of a number (1x) ends with ac......
Then the square of nx will have last digit as the last digit of x^2 i.e c....but second last digit will be (second last digit of 1x^2+(n-1)*x*2).......

Example.....12^2 ends with 44.......Last digit of squares of all numbers ending with 2 will be 4.....now to determine second last digit.....formula is (second last digit of 1x^2+(n-1)*x*2)....here x=2....

For 22^2 (484) it will be 4+1*2*2=8...
For 32^2 (1024) it will be 4+2*2*2=12....ignore everything other than units digit....so 2 remains.....
For 62^2 (3844) it will be 4+5*2*2=24....i.e 4....

Similarly 17^2=289.....x=7......
37^2 (1369) will have second last digit 8+2*7*2=36...i.e. 6....

Also the second last digit of all squares have a cyclicity of 5....other than those ending with 0 of course.....so 12^2 ends with 44.....(12+50)^2 will end with 44......(12+100)^22 will end with 44 and so on.....

Coming back to the question......

First square of five digits is (100)^2......
Now 32^2=1024......so 320^2=102400.....six digits....and 310^2=96100.....

Now all numbers ending with 0 will satisfy the condition......so multiples of 10 between 100 to 310=22.....

Now if a square ends with an odd number then its tens digit will be even.....so odd numbers are ruled out.....

Now consider even numbers......the squares will end with an even number....as said earlier, second last digit has a cyclicity of 5.....also second last digit of nx is even if and only if second last digit of 1x is even.....only 12 and 18 have their second last digit as odd amond even numbers between 10 and 20......

So the numbers satisfying the condition given in question will end with 2 or 8.....

Case 1: Numbers ending with 2.....12^2 ends will 44....so will 62^2....112^2 and so on till 312^2......here you have to check and you will find out that 312^2 has 5 digits......so number of such numbers=7....but we only have to take numbers which have five digits in squares i.e numbers greater that 100....so 12 and 62 are eliminated......so 5 numbers are there......

Case 3: Numbers ending with 18......18^2 ends with 24.....
Consider the number n8......second last digit will be 4 when last digit of (second last digit of 18^2+(n-1)*8*2) is 4......i.e 2nd last digit of (n-1)*8*2 is 2....n-1=2 satisfies this....so n=3......38^2 will have last two digits as 44........next will be 88^2......then 138^2 till 288^2.......338^2 will have six digits and hence is left out......again numbers below 100 i.e 38 and 88 are eliminated.....so total 4 numbers.....

So total number of numbers satisfying condition given in statement are (22+5+4)=31.....

Please verify the answer......
2278
One concept.....suppose the sqaure of a number (1x) ends with ac......
Then the square of nx will have last digit as the last digit of x^2 i.e c....but second last digit will be (second last digit of 1x+(n-1)*x*2).......

Example.....12^2 ends with 44.......Last digit of squares of all numbers ending with 2 will be 4.....now to determine second last digit.....formula is (second last digit of 1x+(n-1)*x*2)....here x=2....

For 22^2 (484) it will be 4+1*2*2=8...
For 32^2 (1024) it will be 4+2*2*2=12....ignore everything other than units digit....so 2 remains.....
For 62^2 (3844) it will be 4+5*2*2=24....i.e 4....

Similarly 17^2=289.....x=7......
37^2 (1369) will have second last digit 8+2*7*2=36...i.e. 6....

Also the second last digit of all squares have a cyclicity of 5....other than those ending with 0 of course.....so 12^2 ends with 44.....(12+50)^2 will end with 44......(12+100)^22 will end with 44 and so on.....

Coming back to the question......

First square of five digits is (100)^2......
Now 32^2=1024......so 320^2=102400.....six digits....and 310^2=96100.....

Now all numbers ending with 0 will satisfy the condition......so multiples of 10 between 100 to 310=22.....

Now if a square ends with an odd number then its tens digit will be even.....so odd numbers are ruled out.....

Now consider even numbers......the squares will end with an even number....as said earlier, second last digit has a cyclicity of 5.....also second last digit of nx is even if and only if second last digit of 1x is even.....only 12 and 18 have their second last digit as odd amond even numbers between 10 and 20......

So the numbers satisfying the condition given in question will end with 2 or 8.....

Case 1: Numbers ending with 2.....12^2 ends will 44....so will 62^2....112^2 and so on till 312^2......here you have to check and you will find out that 312^2 has 5 digits......so number of such numbers=7....but we only have to take numbers which have five digits in squares i.e numbers greater that 100....so 12 and 62 are eliminated......so 5 numbers are there......

Case 3: Numbers ending with 18......18^2 ends with 24.....
Consider the number n8......second last digit will be 4 when last digit of (second last digit of 18+(n-1)*8*2) is 4......i.e 2nd last digit of (n-1)*8*2 is 2....n-1=2 satisfies this....so n=3......38^2 will have last two digits as 44........next will be 88^2......then 138^2 till 288^2.......338^2 will have six digits and hence is left out......again numbers below 100 i.e 38 and 88 are eliminated.....so total 4 numbers.....

So total number of numbers satisfying condition given in statement are (22+5+4)=31.....

Please verify the answer......

quite a lengthy procedure....i wud hv rather left it :p
thnx neway
2279
quite a lengthy procedure....i wud hv rather left it :p
thnx neway


Is the final answer correct???
2280
3^32 mod 50 = 41
So,bx = 82 => x = 2.

3^32 mod 50 = 41
what does MOD mean here? is there any formula ? can anyone help me ??
2282
nishant_88 Says
Is the final answer correct???


i dunno knw mann...should be right....
2283

Hi nishit...I did not get this part of the concept..
as far as i am concerned it shud have been 9 in place of 8 in second line..

Similarly 17^2=289.....x=7......
37^2 (1369) will have second last digit 8+2*7*2=36...i.e. 6..

2284

Find the reminder
3^32/50

2285
Hi nishit...I did not get this part of the concept..
as far as i am concerned it shud have been 9 in place of 8 in second line..

Similarly 17^2=289.....x=7......
37^2 (1369) will have second last digit 8+2*7*2=36...i.e. 6..


nope it will be 8......its the second last digit of 1x^2....x=7....so 17^2....second last digit is 8.....

Find the reminder
3^32/50


3^5=243....

So we can write it as (243)^6*3^2/50=7^6*9/50=49^3*9/50=-1^3*9/50=-9....so remainder is 41.....
2286
Find the reminder
3^32/50


3^32 = 9^16 = (10 - 1)^16

So last two terms of expansion are:-
-160 + 1

Hence last two digits will be 41

=> 3^32 = 41(mod 50)
2287
3^32 = 9^16 = (10 - 1)^16

So last two terms of expansion are:-
-160 + 1

Hence last two digits will be 41

=> 3^32 = 41(mod 50)

Hi chillfactor I couldn't get the concept behind it,can you elaborate? what last 2 digit has to do with reminder??
2288
Asfakul Says
Hi chillfactor I couldn't get the concept behind it,can you elaborate? what last 2 digit has to do with reminder??


Any number abcd.....ef (last two digits are ef) can be written as:-

abce...00 + ef

Now, abcde...00 is divisible by 50. So to get the remainder by 50 its enough to check the remainder for the last two digits.

So, in our case last two digits were 41, so it will be the remainder.

Had it been 95, then the remainder would be 45.
2289

Find the no of 3 digit number that are divisible by 3 and sum of digit of those nos are not divisible by 7?
Ans :28

2290
Find the no of 3 digit number that are divisible by 3 and sum of digit of those nos are not divisible by 7?
Ans :28


If a number is divisible by 3, then its sum of digits should be divisible by 3.

Now, we have - = 300 three digit numbers divisible by 27.

Out of these, numbers having sum of digits divisible by 7 are those numbers which have their sum of digits as 21.

So, digits of such numbers are:-
(7, 7, 7) - 1 such number
(4, 8, 9) - 6 such numbers
(5, 7, 9) - 6 such numbers
(6, 7, 8 ) - 6 such numbers

So, 281 such numbers.
2291
If a number is divisible by 3, then its sum of digits should be divisible by 3.

Now, we have - = 300 three digit numbers divisible by 27.

Out of these, numbers having sum of digits divisible by 7 are those numbers which have their sum of digits as 21.

So, digits of such numbers are:-
(7, 7, 7) - 1 such number
(4, 8, 9) - 6 such numbers
(5, 7, 9) - 6 such numbers
(6, 7, 8 ) - 6 such numbers

So, 281 such numbers.


Saar what about (3,9,9 ),(5,8,8 )etc.? Why these pairs are ruled out?