@saurav205 said:Numbers are 84 and 189
@saurav205 said:bhai question theek se padho...you cannot include the number itself....
han bhai wo typo tha...
3*3*(7*3) and 2*(7*3*2)...
@amresh_maverick said:!> No of zeroes at the end of 25! +26! + 27! + 28! + 30!2> No of zeroes at the end of 27!^27!
For the first one is it 6??
@amresh_maverick said:!> No of zeroes at the end of 25! +26! + 27! + 28! + 30!2> No of zeroes at the end of 27!^27!
1 > 6 zeroes
2 > 6^27! zeroes ?
@amresh_maverick said:!> No of zeroes at the end of 25! +26! + 27! + 28! + 30!2> No of zeroes at the end of 27!^27!
For the first one is it 6??
@amresh_maverick said:!> No of zeroes at the end of 25! +26! + 27! + 28! + 30!2> No of zeroes at the end of 27!^27!
2 one ka toh (10^6)^27! aa raha hai....
@amresh_maverick said:!> No of zeroes at the end of 25! +26! + 27! + 28! + 30!2> No of zeroes at the end of 27!^27!
1) 25!(1+26+27*26+28*27*26+30*29*28*27*26)
2^22*5^6(7+2+6+0)
2^22*5^6(5^1)
hence, 7 zeroes....
2) is it 6^27!??
@saurav205 said:OAs:70Option 4121.07(sorry this question was already discussed earlier and also the options were wrong)2
Virender sehwag wala detail sol post karodo , plz
@amresh_maverick said:!> No of zeroes at the end of 25! +26! + 27! + 28! + 30!2> No of zeroes at the end of 27!^27!
OA : 1> 7 zeroes
2> 6^27!
2> 6^27!
@Logrhythm said:1) 25!(1+26+27*26+28*27*26+30*29*28*27*26)2^22*5^6(7+2+6+0)2^22*5^6(5^1)hence, 7 zeroes....2) is it 6^27!??
Are andar wala 5 nai ara bhai last me, 9 ara hai You missed 29! wala term!
@ScareCrow28 said:Are andar wala 5 nai ara bhai last me, 9 ara hai You missed 29! wala term!
bhai 29! hai hi nahi... 😛
No of zeroes at the end of 25! +26! + 27! + 28! + 30!
To understand this, let us understand the basic idea first
What will be the number of 0s at the end of a + b + c would depend upon the least number of 0s that any one of a or b or c has.
For eg: 300 + 120000 + 17272730 will end in 1 zero
But, if they have the same number of zeroes, we will also have to consider the last non-zero digit.
For eg: 12000 + 161237000 + 1212331000 will not end in 3 zeroes but in 4 zeroes because the last non-zero digits 2, 7 and 1 will add up to generate an extra zero.
If you understood the above part, read on.
We first need to figure out how many zeroes do the factorials individually have
Number of zeroes is given by the sum of the quotients obtained by successive division of n by 5.
Among the ones mentioned,
25!, 26!, 27! and 28! have 6 zeroes each.
30! has 7 zeroes.
We also need to consider the right most digits of 25!, 26!, 27! and 28!
R(n!) = Last Digit of [ 2^a x R(a!) x R(b!) ]
where n = 5a + b
where n = 5a + b
Using this we get, right most non zero digits of 25! as 4
=> 26! will end in 4*6 or 4
=> 27! will end in 4*7 or 8
=> 28! will end in 8*8 or 4
=> 25! + 26! + 27! + 28! will not end in 6 zeroes but in 7 zeroes
We know that 30! ends in 7 zeroes.
So, the overall number 25! + 26! + 27! + 28! + 30! would end in 7 zeroes
You can read more about these ideas in an old PG post of mine: http://www.pagalguy.com/news/ive-got-power-working-with-factorials-cat-2011-quant-a-17445
Footnote: Let me add that this is way too complicated to be asked in CAT. Even in the unlikely scenario that this does get asked, you should avoid it. This is probably something that a teacher would use to impress students or a coaching institute would include in its test-series / question-a-day / material to scare students into buying their products.
Note to self: Start making crazy questions to get more students to join online courses.
Note to self: Start making crazy questions to get more students to join online courses.
@Logrhythm said:bhai 29! hai hi nahi...
Galat baat hai ye! 29! islie chhoda ki kuch laundo ka galat hojaye! @amresh_maverick said:!> No of zeroes at the end of 25! +26! + 27! + 28! + 30!2> No of zeroes at the end of 27!^27!
1) 7 zeroes at the end
@ScareCrow28 said:No of 5s in 27! = 6So It becomes 6^27!
oh thora ghabra gaya the question ko dekhke
anyways thanx
anyways thanx
1>How many perfect squares are the divisors of the product 1!*2!*3!.....8!
2> Last two non-zero digits of 36! - 24!
@amresh_maverick said:2> Last two non-zero digits of 36! - 24!
8
24! has last digit as 2
and 36! will have a zero above 2
when subtracted gives us 8
@amresh_maverick
24! has last digit as 2
and 36! will have a zero above 2
when subtracted gives us 8
@amresh_maverick
Find the number of divisors of 1080 excluding the throughout divisors, which are perfect squares... ??