Number System - Questions & Discussions

MBA CAT 2022
422
the above bold part is wrong!
cant leave 5 o 25.
though the method will be good wen doin for 35 !


Z(30) = L(4^3)*Z(6)

Z(6) = 2 and L(4^3) = 4
so Z(30)= L(4*2) = 8

so z(35)= z (30) *1*2*3*4*5=8*1*2*3*4*5=6

m i wrong somwhere?

it will be 2, check it
last non zero digit of 35! is 2
30! *31*32*33*34*35 = 2
423
it will be 2, check it
last non zero digit of 35! is 2
30! *31*32*33*34*35 = 2

wat is the last non zero digit of 30!?
424
viveknitw Says
wat is the last non zero digit of 30!?

I think it should be 8
425
it will be 2, check it
last non zero digit of 35! is 2
30! *31*32*33*34*35 = 2

y are u nt taking only the unit digit??...i knw ur ans is rt...but did get y to multiply the whole no...! can u tell me y?
426
shanks4mba Says
y are u nt taking only the unit digit??...i knw ur ans is rt...but did get y to multiply the whole no...! can u tell me y?

ok, whenever we have 5&2 as last digit we'll always have 0 as the last digit.
In order to find the non-zero digit we need the 2nd last digits also, that's why we need full number and not the last digits.
eg:-
34!
last digit of 30! = 8
no need to consider full number, last digits will do jst fine
non zero digits of 34! = 8*1*2*3*4 = 2
for 35!, we'll consider the full numbers.
Hope it's clear now.
427
ok, whenever we have 5&2 as last digit we'll always have 0 as the last digit.
In order to find the non-zero digit we need the 2nd last digits also, that's why we need full number and not the last digits.
eg:-
34!
last digit of 30! = 8
no need to consider full number, last digits will do jst fine
non zero digits of 34! = 8*1*2*3*4 = 2
for 35!, we'll consider the full numbers.
Hope it's clear now.


do u mean to say that last non zero gigit of 34! and 35! will b same?
428
ok, whenever we have 5&2 as last digit we'll always have 0 as the last digit.
In order to find the non-zero digit we need the 2nd last digits also, that's why we need full number and not the last digits.
eg:-
34!
last digit of 30! = 8
no need to consider full number, last digits will do jst fine
non zero digits of 34! = 8*1*2*3*4 = 2
for 35!, we'll consider the full numbers.
Hope it's clear now.


how my friend?
and y?
429
viveknitw Says
wat is the last non zero digit of 30!?

It should be 8... we simply need to multiply 1*3*4*6*7*8*9 n c the last digit of this that comes out to be 8...
Can some1 plzz confirm the answer?
430
It should be 8... we simply need to multiply 1*3*4*6*7*8*9 n c the last digit of this that comes out to be 8...
Can some1 plzz confirm the answer?

yeah the answer is 8
but the real question is wats the last non zero digit of 35!
431
yeah the answer is 8
but the real question is wats the last non zero digit of 35!

It will be 6

8*3*4=last digit 6
432
It will be 6

8*3*4=last digit 6


ok just to clear my doubt y did u leave 2 and 5?


wat will be the last non zero digit of 45! and 44 !
:splat:
433

i will be very happy i anybody gives me the full proof value of non zero digit of-
20!, 30!, 40! and 25! , 35! , 45!

434
i will be very happy i anybody gives me the full proof value of non zero digit of-
20!, 30!, 40! and 25! , 35! , 45!

what do you mean by full proof value ?....what do you exactly want bro...
435

hi vivek ! its really boring if u ask the same type of question again and again.. if u see last two pages , its full of last non zero digit wala sum...

i dunno why is this happening.. i understand u need a proof for the formula fused in this type of prob.. pm some one whom u think can help u clarify ur doubt .. have a chat and clear ur doubt ...

hope u take it in right sense and move on !

good luck!

ps: this is again a kind of spammer

436
i will be very happy i anybody gives me the full proof value of non zero digit of-
20!, 30!, 40! and 25! , 35! , 45!



ur qstn is resolved a few pages back--plz invest few minutes in understanding tht
437
wild007love Says
what do you mean by full proof value ?....what do you exactly want bro...



sorry yaar.
actually i want the real values ( 4m ppl who r 100 percent sure of it):splat:
438

There are 8436 steel balls, each with a radius of 1 cm, stacked in a pile, with 1 ball on top, 3 balls in the second layer, 6 balls in the third layer, 10 in the fourth, and so on. The number of horizontal layers in the pile is :

a) 34 b) 38 c) 32 d) 36 e) 40

439
There are 8436 steel balls, each with a radius of 1 cm, stacked in a pile, with 1 ball on top, 3 balls in the second layer, 6 balls in the third layer, 10 in the fourth, and so on. The number of horizontal layers in the pile is :

a) 34 b) 38 c) 32 d) 36 e) 40


answer is 36..i think....

Layer 1 = 1 ball
Layer 2 = 1+2 balls
Layer 3 = 1+2+3 balls
Layer 4 = 1+2+3+4...so on...

Layer n = 1+2+3+4+...+(n-1)+n.

so n*1 + (n-1)*2 + (n-2)*2 + .... + (n-1)*2 + n = 8436....

so n = 36..
440
There are 8436 steel balls, each with a radius of 1 cm, stacked in a pile, with 1 ball on top, 3 balls in the second layer, 6 balls in the third layer, 10 in the fourth, and so on. The number of horizontal layers in the pile is :

a) 34 b) 38 c) 32 d) 36 e) 40


tn term will be n(n+1)/2
sum of tn terms will be
n(n+1)(n+2)/6 = 8436

solving n=36
441
There are 8436 steel balls, each with a radius of 1 cm, stacked in a pile, with 1 ball on top, 3 balls in the second layer, 6 balls in the third layer, 10 in the fourth, and so on. The number of horizontal layers in the pile is :

a) 34 b) 38 c) 32 d) 36 e) 40

The answer is 36.