Number System - Questions & Discussions

first_timer · January 27, 2010, 5:59am

notinuse Says
find the last digit of the lcm of and

The given numbers are consecutive evens, so HCF is 2.
LCM = (3^2003+1)(3^2003-1)/2 = (3^4006-1)/2

Now , the last two digits of 3^4006 is 29.
So last two digits of 3^4006-1 is 28.
So the last digit of (3^4006-1)/2 is 4.

akssps011 · January 27, 2010, 6:03am

LCM = * / 2 --> dividing by 2 becus both are consecutive even nos. hence will have 2 as HCF
to find the last digit:
3^2003 -1 mod100 = 26
3^2003 + 1 mod100 = 28
hence 26*28 / 2 =**364.
hence the last digit shud be 4(if i have not done any mistake)

But my question asks for first digit as 4

suzal001 · February 11, 2010, 4:38pm

hiiii guyz..........
plz hlp me out in dese qus.
Q1.is dis 5 digit no. "abc46" is a prfct square?

Q2.f no. factors f a no. is odd,is it a perfect cube?

options 4 both are yes or no or cant say.

abhinav90 · February 13, 2010, 8:40am

hiiii guyz..........
plz hlp me out in dese qus.
Q1.is dis 5 digit no. "abc46" is a prfct square?

Q2.f no. factors f a no. is odd,is it a perfect cube?

options 4 both are yes or no or cant say.

Plz write in correct language...Question is not understandable..

jaimeen_rana · February 13, 2010, 10:42am

answer 1 )
if the last digit is '6' ; the second last digit must be an odd no if it is a perfect square........
so, ' abc46 ' is surely not a perfect square ......
e.g. 16, 36, 196 , 256 , 576, 676 are perfect squares...........

answer 2 )

If the no. of factors of any number is an odd no. , then that number is definitely a perfect square..........(not perfect cube)

e.g. 4 has total three ( odd no) factors .(1,2,4)
so, 4 is a perf. square.

27 is a perf. cube . it has 1,3,9,27 total 4 factors which is not an odd no.......

i hope my answers are understandeable...........

ask me if u have other querries.....

vivekimmortal · February 13, 2010, 6:18pm

the sum of 20 numbers(may/maynot be distinct)=801. tell minimum possible lcm?

a...36
b...42
c...56
d...60

suzal001 · February 15, 2010, 4:43pm

thanx.........
i undrstud it....
plz hlp me in dese.....

Q1.wat wil b d remaindr of {17^3 +19^3 +21^3 +23^3}/80?
1.10
2.20
3.40
4.60
5.0

Q2.hw many naturl no. below 50 r dere in d cndition satisfyng {n-1}! is not a multiple of n ?
1.15
2.14
3.16
4.17

shanks4mba · February 15, 2010, 4:50pm

thanx.........
i undrstud it....
plz hlp me in dese.....

Q1.wat wil b d remaindr of {17^3 +19^3 +21^3 +23^3}/80?
1.10
2.20
3.40
4.60
5.0

Q2.hw many naturl no. below 50 r dere in d cndition satisfyng {n-1}! is not a multiple of n ?
1.15
2.14
3.16
4.17

1) {17^3 +19^3 +21^3 +23^3}/80
(80)*(17^2 +....23^2} mod80 = 0-->this is of the form (a^3 + b^3) = (a+b)(a^2 - ab + b^2)
option 5

2) (n-1)!
for n = 3,4,5,7,11,13,17,19,23,29,31,37,41,43,47 (all prime nos.) = 15
option 1)

suzal001 · February 16, 2010, 5:01am

1) {17^3 +19^3 +21^3 +23^3}/80
(80)*(17^2 +....23^2} mod80 = 0-->this is of the form (a^3 + b^3) = (a+b)(a^2 - ab + b^2)
option 5

2) (n-1)!
for n = 3,4,5,7,11,13,17,19,23,29,31,37,41,43,47 (all prime nos.) = 15
option 1)

thanx shashank......

bt ans. to Q2. is option no. 3 i.e 16......
which will be d ans. only if v include 1 also as a prime no. too bt again 1 does'nt satisfy d cndition {(n-1)! shud'nt b a multiple f n}

and plz f u cud xplain d Q1. .i m sory bt i m nt familiar wid dis mod concept dat u hv used or direct me to the part f d forum where it is discusd....
plz f u can...........

shanks4mba · February 17, 2010, 9:32am

thanx shashank......

bt ans. to Q2. is option no. 3 i.e 16......
which will be d ans. only if v include 1 also as a prime no. too bt again 1 does'nt satisfy d cndition {(n-1)! shud'nt b a multiple f n}

and plz f u cud xplain d Q1. .i m sory bt i m nt familiar wid dis mod concept dat u hv used or direct me to the part f d forum where it is discusd....
plz f u can...........

well mod is a term that indicates what remainder we get after the division like 7mod2 = 1...
i hope u got it.

pk_gt1 · February 17, 2010, 6:05pm

the sum of 20 numbers(may/maynot be distinct)=801. tell minimum possible lcm?

a...36
b...42
c...56
d...60

ans is 42.. for non distinct numbers

sparkd · March 16, 2010, 9:02am

can someone kindly explain the procedure of solving this

Find the maximum value of 'n' such that 157! is perfectly divisible by 18^n.

sparkd · March 16, 2010, 12:45pm

no need to answer my first question, i got it myself.
it will b nice if sumwan can explain the working of the following sum

find the max value of 'n' such that
77*42*37*57*30*90*70*2400*2402*243*343 is perfectly divisible by 21^n.

kinji.at.pg1 · March 16, 2010, 3:52pm

can someone kindly explain the procedure of solving this

Find the maximum value of 'n' such that 157! is perfectly divisible by 18^n.

I hope the answer is 37 for the above question.

no need to answer my first question, i got it myself.
it will b nice if sumwan can explain the working of the following sum

find the max value of 'n' such that
77*42*37*57*30*90*70*2400*2402*243*343 is perfectly divisible by 21^n.

21 = 3*7

we need to break N = 77*42*37*57*30*90*70*2400*2402*243*343
into factors of 3 and 7.

if we break N (into factors of 3 and 7 ONLY) we get 7(out of 77)*3*7(both out of 42)*3(out of 57)*3(out of 30)*3^2 (out of 90) * 7(out of 70) * 3 (out of 2400) * 3^5 (out of 243) * 7^3(out of 343)

Therefore only considering for 3 and 7 = 3^11 * 7^6. As 7s are least abundant than 3, the max value of n = 6.

Hope it helps.

BTW Try to post the Quant problems in CAT 2010 official Quant Thread for instant response here Official Quant Thread for CAT 2010 - Page 122 - PaGaLGuY.com - The Everything of MBA in India and Abroad, CAT 2009, GMAT, XAT, MAT

This thread is more active. All the best

akrosh007 · March 16, 2010, 4:08pm

can someone kindly explain the procedure of solving this

Find the maximum value of 'n' such that 157! is perfectly divisible by 18^n.

i think the ans shuld be 39..
is it..?
..................................................................................................
never say die attitude..:grin:

akrosh007 · March 16, 2010, 4:17pm

no need to answer my first question, i got it myself.
it will b nice if sumwan can explain the working of the following sum

find the max value of 'n' such that
77*42*37*57*30*90*70*2400*2402*243*343 is perfectly divisible by 21^n.

the ans is 6..
find the total no of 3 and 7..(21=3*7)
and make set of 3 and 7's to get the value of n..
=========================================================
never say die attitude..

abhinav90 · March 16, 2010, 4:24pm

can someone kindly explain the procedure of solving this

Find the maximum value of 'n' such that 157! is perfectly divisible by 18^n.

As 18 = 3^2 * 2

Hence check the number of 3s in 157!...It will come out to b 75

Hence no. of 9 = 75/2...that is 37

Hence n=37...

sparkd · March 17, 2010, 12:33pm

thank ypu all for ur invaluable support,yes the answer for the second question is 6.thank u guys once again ...btw the answer for the first was 37.thank u guys.will post my furthure queries in the thread as given.

harry4u9 · March 17, 2010, 1:11pm

wanted to join in too looks v v interesting i just finished Number system prep

kinji.at.pg1 · March 17, 2010, 4:19pm

harry4u9 Says
wanted to join in too looks v v interesting i just finished Number system prep

Simply post and solve others question. There are quite good problems in http://www.pagalguy.com/discussions/official-quant-thread-for-cat-2010-25047731