Number System - Questions & Discussions

rituraj123 · August 28, 2009, 6:19am

Q1)Let N be the product of five different odd prime numbers. If N is the five-digit number abcab, 4
a. 1
b. 2
c. 3
d. 4
e. more than 4

Q2) My grandfather said he was 84 years old but he was not counting the Sundays. How old my grandfather really was?
a. 96
b. 97
c. 98
d. 64

1..e.More than 4..
Nos will be=3*5*7*11*13=15015
3*17*7*11*13=51051
3*19*7*11*13=57057
3*23*7*11*13=79079
3*29*7*11*13=87087
3*31*7*11*13=93093

vyomconfused · August 28, 2009, 6:42am

1..e.More than 4..
Nos will be=3*5*7*11*13=15015
3*17*7*11*13=51051
3*19*7*11*13=57057
3*23*7*11*13=79079
3*29*7*11*13=87087
3*31*7*11*13=93093

it will be three since condition is mentioned there for a ie 4

rituraj123 · August 28, 2009, 7:17am

vyomconfused Says
it will be three since condition is mentioned there for a ie 4

Yup missed that completely...:banghead::banghead::banghead::banghead:

sharingan · August 28, 2009, 2:49pm

Normal year 52 weeks + 1 day(84/7)
Leap year 52 weeks + 2 days(84/4*7)

/365 = 12 years and 3 days, hence 96 years

kindly let me know, if you have any observation..

The answer should be 98 yrs.

What you people are doing wrong is:

You were calculating the number of Sundays on 84yrs,not in his actual age (which we need to find out).
Actual age - (no. of yrs arising out general Sundays)-(no. of yrs arising out of leap Sundays)=84
u will get 98.

PS: Please refer to the posts above someone has posted soln:
6days-->84 yrs
7days-->98 yrs.

:-P

pk_gt1 · August 28, 2009, 3:35pm

1..e.More than 4..
Nos will be=3*5*7*11*13=15015
3*17*7*11*13=51051
3*19*7*11*13=57057
3*23*7*11*13=79079
3*29*7*11*13=87087
3*31*7*11*13=93093

i don't think both these will be taken as condition is 4

pk_gt1 · August 28, 2009, 3:50pm

divishth Says
Is it 96 years ???

i'm also getting 96 days..
84 *365=30660
less-84*313=26292

4368 days +21 days as leap one=4389
4389/365=12 appx

84+12=96 yrs...

how did ya people getting 98 yrs

plz explain lucidly

anshuldce · August 28, 2009, 4:07pm

i'm also getting 96 days..
84 *365=30660
less-84*313=26292

4368 days +21 days as leap one=4389
4389/365=12 appx

84+12=96 yrs...

how did ya people getting 98 yrs plz explain lucidly

me also gettin 96 same approach as above
kindly explain how 98 is coming m also not gettin....

tac007 · August 29, 2009, 5:05pm

after 5!,,each term will end with 0..so we'll add 1!+2!+3!+4!= 1+2+6+24=33
5789^33/10
(5780+9)^33/10
9^33/10=9 remainder....

How can u be sure that 5!+6!+7!+.....+1000! has TENS DIGIT as zero?

avinav2712 · August 29, 2009, 5:17pm

tac007 Says
How can u be sure that 5!+6!+7!+.....+1000! has TENS DIGIT as zero?

Are you talking of the ten's digit or the unit's one? If you're talking of why the unit's place digit will be 0, then the answer is because by the time you reach 5!, you already get one 2 and one 5 to make a 10, which thus adds up at the end of the number. So, no matter what you multiply to it later, you'll always get a 0 in the unit's place.

jain_ashu · August 29, 2009, 6:08pm

tac007 Says
How can u be sure that 5!+6!+7!+.....+1000! has TENS DIGIT as zero?

If its 1!+2!+...+1000! then the ten's digit is 0.From 10! onwards no number contributes to 10's digit.
If its 5!+6!+7!+.....+1000! then the ten's digit is 8.

fuzzyl0gik · August 29, 2009, 6:08pm

Simple number problems ..

1.find remainder when 36^41+41^36 divided by 77.

2. when 41!^2 divided by 83.

Thanks

divishth · August 29, 2009, 6:29pm

Simple number problems ..

1.find remainder when 36^41+41^36 divided by 77.

Thanks

(36^41)/7*11

(36^41) mod 7 = 1
(36^41) mod 11 = 3

so remainder
7x+1 = 11y+3
x=(11y+2)/7
put y = 3
x = 5

so (36^41) mod 77 = 36

(41^36) mod 7*11

(41^36) mod 7 = 1
(41^36) mod 11 = 3

so from here also remainder = 36
now 2*36 mod 77 = 72..correct me if i am wrong

jain_ashu · August 29, 2009, 6:33pm

Simple number problems ..

1.find remainder when 36^41+41^36 divided by 77.

Thanks

Answer is 72.

jain_ashu · August 29, 2009, 6:34pm

(36^41)/7*11

(36^41) mod 7 = 1
(36^41) mod 11 = 3

so remainder
7x+1 = 11y+3
x=(11y+2)/3
put y = 5
x = 19
so (36^41) mod 77 = 58

(41^36) mod 7*11

(41^36) mod 7 = 1
(41^36) mod 11 = 3

so from here also remainder = 58
now 2*58 mod 77 = 37..correct me if i am wrong

x=(11y+2)/7
Put y = 3.

divishth · August 29, 2009, 6:38pm

x=(11y+2)/7
Put y = 3.

got the mistake b4 u posted...anyways..thanks.....

fuzzyl0gik · August 29, 2009, 6:39pm

(36^41)/7*11

(36^41) mod 7 = 1
(36^41) mod 11 = 3

so remainder
7x+1 = 11y+3
x=(11y+2)/3
put y = 5
x = 19

so (36^41) mod 77 = 58

(41^36) mod 7*11

(41^36) mod 7 = 1
(41^36) mod 11 = 3

so from here also remainder = 58
now 2*58 mod 77 = 37..correct me if i am wrong

doubt on red colored

i got answer 72

correction invited 😃

divishth · August 29, 2009, 6:41pm

doubt on red colored

i got answer 72

correction invited :)

sorry for spamming...corrected my solution already...

:

docsmile · August 29, 2009, 6:42pm

1) For which of the following functions of f is f(x)=f(1-x) for all x?

OA is f(x)=x^2(1-x)^2
Can some1 explain how?

2) If n is positive integer and product of all integers from 1 to n inclusive is multiple of 990 ,then what is the least possible value of n?

divishth · August 29, 2009, 6:44pm

2) If n is positive integer and product of all integers from 1 to n inclusive is multiple of 990 ,then what is the least possible value of n?

is the answer 11 ?

docsmile · August 29, 2009, 6:47pm

divishth Says
is the answer 11 ?

Yes .Can u explain?