Simple number problems ..

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

Thanks

Answer is 72.

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

Simple number problems ..

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

2. when 41!^2 divided by 83.


Thanks

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.

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.

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?

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....

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

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<8

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

vyomconfused Says

it will be three since condition is mentioned there for a ie 4 <8 and only 1st , 2nd and 3rd satisfies the condition

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

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 <8 and only 1st , 2nd and 3rd satisfies the condition

Q1)Let N be the product of five different odd prime numbers. If N is the five-digit number abcab, 4 < a < 8, how many values of N are possible?
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

guys hav a question here: Progressions.

If a times ath term of an AP equals b times bth term. find the (a+b)th term.

- What i wanna ask is, is there a non-algebraic way of cracking this? A short-cut... u go thru the formulae, things work out fine.. but takes a lil time...

Thanks much!

TAKE An A.P = 2 1 0 -1...
let a = 1 and b = 2
the case is satisfied...so a+bth term is the 3rd term which is 0:)

Another approach...
given a(x + (a-1)d) = b(x + (b-1)d)
take a = 1 and b = 2 then
x = 2x + 2d, assume d = -1
then x = - 2d = 2
so u get A.P = 2,1 0 -1... hence the third term is 0:)

PS-always try to assume values when it comes to Progressions, thn u can find out the ans quickly. most of the time this works:). Same wit Time and work too:)

Q1)Let N be the product of five different odd prime numbers. If N is the five-digit number abcab, 4 < a < 8, how many values of N are possible?
a. 1
b. 2
c. 3
d. 4
e. more than 4

1)N = abcab where 4 < a < 8
N has to be in the form 1001* P1*P2
where 1001 = 7*11*13
N = 1001*P1*P2
as 4 < a < 8
so we have set p1,p2 = (3*17) , (3*19) and (3*23)
hence total 3 values of N are possible
option c

guys hav a question here: Progressions.

If a times ath term of an AP equals b times bth term. find the (a+b)th term.

- What i wanna ask is, is there a non-algebraic way of cracking this? A short-cut... u go thru the formulae, things work out fine.. but takes a lil time...

Thanks much!

Cool thread..!!

I will posting my doubts and any new thing that i will be coing across..!!

Lets contribute.

divishth Says

Is it 96 years ???

jain_ashu Says

Don't have the OA but I am too getting the same answer. But would like to know your approach especially regarding leap year part.

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..

Q1)Let N be the product of five different odd prime numbers. If N is the five-digit number abcab, 4 < a < 8, how many values of N are possible?
a. 1
b. 2
c. 3
d. 4
e. more than 4

option c) 3
abcab
there is a pattern there
3*7*11*13 = 3003
3003*17 = 51051
3003*19 = 57057
3003*23 = 69069

as 4<8, we stop here

for grandfather 7 days=6 days..

so no. of yrs=84*7/6=98

In 84 years he is missing out almost 84*52 Sundays plus max. 21 for leap years. Which is equal to 4368+21 = 4389 = approx. 12 years.
Where are we going wrong???

find the remainder when (38^16!)^1777 is divided by 17

a. 1
b. 16
c. 8
d. 13

Is the ans 1?
i think we can do it by Euler's..plz comment

find the remainder when (38^16!)^1777 is divided by 17

a. 1
b. 16
c. 8
d. 13

38^16! mod 17 = 4^16! mod 17 = (4^2)^even no mod 17 = (-1)^even no mod 17 = 1
thus 1^ 17777 mod 17 = 1

jain_ashu Says

Don't have the OA but I am too getting the same answer. But would like to know your approach especially regarding leap year part.

for grandfather 7 days=6 days..

so no. of yrs=84*7/6=98

divishth Says

Is it 96 years ???

Don't have the OA but I am too getting the same answer. But would like to know your approach especially regarding leap year part.

Q1)
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

Is it 96 years ???

Q1)Let N be the product of five different odd prime numbers. If N is the five-digit number abcab, 4 < a < 8, how many values of N are possible?
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

yes the ans is one

will try to be regular from now onwards..(in this thrd)

okay one qstn from my side

5789^(1!+2!+3!-------+1000!)has the unit digit of ?

after 5!,,each term will end with 0..so we'll add 1!+2!+3!+4!= 1+2+6+24=33

and last two digits of 5!+6!-----9!=80..adding 80+33=13..

5789^13/10
(5780+9)^13/10
9^13/10=9 unit digit

find the remainder when (38^16!)^1777 is divided by 17

a. 1
b. 16
c. 8
d. 13

ans:1

(38^16!)^1777=(34+4)^16!^1777/17

4^16!^1777/17

16! is divisbile by 4 so (4^2)4k^1777/17

(16+1)^4k^1777/17..

so 1 remainder.

this is fermat little theorem
this can be seen rem(M^(N-1)/N) = 1 where N is a prime number
so the answer is 1*1*...............1 =

sorry for the wrong explanation
I studied post wrong
will solve and get back 2 u

find the remainder when (38^16!)^1777 is divided by 17

a. 1
b. 16
c. 8
d. 13

(38^16!)^1777mod 17 is 1
option a)

yes the ans is one

will try to be regular from now onwards..(in this thrd)

okay one qstn from my side

5789^(1!+2!+3!-------+1000!)has the unit digit of ?

unit digit is 9 as the power is odd ( coz except 1! all other factorials are even)

yes the ans is one

will try to be regular from now onwards..(in this thrd)

okay one qstn from my side

5789^(1!+2!+3!-------+1000!)has the unit digit of ?

last digit is 9 as power is odd.

yes the ans is one



5789^(1!+2!+3!-------+1000!)has the unit digit of ?

1! +2!+3!+4! = 33
so, 9^33= 9
hence, answer is 9

Let's start things afresh
Q
If n is an odd integer then the last two digits in base ten of 2^2n * (2^(2n+1) -1) are
I. 18 II. 28 c. 38
a. Only I
b. Only I & II
c. Only II
d. Only I, II & III
e. None of the above

we can write expression as 2^(4n+1)-2^2n
Last digits of 2^(4n+1) are in cycle 32,92,52,12,72
Last digits of 2^2n are in cycle 04,64,24,84,44
Last 2 digits are always 28

yes the ans is one

will try to be regular from now onwards..(in this thrd)

okay one qstn from my side

5789^(1!+2!+3!-------+1000!)has the unit digit of ?

1+2+6+24+20+20+40+20+80 = xx13
5789^xx13
unit digit is 9

yes the ans is one

will try to be regular from now onwards..(in this thrd)

okay one qstn from my side

5789^(1!+2!+3!-------+1000!)has the unit digit of ?

the answer is 9

as 1!+ 2! + 3! +4! ends in 3

9^odd gives 9 as unit digit

yes the ans is one

will try to be regular from now onwards..(in this thrd)

okay one qstn from my side

5789^(1!+2!+3!-------+1000!)has the unit digit of ?

Hi Pk,

As the cyclic form will give only two possibility as 9 or 1 as last digit depending on the power (if odd then 9, if even then 1)

So let us work on power. After the 5! all the sum will have last digit as 0, so we have to work only upto 4!
1!=1, 2!=2, 3!=6, 4!=24 sum of all will give 33. So we are concerned only with 3 as last digit of power which is ODD. So the answer will be 9.
Please confirm

yes the ans is one

will try to be regular from now onwards..(in this thrd)

okay one qstn from my side

5789^(1!+2!+3!-------+1000!)has the unit digit of ?

find the remainder when (38^16!)^1777 is divided by 17

a. 1
b. 16
c. 8
d. 13

Answer should be 1.
E(17) = 16, (38^16!)^1777 mod 17 = 38^16k mod 17 = 1

find the remainder when (38^16!)^1777 is divided by 17

a. 1
b. 16
c. 8
d. 13

E(17) = 16
(38^16!)^1777 mod 17=
38^16! mod 17 = 1
remainder = 1

janvats Says

is t answer c only II?

naga25french Says

c) only II

Answer is 28 only but prove it guys.

find the remainder when (38^16!)^1777 is divided by 17

a. 1
b. 16
c. 8
d. 13

option II only

Let's start things afresh
Q
If n is an odd integer then the last two digits in base ten of 2^2n * (2^(2n+1) -1) are
I. 18 II. 28 c. 38
a. Only I
b. Only I & II
c. Only II
d. Only I, II & III
e. None of the above

take n = 1
then we have 2^2n * (2^(2n+1) -1) = 4*(7) = 28
take n = 3
then we have 64*127 = 28
basically its even*odd = (x4)*(y7) = x28
shud be opption C only II

Let's start things afresh
Q
If n is an odd integer then the last two digits in base ten of 2^2n * (2^(2n+1) -1) are
I. 18 II. 28 c. 38
a. Only I
b. Only I & II
c. Only II
d. Only I, II & III
e. None of the above

answer is c only i.e 28

Let's start things afresh
Q
If n is an odd integer then the last two digits in base ten of 2^2n * (2^(2n+1) -1) are
I. 18 II. 28 c. 38
a. Only I
b. Only I & II
c. Only II
d. Only I, II & III
e. None of the above

c) only II

Let's start things afresh
Q
If n is an odd integer then the last two digits in base ten of 2^2n * (2^(2n+1) -1) are
I. 18 II. 28 c. 38
a. Only I
b. Only I & II
c. Only II
d. Only I, II & III
e. None of the above

Is the answer option c) Only II ??
And the answer to the question you had earlier posted is Rs.50/-

Let's start things afresh
Q
If n is an odd integer then the last two digits in base ten of 2^2n * (2^(2n+1) -1) are
I. 18 II. 28 c. 38
a. Only I
b. Only I & II
c. Only II
d. Only I, II & III
e. None of the above

is t answer c only II?

1. 3^32/50 give remainder and {.}denotes the fractional part.the fractional part is of the form(0.bx) value of x is??

3^32 mod 50 = 41
So,bx = 82 => x = 2.

Let's start things afresh
Q
If n is an odd integer then the last two digits in base ten of 2^2n * (2^(2n+1) -1) are
I. 18 II. 28 c. 38
a. Only I
b. Only I & II
c. Only II
d. Only I, II & III
e. None of the above