Question

Count the \(\LARGE{ ◊ } \)

Answer 1

There are 7 thieves..

They steal diamonds from diamond merchant & run away in the jungle..

While running, Night sets in & they decide to rest in. the jungle..

When everybody's asleep two of them who are closest friends get up & decide to divide the diamonds among themselves & run away..

So they start distributing but they find that one diamond was extra..

So they decide to wake up 3rd one & divide diamonds ägäin.. Only to their surprise they still find one diamond extra..

So they decide to wake up fourth..

Ägäin one diamond is spare..

5th woken up.. Still one extra!!

6th.. Still one extra..

Now they wake up 7th one & diamonds are distributed equally..

How many diamonds in total did they steal??

Answer 2

It's an easy question, why no answer yet?

Answer 3

still 7 huh?

Answer 4

:D

Answer 5

isnt it 7?

Answer 6

7/4, its remainder not 1 but 3

Answer 7

91 = 3 mod4 :)

Answer 8

wt

Answer 9

No,divide 91 by 4 and you get remainder as 3 but you want it to be 1 in all cases

Answer 10

7

Answer 11

like wat else could be it

Answer 12

No^ ...

Answer 13

watt

Answer 14

wat how 91

Answer 15

No. "No,divide 91 by 4 and you get remainder as 3 but you want it to be 1 in all cases "

Answer 16

y divide by 91/4 im confused here

Answer 17

"So they decide to wake up fourth.. "

This is one of the conditions, and from here you have to find the no. of diamonds such that when 4 persons try to divide it equally , they get 1 spare diamond.

That is, let the no. of diamonds be "x" .

Now, x must give remainder as 1 when divided by :

2 , 3 , 4, 5 , 6 .

And finally , remainder = 0 when divided by 7.

If you say, x = 91 then :

91 must give remainder = 1 when divided by 4

91 divided by 4 gives remainder as 3 and quotient as 22 .

Answer 18

That's 50% right. Can you tell me the "Smallest" possible no. of diamonds ?

Answer 19

301

Answer 20

try 301

Answer 21

i think 301 is correct

Answer 22

Can I have logic of this? I realize that 301 is a correct one but don't know how to get it.

Answer 23

my logic is built up from mod2, mod3,....

Answer 24

Pleassse, don't guesssss, but logic.

Answer 25

LCM(1,2,3,4,5,6) = 60
let out required answer be x.
x should be divisible by 7, and also leave remainder 1 when divided by 1,2,3,4,5,6

So clearly, our required answer will be when
60n + 1 is divisible by 7
hit an trial shows n=5 is what we require
hence ans is 301

Answer 26

There are several methods for this quest. :

i) Check the table of 7, and chose those who're odd , and end with "1" ( unit place = 1) :

21,91,161 , 231 , 301 ....

301 is the least no. satisfying the condition .

ii) "Answer should be of the form 60k+1(for only those values of k for which it's divisible by 7) which leads to (56k+4k+1).. Hence 4k+1 should be divisible by 7.. So possible answers can be 301, 721, 1141, 1561 and so on for k=5,12,19.26 and so on." -FB

Using modulo may also work...

Answer 27

Thank you, I know what my mistake is.

Answer 28

It was quite easy quest. ! :)

Answer 29

You're welcome @Loser66 . Getting to know about the mistakes and then improving them in future is good ....

Also, thanks to FB for the quest.

Answer 30

Really, FB is for education too. Depends on the student how he/she uses it.

Answer 31

what does FB mean?

Answer 32

Facebook ;)

Answer 33

:)

Answer 34

crt

Answer 35

crt?

Answer 36

correct, I guess!! :)