Math, asked by Angelene7691, 10 months ago

There are 7 thieves. They steal diamonds from diamond merchant and run away in jungle. While running, night sets in and they decide to rest in jungle. When everybody's sleeping, two of them who are closest friends get up and decide to divide the diamonds among them and run away. So they start distributing but they found that one diamond was extra. So they decide to wake up 3rd one and divide diamonds again. Only to their surprise they still find one diamond extra. So they decide to wake up fourth. Again one diamond is spare.5th woken up. Still one extra!!6th.Still one extra. Now they wake up 7th one and diamonds are distributed equally. How many minimum diamonds they steal? *

Answers

Answered by BrainlyTornado

ANSWER:

The minimum diamonds they steal = 301

ASSUMPTION:

Let us take the number of diamonds as x

GIVEN:

If x/2 then Remainder = 1

If x/3 then Remainder = 1

If x/4 then Remainder = 1

If x/5 then Remainder = 1

If x/6 then Remainder = 1

If x/7 then Remainder = 0

TO FIND:

Minimum diamonds they steal

EXPLANATION:

The number should be divisible by 2, 3, 4, 5, 6

Take L.C.M

We will get 60

Add + 1 to 60 and the result will be 61

But 61 is not divisible by 61

Try by Trial or error method

(2 × 60) + 1 = 121 ( Not divisible by 7)

(3 × 60) + 1 = 181 ( Not divisible by 7)

(4 × 60) + 1 = 241 ( Not divisible by 7)

(5 × 60) + 1 = 301 ( divisible by 7)

Hence the minimum diamonds they steal = 301.

VERIFICATION:

301/2 = 150 and Remainder = 1

301/3 = 100 and Remainder = 1

301/4 = 75 and Remainder = 1

301/5 = 60 and Remainder = 1

301/6 = 50 and Remainder = 1

301/7 = 43 and Remainder = 0

HENCE VERIFIED.

Answered by poojan

The total diamonds stolen by the thieves are 301 in number

Explanation:

There were seven thieves in all.

Suppose they stole x diamonds in total.

It says that if 7 of them distribute the diamonds among them equally, no diamond will be left. i.e., x is exactly divisible by 7.

If 2 thieves share the diamonds between them, one diamond will be left.

i.e, x = 1 + T 1 * 2 --------------- 1

If 3 thieves share the diamonds among them, one diamond will be left.

i.e., x = 1 + T 2 * 3 --------------- 2

If 4 thieves share the diamonds among them, one diamond will be left.

i.e., x = 1 + T 3 * 4 --------------- 3

If 5 thieves share the diamonds among them, one diamond will be left.

i.e., x = 1 + T 4 * 5 --------------- 4

If 6 thieves share the diamonds among them, one diamond will be left.

i.e., x = 1 + T 5 * 6 --------------- 5

If all 7 thieves distribute the diamonds among them, no diamond will be left. i.e, x = T 6 * 7 --------------- 6

If we see the first 5 equations, we can observe that in all the situations only 1 diamond left.

So, we can say that T1*2 = T2*3 = T3*4 = T4*5 = T5*6 = y

From this, we get to know that 2, 3, 4, 5, 6 are the common multiples of y.

The least common factor (LCM) here is : LCM(2,3,4,5,6) = 60

As we know that one diamond will be left,

The answer would be the common multiple + 1 and it should be divisible by 7.

60 + 1 = 61 => No, because it is not divisible by 7.

60*2 + 1 = 121 => Not divisible by 7.

60*3 + 1 = 181 => Not divisible by 7

60*4 + 1 = 241 => Not divisible by 7

60*5 + 1 = 301 -> DIVISIBLE!

So, the diamonds that are stolen by the 7 thieves are 301 in number.

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