farmer's egg problem

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farmer's egg problem

A farmer took a basket of eggs to sell, a horse rider hit the basket and broke all eggs, then he told the farmer how many eggs you had in the basket, I will pay for them. The farmer said: I don't know exactly, but when I was taking the eggs out of my basket two by two, the last one was one egg, when I took them 3 at a time, again one was left, same thing happened when I took them out 4 at a time, 5 at a time, 6 at a time, but when I took them out 7 at a time, nothing was left in the basket.
Can you tell the minimum number of eggs in the basket?

Posted 235 day ago

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Answers (2)

teacher1320

1st time took out 2. 2 & 2 & 2 =6 One eggs still remained.
2nd time took out 3 3 & 3= 6 One egg left.
you took out 7 eggs at one time. no eggs left. So you answer would be 7.

Posted 224 day ago

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math_fanatic

The answer is 301.

The way I solved this problem was by finding a common multiple of 2, 3, 4, 5, 6, and adding one until I found a number that was divisible by 7.

The least common multiple of 2, 3, 4, 5, 6 has a prime factorization of 2*2*3*5 = 60. 61 is not divisible by 7 however so this is not the answer.

The next number I got by multiplying the LCM by another prime factor of 2 since it's the smallest prime:
2*2*2*3*5 = 120. 121 is not divisible by 7

Next I tried multiplying the LCM by another prime factor of 3 since it's the next smallest prime after 2:
2*2*3*3*5 = 180. 181 is not divisible by 7

Next I tried multiplying the LCM by another prime factor of 5 since it's the next smallest prime after 3:
2*2*3*5*5 = 300. 301 IS divisible by 7

To make sure this is indeed the smallest number I multiplied the LCM by a fourth prime factor of 2:
2*2*2*2*3*5 = 240. 241 is not divisible by 7 and if I multiply by another prime factor of 2 the result will be greater than 301 so I can now rule out multiplying by any more prime factors of 2.

If I were to multiply the LCM by a third prime factor of 3 I would get 2*2*3*3*3*5 = 540 which is greater than 301 so I need go no further down this road.

The only other possibility left would be to multiply the LCM by one prime factor of 2 and one prime factor of 3 however this would be the same as multiplying the LCM by 6 which will give a product greater than if I had multiplied it by a single prime factor of 5. Therefore, I can rule out this result without doing the calculation.

Thus, the minimum number of eggs in the basket is 301.

Posted 235 day ago

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