Question

If eggs in the basket are removed two at a time, one egg will remain. If the eggs are removed three at a time, 2 eggs will remain. If the eggs are removed 4, 5, or 6 at a time 3, 4, and 5 eggs will remain, respectfully. If they are taken out seven at a time, however, no eggs will be left over. Find the smallest number of eggs that could be in the basket.

Answer 1

Answer:

Smallest number of eggs in the basket = 119

Step-by-step explanation:

In the question,

We know that the number of eggs in the basket should be a a multiple of 7.

But it can not be a multiple of 2, 3, 4, 5 and 6 because every time we pick up the eggs 2, 3, 4, 5 or 6 at a time we are left with some eggs with us.

Therefore, the number of eggs can not be a multiple of these numbers.

Now,

Let us say the number of eggs in the basket be 7x.

So,

Let us take the LCM of 2, 3, 4, 5 and 6.

So,

LCM = 60

Now, the number would be greater than 60 and the multiple of 7.

So, checking on all the multiples of 7 above 60 and checking the condition that, the remainder left on dividing by,

2 is 1. (2x + 1)

by 3 is 2. (3x + 2)

by 4 is 3. (4x + 3)

by 5 is 4. (5x + 4)

by 6 is 5. (6x + 5)

So,

On Checking the multiples of 60 which are divisible by 7 are ,

60 + 1 , 120 + 1, 60 - 1, 120 - 1.

So,

For 61 it is not satisfying all the conditions.

121 is also not satisfying all the conditions.

59 is also not satisfying all the conditions.

But,

The number 119 on checking its divisibility by 2. Leaves remainder 1 as, (2(59) + 1).

Divisibility by 3 leaves remainder 2 as, {3(39) + 2}.

Divisibility by 4 leaves remainder 3 as, {4(29) + 3}.

Divisibility by 5 leaves remainder 4 as, {5(23) + 4}.

Divisibility by 6 leaves remainder 5 as, {6(19) + 5}.

Divisibility by 7 leaves remainder 0 as, {7(17) + 0}.

Therefore, the minimum number of eggs in the basket are 119.

Answer 2

Answer: 119 is the smallest number of eggs in the  basket

Step-by-step explanation:

let k be the number

• If eggs in the basket are removed two at a time, one egg will remain

so k is less than a multiple of 2 by 1 so we have k = 2A - 1 for a positive integer  A

• If the eggs are removed three at a time, 2 eggs will remain

so k is 2 more and thus less with 1, for a multiple of 3, k = 3B - 1, for some positive integer B

• If the eggs are removed 4  at a time,  3 eggs will remain

so k is 3 more and thus less with 1, with a multiple of 4, so k = 4C -1 for some positive integer

• If the eggs are removed 5 at a time,  4 eggs will remain

so k is 4 more and thus less with 1, with a multiple of 5, so k = 5D -1 for some positive integer

• If the eggs are removed 6 at a time,  5 eggs will remain

so k is 5 more and thus less with 1, with a multiple of 6, so k = 6E -1 for some positive integer

• If they are taken out seven at a time, however, no eggs will be left over

thus k is a multiple of 7, so k = 7F for some positive integer F

THUS K = 2A - 1 = 3B - 1 = 4C -1 = 5D -1 = 6E -1 = 7F

ADD 1 to k above gives k + 1 = 2A = 3B = 4C = 5D = 6E = 7F + 1

of k + 1 = 7F + 1 has to be a common multiple of 2, 3, 4, 5, 6. The least common multiple  of 2, 3, 4, 5, 6 is 60

thus k + 1 is a multiple of 60 so k = 60 -1 = 59

again  find the least multiple of 60 that is 1 greater  than a multiple of  60

the first multiple of 60 is 60 and 60 - 1 = 59, but 59 is not a multiple of 7

the second multiple of 60 is 120 and 120 -1 = 119

and 7 is a multiple of 119 because 7 * 17 = 119