The least number of eggs that could be in the basket is **119.**

**Multiple of the number:**

The **multiple **is the numbers you get when you multiply a certain number by an integer.

Given,

If the eggs in a basket are** removed 2** at a time, **1 egg** will remain.

If the eggs are **removed 3 at a time, 2 eggs** will remain.

If the eggs are removed **4, 5, or 6 **at a time, then **3, 4, and 5** eggs will remain, respectfully.

But if they are taken out** 7 **at a time, **no eggs** will be left over.

Here we need to find the **least number of eggs** in the basket.

Let the number of eggs in the basket be N.

We know that,

if the eggs in a basket are removed 2 at a time, one eggs will remain.

So N is 1 less than a multiple of 2, so

=> N=2A-1

If they are removed 3 at a time, 2 eggs remain.

So N is 2 more, and therefore 1 less than, a multiple of 3, so

=> N=3B-1

If the eggs are removed 4...at a time, then 3...eggs remain,

So N is 3 more, and therefore 1 less than, a multiple of 4, so

=> N=4C-1

Similarly,

For 5,

=> N = 5D - 1

For 6,

=> N = 6E - 1

So, if they are taken out 7 at a time, no eggs will be left over.

So,

N => 7F

Where A,B,C,D,E, and F are any **positive integer**.

Therefore,

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

Add 1 to all those:

N+1 = 2A = 3B = 4C = 5D = 6E = 7F+1

Through this,

N + 1 = 7F + 1

has to be a **common multiple** of 2,3,4,5,6.

So, the **LCM **of 2,3,4,5,6 is 60.

So N+1 is a multiple of **60 **which means N is 1 **less than** a multiple of 60.

So we find the least multiple of 60 that is 1 more than a multiple

of 60.

The multiples are,** 60, 59, ....**but 59 is **not a multiple **of 7.

The next multiple of 60 is **120**.

1 less than 120 is **119**, and sure enough, 119 is a **multiple **of 7.

So,

=> 119 = 17*7.

Therefore, the least number of eggs that could be in the basket is **119.**

To know more about **Multiple **here

brainly.com/question/5992872

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