(Brahmagupta, 7th century A. D.). When eggs in a basket are removed two, three, four, five or six at a time, there remain, respectively, one, two, three, four, or five eggs. When they are taken out seven at a time, none are left over. Find the smallest number of eggs that could have been contained in the basket answer

119 eggs

Step-by-step explanation:

This is a very interesting and old problem. It involves the concept of multiples of an integer number and the remainder of divisions between integers.

Let's start by using the last condition: "When the eggs are taken out seven at a time, none are left over"

It means that the number of eggs N must be a multiple of 7:

N={7,14,21,28,...}

Now we use the fact that when taken 2, 3, 4, 5 or 6 at a time there are remainders, which means the number N cannot be a multiple of any of those numbers. We must find numbers in the sequence

N=7k, such as k is not a multiple of

m={2,3,4,5,6}

The first values for k are:

k={7,11,13,17}

Let's try any of these in order.

For k=7, N=49. The remainders of the divisions N/m, where

m={2,3,4,5,6} are

{1,1,1,4,1}

The remainders don't meet the conditions. Let's try k=11, N=77. The remainders are now

{1,2,1,2,5}

They don't meet the conditions either. Now for k=13, N=91. The remainders are

{1,1,3,1,1}

Let's try for k=17, N=119. The remainders are now exactly as stated in the problem:

{1,2,3,4,5}

Thus the smallest number of eggs contained in the basket is 119

16

step-by-step explanation:

2 x 40 = 80

80/5

= 16

1)is not similar since corresponding sides are not in the same ratio

2)are similar since corresponding sides are in the same ratio and corresponding angles are equal

hope it will

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