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Suppose we have an unknown number of objects.
When counted in threes, 2 are left over, when counted in fives, 3 are left over, and when counted in sevens, 2 are left over.
How many objects are there?

 Mar 24, 2020
 #1
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Hey guest! In this problem, we can make use of some number theory.

First we can write the following equivalences:

\(x \equiv 2 \pmod3\)

\(x \equiv 3 \pmod5\)

\(x \equiv 2 \pmod 7\)

Here, let's assume the first two conditions are true. 

We then get:

x = 3k + 2

x = 5j + 3

Let's start with x = 0, 1, 2, 3.....

 

for 3k + 2:

2, 5, 8, 11, 14, 17, 20, 23...... 38 etc.

for 5j + 3:

3, 8, 13, 18, 23, ....... 38, etc.

Do you see what's going on here now?

Every 15, starting from 8, the two sequences share a common element. This is because 15 is the lcm of 3 and 5, which makes sense in the context of this problem. We can then write that x would be of the form:

x = 15w + 8

Now let's look at the third and last equation.

We can write:

x = 7h + 2

Let's do the same sequences pattern. 

for 15w + 8:

8, 23, 38, 53....... 128

for x = 7h + 2:

2, 9, 16, 23, 30, 37, 44, 51.... 128

 

Here, the lcm of 15 and 7 is 105. 

The shared elements are of the form:

This is a combination of all three modulo equations.

105k + 23.

All solutions that hold true for this equation will satisfy the three modulo equations. Assuming that we are trying to find the smallest one(the problem doesn't say what to find exactly), if k =1, we get:

128 as our final answer. 

 Mar 24, 2020
 #2
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Excellent work, jfan17 ! However, the minimum number of objects is 23, since it satisfies all 3 modular equations. This is because your generalized solution of: 105k + 23 applies, because k =0, 1, 2, 3.....etc.

 Mar 24, 2020
 #3
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Oh right, oops on my part! The problem didn't say whether we have to find the smallest one tho, but I'm assuming it does :)

jfan17  Mar 24, 2020

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