# How many chocolates are there?

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A Tough One.

Here’s a way:
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Try first!!.
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We’re finding the smallest multiple of 7 (let’s call it 7k) such that:

7k = 1 (mod 2)
7k = 2 (mod 3)
7k = 3 (mod 4)
7k = 4 (mod 5)
7k = 5 (mod 6)

Begin:
7k – 4 = 0 (mod 5) means last digit is either 0 or 5 implying that:

The last digit for 7k is either 4 or 9. But we’re given 7k = 1 (mod 2) —> means 7k is odd.

Hence, 7k has last digit of 9. Now, 7k with last digit 9 is only possible if k = an integer ending with last digit 7.

Trying k = 7, 17, 27,…
We find the smallest such k satisfying 7k = 5 (mod 6) is
k = 17 since

7 (17) = 119 = 114 + 5 = 19 (6) + 5 = 5 (mod 6)

Next, we check if k= 17 satisfies all of the required congruence as follows:

7 (17) = 59(2) + 1 = 1 (mod 2)

7 (17) = 39(3) + 2 = 2 (mod 3)

7 (17) = 29(4) + 3 = 3 (mod 4)

7 (17) = 23(5) + 4 = 4 (mod 5)

7 (17) = 19(6) + 5 = 5 (mod 6)

Since k = 17 satisfies all the required congruence, the smallest such number is

7k = 7 x 17 = 119 chocolates (Answer).