My first logic problem

Number Theory Level 3

I have a pack of pens to distribute. If I keep 4, 5 or 6 in a pack, I am left with 3 pens. If I keep 7 in a pack, I am left with none. What is the minimum number of pens I have to pack and distribute?


The answer is 63.

Raven Herd by Raven Herd , 21, India

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2 solutions

Sravanth C.
Jun 27, 2015

Let's consider the total nuber of pens to be x x .

According to the question, we have the following conditions: x ≡ 3 ( m o d 4 ) x ≡ 3 ( m o d 5 ) x ≡ 3 ( m o d 6 ) x ≡ 0 ( m o d 7 ) x\equiv 3\pmod{4} \\ x\equiv 3\pmod{5} \\ x\equiv 3\pmod{6} \\ x\equiv 0\pmod{7}

Simplifying it we get, x − 3 ≡ 0 ( m o d 4 ) x − 3 ≡ 0 ( m o d 5 ) x − 3 ≡ 0 ( m o d 6 ) x ≡ 0 ( m o d 7 ) x-3\equiv 0\pmod{4} \\ x-3 \equiv 0\pmod{5}\\ x-3\equiv 0\pmod{6} \\ x\equiv 0\pmod{7}

Now, by trial and error, we find that 63 \boxed{63} is the least number satisfying all the conditions above.

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Moderator note:

Is there a better way than "trial and error"?

Look up the Chinese Remainder Theorem .

Chitty Gadiparthi
Aug 19, 2015

I found it a different way. I found the LCM of 4,5,6, which is 60. Then I added three. The result is 63, which is divisible by 7.

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