[Pri20150303CRT] Three Pencils in the Haystack?

Question

A man had fewer than 500 pencils.  If he packed them into 28 packets, there was a remainder of 3 pencils.  If he packed them into 20 packets, he had 7 pencils left over.  How many pencils had he?

Solution

     Let us “shift” the problem [Heuristic H09].  Imagine if there were 3 pencils less.  Then the remainders would be 0, 4 when divided by 28, 20 respectively.  Noting that 4 is also a common factor of 28 and 20, we divide through by 4 and get a simplified problem [Heuristic H10].  We now look for a number whose remainders are 0, 1 when divided by 7, 5 respectively. [Heuristic H11]

We draw tables represent the problem. [Heuristic H02]

     Obviously, the desired number must be a multiple of 7.  First, let us try 7 [Heuristic H07].  It does not work.   Trying the higher multiples 14, 21, ..., we find that 21 fits the bill.  Now just multiply everything back by 4 and we find that 84 leaves a remainder of 0, 4 when divided by 28, 20 respectively.  This is the number 3 less than what we want.  So we “shift” back, and get 87 as an answer.

     However, the is not the only possible answer.  Note that the LCM of 28 and 20 is 140.  Since 140 is divisible by both 28 and 20, we know that if we keep on adding 140, there will not be any change in the respective remainders.  We do just that, giving us two other possible answers below 500, namely 227 and 367.

Ans: 87, 227 or 367

List of Heuristics Used

H02. Use a diagram / model

H07. Use guess and check

H09. Restate the problem in another way

H10. Simplify the problem

H11. Solve part of the problem

Commentary

     This is another question involving remainders.  It is quite a tough job juggling the requirements when we have different remainders and the divisors have a common factor.  We solved this problem by “shifting” (imagining that there were 3 pencils less) and reducing the problem (via division by 4).  Thus this problem can be solved in a relatively easy manner, without “cheating”by using advanced knowledge and without searching the whole haystack for the proverbial needle, so to speak.

     The person who set this question overlooked the fact that there could be multiple answers.  He/she listed the answer as 367 whereas a pupil who attempted this question got 227 which could be verified to be a correct answer.  Question setters should be cognisant of LCM and multiple answers, and specify a suitable range for the expected answer, if it is meant to be unique.  As “consumers” we should also be critical and creative thinkers not to blindly rely on the author’s “correct answer”.