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
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”.
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