[Pri20150306MOD] A Telescope Too Far?

Question

Introduction

     This is a nasty multiple-choice question to be set for a primary 6 (~ grade 6) pupil for a test.  It could also be a time trap as the pupil might spend a lot of time to no avail just to try to score that miserable mark.  With advanced knowledge, we can solve this using telescoping sums (a.k.a. the method of differences), where many terms cancel and the sum can be shortened, much like a telescope.

     However, this solution looks like over-kill.  Is there a solution that is more accesible to a primary 6 pupil?  Read on!

A Simpler Solution?

     We observe that a denominator of  5  is common.  In fact, since every term has an even factor, we have a common factor of 2 in the denominator also.  That means every term in the sum can be expressed as ¹/₁₀ of something [ Heuristic H09 ].  We can quickly work out the first few partial sums and express each of them as  ¹/₁₀ of something.  [ Heuristic H04 ]

     Tabulating the results [ H02 & H03 ], we observe that each partial sum is  ¹/₁₀ of something slightly less than ½.  So we try to express them as ½ minus something.  How?  We can subtract to find out.  [ H05 ]  For example,  if  ⁵/₁₂ = ¹/₂ – ???   then   ??? = ¹/₂ – ⁵/₁₂ = ¹/₁₂.

     Later on, we notice   [ H04 again ] another pattern: the factors in the denominators of the subtracted quantity matches the last two denominators of the last term.  If we considered the full sum, then what would those factors be?  (See the part highlighted in yellow).  These would be 20 and 21, as per the last term of the series.  Using this pattern, we work out the required sum as follows

That is the answer!

Heuristics Used

H02. Use a diagram / model

H03. Make a systematic list

H04. Look for pattern(s)

H05. Work backwards
H09. Restate the problem in another way

[Pri20150303LIH] Lollipop in the Haystack

Question

The number of lollipops in a box is between  60  and  100.  If they are put into packets of 3, there will be 1 lollipop left.  If they are put into packets of 5, there will be 1 lollipop left.  If they are put into packets of 7, there will be no lollipop left.  How many lollipops are there in the box?

Introduction

     This question seems to involve writing lists of numbers and finding the elusive common number.  Some school teacher did just that starting from 3, 4, 5, 6 and 7, when the question already mentions that the number is between  60  and  100.  To add insult to injury, the boy who got this wrong in a test copied this teacher’s “model solution” as corrections!  Sigh!

     Is there a simple way to solve this mathematics problem without rummaging through the entire haystack, as it were?  The answer is: thankfully yes!

Solution

Ans: 91 lollipops

Explanation

First, I draw a table to analyse the remainders when divided by 3, 5 and 7.  [using heuristics H02 and H03].  The question asks for a number with remainders (1, 1, 0).  However, we can SHIFT the problem.  [H09]  If we consider one less than the required number, the remainders are (0, 0, 6).  We look for a number with such a profile.

Having zero remainders under division by 3 and by 5, this number must be a multiple of 15.  Trying 15 [H07], I get  (0, 0, 1) because 15 = 2´14+1.  To get a remainder-profile of (0, 0, 6), I multiply by 6, and I get 15´6 = 90. Now I just SHIFT back 1 to get the required answer!

List of Heuristics Used

H02. Use a diagram / model

H03. Make a systematic list

H07. Use guess and check

H09. Restate the problem in another way

For Your Information

     In solving this question, I did not really use any Chinese Remainder Theorem or advanced mathematics.  I merely used heuristics that can be understood by most people, including the parents helping them and the teachers marking the test scripts.  J

The kids?  Oh!  They will be fine.  They will learn well if we equip them with powerful thinking skills but do not interfere too much.  Shift happens!   J

Related problem here.