Test of Mathematics Solution Subjective 36 - Invariance Principle

Test of Mathematics Solution Subjective 36 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

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Problem

Let $ a_1 , a_2 , ... , a_n $ be n numbers such that each $ a_i $ is either 1 or -1. If $ a_1 a_2 a_3 a_4 + a_2 a_3 a_4 a_5 + ... + a_n a_1 a_2 a_3 = 0 $ then prove that 4 divides n.

Solution

Let us denote each product of four numbers as $ b_i $ . For example $ b_1 = a_1 a_2 a_3 a_4 $. Note that the last one begins with $ a_n $ indicating there are n $ b_i $'s.

We check the equation modulo 4 (that is in each step we change 'something' in the equation and check what happens to the remainder of the sum when divided by 4).In the first step we note that equation is 0 mod 4 (since 0 when divided by 4 gives 0 as the remainder).

Next we convert one of the a's which is -1 into +1. Note that each $ a_i $ appears in exactly 4 $ b_j $. Hence converting the one $ a_i $ from -1 to +1 will alter the values of 4 $ b_j $ Thus we may have five cases:

All the four b's were -1. Hence after conversion all of their values will change into +1. Therefore the value of the total sum will increase by +8 (as previously the four b's added up to -4 and now they add up to +4). Thus no change of the sum modulo 4.
Exactly three b's were -1 and the remaining was +1. Hence after conversion three of their values will change into +1 and the remaining one changes from +1 to -1. Therefore the value of the total sum will increase by +4 (as previously the four b's added up to -2 ((-1) + (-1) + (-1) + (+1)) and now they add up to +2). Thus no change of the whole sum modulo four.
Exactly two b's were -1 and the remaining two were +1. Hence after conversion two of their values will change into +1 and the remaining two will change from +1 to -1. Therefore the value of the total sum will increase by 0 (as previously the four b's added up to 0 ((-1) + (-1) + (+1) + (+1)) and now they add up to 0). Thus no change of the whole sum modulo four.
Exactly one b was -1 and the remaining were +1. Hence after conversion three of their values will change into -1 and the remaining one changes from -1 to +1. Therefore the value of the total sum will decrease by 4 (as previously the four b's added up to 2 ((+1) + (+1) + (+1) + (-1)) and now they add up to -2). Thus no change of the whole sum modulo four.
All the four b's were +1. Hence after conversion all of their values will change into -1. Therefore the value of the total sum will decrease by 8 (as previously the four b's added up to +4 and now they add up to -4). Thus no change of the sum modulo 4.

Thus we see that changing one $ a_i $ -1 to +1 does not change the value of the total sum modulo 4. Similarly we change all negative one's into positive one's (and in each step the sum remains invariant in modulo 4) to get n. Thus $ n \equiv 0 \mod 4 $ . This implies 4 divides n.

Key Idea

Modular arithmetic, invariance principle.

Numbers on a blackboard

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Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.22.4" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px"]Azambuja writes a rational number $q$ on a blackboard. One operation is to delete $q$ and replace it by $q+1$ ; or by $q-1$ ; or by $\frac{q-1}{2q-1}$ if $q \neq \frac{1}{2}$ . The final goal of Azambuja is to write the number $\frac{1}{2018}$ after performing a finite number of operations. Show that if the initial number written is $0$ , then Azambuja cannot reach his goal.

[/et_pb_text][et_pb_text _builder_version="3.22.4" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px"][/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.22.4"][et_pb_column type="4_4" _builder_version="3.22.4"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.23.3" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="3.23.3" title_text_shadow_horizontal_length="0em" title_text_shadow_vertical_length="0em" title_text_shadow_blur_strength="0em" closed_title_text_shadow_horizontal_length="0em" closed_title_text_shadow_vertical_length="0em" closed_title_text_shadow_blur_strength="0em"]Brazilian national mathematical olympiad 2018[/et_pb_accordion_item][et_pb_accordion_item title="Topic" open="off" _builder_version="3.23.3" title_text_shadow_horizontal_length="0em" title_text_shadow_vertical_length="0em" title_text_shadow_blur_strength="0em" closed_title_text_shadow_horizontal_length="0em" closed_title_text_shadow_vertical_length="0em" closed_title_text_shadow_blur_strength="0em"]Invariance[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" open="off" _builder_version="3.23.3" title_text_shadow_horizontal_length="0em" title_text_shadow_vertical_length="0em" title_text_shadow_blur_strength="0em" closed_title_text_shadow_horizontal_length="0em" closed_title_text_shadow_vertical_length="0em" closed_title_text_shadow_blur_strength="0em"]Easy[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" open="off" _builder_version="3.23.3" title_text_shadow_horizontal_length="0em" title_text_shadow_vertical_length="0em" title_text_shadow_blur_strength="0em" closed_title_text_shadow_horizontal_length="0em" closed_title_text_shadow_vertical_length="0em" closed_title_text_shadow_blur_strength="0em"]Problem Solving Strategies by Arthur Engel[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px"]

Start with hints

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Do you really need a hint? Just try it yourself! [/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.23.3"]It is always a good idea to try using the invariance principle in such problems. [/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.23.3"]Make a change of variables to see patterns. [/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.23.3"]Note that the operation is restricted to rational numbers. Hence, writing $latex q=\frac{r}{s}=(r,s)$ could help. [/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.23.3"]Let us denote by $latex q_n$ the number on the board at the $latex n$th step. We shall use the new variable $latex a_n=2q_n-1$ (just to simplify the denominator). Clearly, $latex a_{n+1}$ is either $latex a_n\pm 2$ or $latex -\frac{1}{a_n}$. Writing $latex a_n=(r_n,s_n)$, this means that $latex (r_{n+1},s_{n+1})$ is either $latex (r_n \pm 2s_n,s_n)$ or $latex (-s_n,r_n)$. Thus, we need to find out if $latex (-1008,1009)$ is reachable starting from $latex (-1,1)$. However, (odd, odd) pairs can produce only other (odd,odd) pairs under this operation, and $latex (-1008,1009)$ is an (even, even) pair. Hence $latex \frac{1}{2018}$ cannot be reached starting from 0. [/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px"]

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