Combinatorics, Israel MO 2014, Problem 4

[et_pb_section fb_built="1" _builder_version="3.22.4"][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27.4" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]

The three-digit number 999 has a special property: It is divisible by 27, and its digit sum is also divisible by 27. The four-digit number 5778 also has this property, as it is divisible by 27 and its digit sum is also divisible by 27. How many four-digit numbers have this property?

[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.22.4" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" hover_enabled="0"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="4.0" hover_enabled="0"]Israel 2014, Problem 4 [/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="4.0" hover_enabled="0" open="off"]Combinatorics, Number Theory [/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.0" hover_enabled="0" open="off"]6/10 [/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.0" hover_enabled="0" open="off"]Excursion in Mathematics by Bhaskarcharya Prathistan [/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Start with hints

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.22.4" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff" hover_enabled="0"][et_pb_tab title="Hint 0" _builder_version="3.22.4"]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="4.0" hover_enabled="0"]Let's write the problem mathematically, i.e. in terms of the equations. Let's write the condition mathematically. Let $abcd$ be a four-digit number, with $1\le a\le9$ and $0\le b,c,d\le 9$ , and $a,b,c,d$ positive integers. Then we need to have $1000a+100b+10c+d=27n$ and $a+b+c+d=27m$ , where $n,m$ are positive integers. Now, we have to count the number of such solutions. Observe that the sum of the digits can be at most 36. So, m = 1. Hence, a+b+c+d = 27. This leads to $111a+11b+c=3(n-1)$ . This implies $2b+c$ being a multiple of $3$ . Now, this implies we have quite a number of cases to investigate. Let's do them one by one patiently. [/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.0" hover_enabled="0"]

b	c	a+d
0	3,6,9	27,24,21,18

Out of this b = 0, c = 9, a+d = 18, a = 9, c =9 is the only possibility considering the maximum possibble value of a+d being 18.

b	c	a+d
1	1,4,7	25,22,19

None of the cases is fine as the maximum possible value of a+d is 18. [/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.0" hover_enabled="0"]

b	c	a+d
2	2,5,8	23,20,17

This gives rise to b = 2, c = 8, a+d = 17. Hence two solutions 8289 and 9288.

b	c	a+d
3	0,3,6,9	24,21,18,15

Thus you can understand this gives rise to 4 solutions: ( $6399,7398,8397$ and $9396$ ) [/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.0" hover_enabled="0"]

b	c	a+d
4	1,4,7	22,19,16

This gives rise to three solutions: ( $7479,8478,$ and $9477$ ).

b	c	a+d
5	2,5,8	20,17,14

Only the last two choices are acceptable; the former gives us two solutions, and the latter 5 (for a total of seven solutions). [/et_pb_tab][et_pb_tab title="HInt 5" _builder_version="4.0" hover_enabled="0"]

b	c	a+d
6	0,3,6,9	21,18,15,12

The last three choices are acceptable and give us $1+4+7=12$ solutions.

b	c	a+d
7	1,4,7	19,16,13

The last two choices are acceptable and give us $3+6=9$ solutions.

b	c	a+d
8	2,5,8	17,14,11

All choices are acceptable and give us $2+5+8=15$ solutions.

b	c	a+d
9	0,3,6,9	18,15,15,9

All choices are acceptable and give us $1+4+7+9=21$ solutions. Therefore, there are $1+2+5+3+7+12+9+15+21=75$ four-digit numbers with this property. [/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Watch video

[/et_pb_text][et_pb_code _builder_version="3.26.4"]

[/et_pb_code][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" min_height="12px" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Connected Program at Cheenta

[/et_pb_text][et_pb_blurb title="Math Olympiad Program" url="https://cheenta.com/matholympiad/" url_new_window="on" image="https://cheenta.com/wp-content/uploads/2018/03/matholympiad.png" _builder_version="3.23.3" header_font="||||||||" header_text_color="#e02b20" header_font_size="48px" link_option_url="https://cheenta.com/matholympiad/" link_option_url_new_window="on"]

Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year. Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.[/et_pb_blurb][et_pb_button button_url="https://cheenta.com/matholympiad/" url_new_window="on" button_text="Learn More" button_alignment="center" _builder_version="3.23.3" custom_button="on" button_bg_color="#0c71c3" button_border_color="#0c71c3" button_border_radius="0px" button_font="Raleway||||||||" button_icon="%%3%%" background_layout="dark" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3"][/et_pb_button][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Combinatorics, Israel MO 2014, Problem 4

Understand the problem

Start with hints

Watch video

Connected Program at Cheenta

Similar Problems

Leave a Reply Cancel reply