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AMC 10 USA Math Olympiad

AMC 10A Year 2006 Problems 21 Sequential Hints

A beautiful problem form American Math Contest (AMC 10A) 2006, Problem 21 solution. Key idea is to use basic calculation.

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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″]How many 4 digit positive numbers have at least that is a 2 or a 3?

(a)2439.     (b)4096.     (c)4903.     (d)4904.     (e)5416

[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version=”4.0″][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=”4.0″ 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=”4.0″]American Mathematical Contest 2006 10 A Problem 21

[/et_pb_accordion_item][et_pb_accordion_item title=”Topic” _builder_version=”4.0″ open=”off”]Combinatorics 

[/et_pb_accordion_item][et_pb_accordion_item title=”Difficulty Level” _builder_version=”4.0″ open=”off”]4/10

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Challenges and Thrills in Pre College Mathematics Excursion Of Mathematics 

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Step 1. After having a look into this problem you can see our main aim here is to find out the total number of 4 digit positive numbers with at least one 2 or one 3 in it. Now we can do this by subtracting the number of 4 digit positive number which do not have any 2’s or 3’s from the total number of 4 digit positive numbers. Its easy, give it a try.

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Step 2. Now lets’ proceed forward and try to find out the total number of 4 digit positive integers . We can use all digit from 0 to 9 as we do not have any restriction . Now to make a table with this

So that you can fill the boxes with number of options in order. As the first box from the right is of unit place ,the next box is of the tenth place, the next box is of the hundredth place and the last box in the extreme left is of the thousandth place. Now you can fill these boxes with the possibilities. (Like here the thousandth place can take 9 options as we can not put 0 there then it will be a 3 digit number). As simple as that, give it a try!!!.

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Step 3.  Now for the 2nd part. So try to find out the total number of 4 digit positive integers which don’t have any 2’s and 3’s in it  . So now in this case we have 8 digits(2 and 3 are excluded). Now to make a table with this 

.So that you can fill the boxes with number of options in order. As the first box from the right is of unit place ,the next box is of the tenth place, the next box is of the hundredth place and the last box in the left is of the thousandth place. Now you can fill these boxes with the possibilities in this case. (Like here the thousandth place can take 7 options as we can not put 0 there then it will be a 3 digit number)

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Step 4. Now when you have found the number of positive 4 digit numbers in both the cases now 

(total number of 4 digit positive integers)-( total number of 4 digit positive integers which don’t have any 2’s and 3’s in it )=(total number of 4 digit positive numbers with at least one 2 or one 3 in it). Very close to the solution just about to crack it !!!.

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Total number of 4 digit positive integers(Table 1) = 9000 Total number of 4 digit positive integers which don’t have any 2’s and 3’s in it(Table 2) = 3584 Total number of 4 digit positive numbers with at least one 2 or one 3 in it = 5416 that’s your required answer.

 

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Start with hints

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Watch video

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Similar Problems

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