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Difference between revisions of "2007 AMC 8 Problems/Problem 18"

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<math>\mathrm{(A)}\ 3 \qquad \mathrm{(B)}\ 5 \qquad \mathrm{(C)}\ 6 \qquad \mathrm{(D)}\ 8 \qquad \mathrm{(E)}\ 10</math>
 
<math>\mathrm{(A)}\ 3 \qquad \mathrm{(B)}\ 5 \qquad \mathrm{(C)}\ 6 \qquad \mathrm{(D)}\ 8 \qquad \mathrm{(E)}\ 10</math>
  
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==Video Solution==
 
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https://youtu.be/7an5wU9Q5hk?t=2085
 
  
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==Solution==
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We can first make a small example to find out <math>A</math> and <math>B</math>. So,
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<math>303\times505=153015 </math>
  
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==Solution==
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The ones digit plus thousands digit is <math>5+3=8</math>.
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We can first make a small example to find out A and B So,
 
  
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303*505=153015
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Note that the ones and thousands digits are, added together, <math>8</math>. (and so on...) So the answer is <math>\boxed{\textbf{(D)}\ 8}</math>
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This is a direct multiplication way.
  
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The ones digit plus thousands digit is 5+3=8
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==Video Solution by OmegaLearn==
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https://youtu.be/7an5wU9Q5hk?t=2085
  
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Note that the ones and thousands digits are, added together 8 (and so on...) So the answer is D
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~ pi_is_3.14
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This is a direct multlipication way.
 
  
 
==Video Solution by WhyMath==
 
==Video Solution by WhyMath==

Latest revision as of 04:43, 31 December 2022

Problem

The product of the two $99$-digit numbers

$303,030,303,...,030,303$ and $505,050,505,...,050,505$

has thousands digit $A$ and units digit $B$. What is the sum of $A$ and $B$?

$\mathrm{(A)}\ 3 \qquad \mathrm{(B)}\ 5 \qquad \mathrm{(C)}\ 6 \qquad \mathrm{(D)}\ 8 \qquad \mathrm{(E)}\ 10$


Solution

We can first make a small example to find out $A$ and $B$. So,

$303\times505=153015$

The ones digit plus thousands digit is $5+3=8$.

Note that the ones and thousands digits are, added together, $8$. (and so on...) So the answer is $\boxed{\textbf{(D)}\ 8}$ This is a direct multiplication way.

Video Solution by OmegaLearn

https://youtu.be/7an5wU9Q5hk?t=2085

~ pi_is_3.14

Video Solution by WhyMath

https://youtu.be/_goaFuScO6M

~savannahsolver

See Also

2007 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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