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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==Solution==
 
==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>
 
<math>303\times505=153015 </math>
  
 
The ones digit plus thousands digit is <math>5+3=8</math>.
 
The ones digit plus thousands digit is <math>5+3=8</math>.
  
<math> 30303\times50505=1530453015</math>
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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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==Video Solution by OmegaLearn==
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https://youtu.be/7an5wU9Q5hk?t=2085
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~ pi_is_3.14
  
Note that the ones and thousands digits are, added together, <math>8</math>. (and so on...) So the answer is <math> \boxed{D} </math> <math> (8) </math>.
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==Video Solution by WhyMath==
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https://youtu.be/_goaFuScO6M
  
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~savannahsolver
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2007|num-b=17|num-a=19}}
 
{{AMC8 box|year=2007|num-b=17|num-a=19}}
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{{MAA Notice}}

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 (ProblemsAnswer KeyResources)
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

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