Difference between revisions of "2021 AMC 12A Problems/Problem 3"
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==Solution== | ==Solution== | ||
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| + | The units digit of a multiple of <math>10</math> will always be <math>0</math>. We add a <math>0</math> whenever we multiply by <math>10</math>. So, removing the units digit is equal to dividing by <math>10</math>. | ||
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| + | Let the smaller number (the one we get after removing the units digit) be <math>a</math>. This means the bigger number would be <math>10a</math>. | ||
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| + | We know the sum is <math>10a+a = 11a</math> so <math>11a=17402</math>. So <math>a=1582</math>. The difference is <math>10a-a = 9a</math>. So, the answer is <math>9(1582) = 14238</math>. | ||
==See also== | ==See also== | ||
{{AMC12 box|year=2021|ab=A|num-b=2|num-a=4}} | {{AMC12 box|year=2021|ab=A|num-b=2|num-a=4}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 14:39, 11 February 2021
Problem
The sum of two natural numbers is 
. One of the two numbers is divisible by 
. If the units digit of that number is erased, the other number is obtained. What is the difference of these two numbers?
Solution
The units digit of a multiple of will always be 
. We add a whenever we multiply by . So, removing the units digit is equal to dividing by .
Let the smaller number (the one we get after removing the units digit) be 
. This means the bigger number would be 
.
We know the sum is 
so 
. So 
. The difference is 
. So, the answer is 
.
See also
| 2021 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 2 |
Followed by Problem 4 |
| 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 AMC 12 Problems and Solutions | |
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