Difference between revisions of "1999 AHSME Problems/Problem 4"
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| − | Find the sum of all prime numbers between 1 and 100 that are simultaneously 1 greater than a multiple of 4 and 1 less than a multiple of 5. | + | ==Problem== |
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| − | + | Find the sum of all prime numbers between <math>1</math> and <math>100</math> that are simultaneously <math>1</math> greater than a multiple of <math>4</math> and <math>1</math> less than a multiple of <math>5</math>. | |
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| − | + | <math> \mathrm{(A) \ } 118 \qquad \mathrm{(B) \ }137 \qquad \mathrm{(C) \ } 158 \qquad \mathrm{(D) \ } 187 \qquad \mathrm{(E) \ } 245</math> | |
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| + | ==Solution== | ||
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| + | Numbers that are <math>1</math> less than a multiple of <math>5</math> all end in <math>4</math> or <math>9</math>. | ||
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| + | No prime number ends in <math>4</math>, since all numbers that end in <math>4</math> are divisible by <math>2</math>. Thus, we are only looking for numbers that end in <math>9</math>. | ||
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| + | Writing down the ten numbers that so far qualify, we get <math>9, 19, 29, 39, 49, 59, 69, 79, 89, 99</math>. | ||
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| + | Crossing off multiples of <math>3</math> gives <math>19, 29, 49, 59, 79, 89</math>. | ||
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| + | Crossing off numbers that are not <math>1</math> more than a multiple of <math>4</math> (in other words, numbers that are <math>1</math> less than a multiple of <math>4</math>, since all numbers are odd), we get: | ||
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| + | <math>29, 49, 89.</math> | ||
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| + | Noting that <math>49</math> is not prime, we have only <math>29</math> and <math>89</math>, which give a sum of <math>118</math>, so the answer is <math>\boxed{A}</math>. | ||
==See Also== | ==See Also== | ||
{{AHSME box|year=1999|num-b=3|num-a=5}} | {{AHSME box|year=1999|num-b=3|num-a=5}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 16:15, 19 December 2018
Problem
Find the sum of all prime numbers between 
and 
that are simultaneously greater than a multiple of 
and less than a multiple of 
.
Solution
Numbers that are less than a multiple of all end in or 
.
No prime number ends in , since all numbers that end in are divisible by 
. Thus, we are only looking for numbers that end in .
Writing down the ten numbers that so far qualify, we get 
.
Crossing off multiples of 
gives 
.
Crossing off numbers that are not more than a multiple of (in other words, numbers that are less than a multiple of , since all numbers are odd), we get:
Noting that 
is not prime, we have only 
and 
, which give a sum of 
, so the answer is 
.
See Also
| 1999 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 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 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
