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.
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==Problem==
'''(A) 118'''
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'''(B) 137'''
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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>.  
'''(C) 158'''
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'''(D) 187'''
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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>
'''(E) 245'''
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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>.
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==See Also==
 
==See Also==
  
 
{{AHSME box|year=1999|num-b=3|num-a=5}}
 
{{AHSME box|year=1999|num-b=3|num-a=5}}

Revision as of 13:21, 8 August 2011

Problem

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 $5$.

$\mathrm{(A) \ } 118 \qquad \mathrm{(B) \ }137 \qquad \mathrm{(C) \ } 158 \qquad \mathrm{(D) \ } 187 \qquad \mathrm{(E) \ } 245$

Solution

Numbers that are $1$ less than a multiple of $5$ all end in $4$ or $9$.

No prime number ends in $4$, since all numbers that end in $4$ are divisible by $2$. Thus, we are only looking for numbers that end in $9$.

Writing down the ten numbers that so far qualify, we get $9, 19, 29, 39, 49, 59, 69, 79, 89, 99$.

Crossing off multiples of $3$ gives $19, 29, 49, 59, 79, 89$.

Crossing off numbers that are not $1$ more than a multiple of $4$ (in other words, numbers that are $1$ less than a multiple of $4$, since all numbers are odd), we get:

$29, 49, 89$

Noting that $49$ is not prime, we have only $29$ and $89$, which give a sum of $118$, so the answer is $\boxed{A}$.


See Also

1999 AHSME (ProblemsAnswer KeyResources)
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
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