Difference between revisions of "1994 AJHSME Problems/Problem 6"

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==Solution==
 
==Solution==
 
Within six consecutive integers, there must be a number with a factor of <math>5</math> and an even integer with a factor of <math>2</math>. Multiplied together, these would produce a number that is a multiple of <math>10</math> and has a units digit of <math>\boxed{\text{(A)}\ 0}</math>.
 
Within six consecutive integers, there must be a number with a factor of <math>5</math> and an even integer with a factor of <math>2</math>. Multiplied together, these would produce a number that is a multiple of <math>10</math> and has a units digit of <math>\boxed{\text{(A)}\ 0}</math>.
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==Solution 2==
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We can easily compute the product of the first 6 positive integers:
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<math>(1*2*3*4*5*6)=6!=720</math>
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Therefore the units digit must be <math>\boxed{\text{(A)}\ 0}</math>.
  
 
==See Also==
 
==See Also==
 
{{AJHSME box|year=1994|num-b=5|num-a=7}}
 
{{AJHSME box|year=1994|num-b=5|num-a=7}}
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{{MAA Notice}}

Latest revision as of 14:25, 12 January 2014

Problem

The unit's digit (one's digit) of the product of any six consecutive positive whole numbers is

$\text{(A)}\ 0 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8$

Solution

Within six consecutive integers, there must be a number with a factor of $5$ and an even integer with a factor of $2$. Multiplied together, these would produce a number that is a multiple of $10$ and has a units digit of $\boxed{\text{(A)}\ 0}$.

Solution 2

We can easily compute the product of the first 6 positive integers: $(1*2*3*4*5*6)=6!=720$ Therefore the units digit must be $\boxed{\text{(A)}\ 0}$.

See Also

1994 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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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