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

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<math>\text{(A)}\ 0 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8</math>
 
<math>\text{(A)}\ 0 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8</math>
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==Solution==
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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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==See Also==
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{{AJHSME box|year=1994|num-b=5|num-a=7}}

Revision as of 00:18, 23 December 2012

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 $\boxed{\text{(A)}\ 0}$.

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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