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== | ||
| + | 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>. | ||
| + | |||
| + | ==See Also== | ||
| + | {{AJHSME box|year=1994|num-b=5|num-a=7}} | ||
Revision as of 23:18, 22 December 2012
Problem
The unit's digit (one's digit) of the product of any six consecutive positive whole numbers is
Solution
Within six consecutive integers, there must be a number with a factor of 
and an even integer with a factor of 
. Multiplied together, these would produce a number that is a multiple of 
and has a units digit of 
.
See Also
| 1994 AJHSME (Problems • Answer Key • Resources) | ||
| 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 | ||
