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> | ||
| + | |||
| + | ==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>. | ||
| + | |||
| + | ==Solution 2== | ||
| + | |||
| + | We can easily compute the product of the first 6 positive integers: | ||
| + | <math>(1*2*3*4*5*6)=6!=720</math> | ||
| + | Therefore the units digit must be <math>\boxed{\text{(A)}\ 0}</math>. | ||
| + | |||
| + | ==See Also== | ||
| + | {{AJHSME box|year=1994|num-b=5|num-a=7}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 13:25, 12 January 2014
Contents
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 
.
Solution 2
We can easily compute the product of the first 6 positive integers:
Therefore the units digit must be .
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
