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December 22, 2014

December 22, 2014

Posted by **Debbie** on Thursday, October 25, 2012 at 7:14pm.

- math -
**Reiny**, Thursday, October 25, 2012 at 7:28pmto be multiples of 4 and 6, they must be multiples of 12, the LCM

The smallest 3-digit multiple of 12 is

108 , then 120 , 132 .. .996

consider this to be an arithmetic sequence

a = 108, d = 12 , n = ? -- the number of terms

t(n) = a + (n-1)d

996 = 108 + 12(n-1)

888 = 12n - 24

12n = 912

n = 76

There are 76 of them

- math -
**Debbie**, Friday, October 26, 2012 at 7:26pmSince 12n = 12(5m+1)= 60m +12, our goal is to now count integers m such that 100 less than or equal to 60m +12<1000. Subtracing 12 from each part of the inequality, we get 88 lees than or equal to 60m<988. Dividing by 60, we get 2 less than or equal to m less than or equal to 16, so there are 16-1=15 values for the integer m that give us the three-digit multiples of 4 and 6 that have units digits of 2.

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