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help!
pAtRiCk_sTar
post Dec 7 2004, 11:31 PM
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aaacch i don't get this math problem, i'm really stupid. it's a junior high problem, so maybe someone could help?

Claire finished her assignment early, so her teacher suggested that she find the last digit of the product 3 x 3 to third power x 3 to fifth......x 3 to nineteenth. The entire product was too big for her calculator, but she still found the last digit. What is the last digit? Explain how she determined her answer

i know that if you add all the exponents, you'll get 3 to the 100th power, since
3+5+7+9+11+13+15+17+19=100 right (those are the exponent values...), so it'd be 3 to the 100th power. on my calc i got 5.153775207 to the 47th power.....what next?!?!!?

sorry if this is in the wrong board...i saw someone ask for math help awhile back, so i figured it'd be OK whistling.gif
 
 
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whomps
post Dec 8 2004, 02:16 AM
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QUOTE(pAtRiCk_sTar @ Dec 7 2004, 11:02 PM)
ooooh i think i see what you're saying. so the 3 to the 100th power ends with a one right? but did you just write it all out or use some sort of technique??

If the exponenent is divisble by 4, then the last digit will be 1. In this case, 100 is divisible by 4, and therefore the last digit of 3^100 is 1.
 

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